The maximum stress in the beam's cross-section is determined using the formula max stress = [tex](M * c) / I[/tex], where M is [tex]43 lb-ft[/tex], c is 2 inches, and I is [tex]10.67 in^4[/tex] , resulting in a value of 8.07 psi.
To determine the maximum stress in the beam's cross-section, we can use the formula:
[tex]max stress = (M * c) / I[/tex]
Where M is the bending moment (given as [tex]43 lb-ft[/tex]), c is the distance from the neutral axis to the outermost point in the cross-section (in this case, the distance from the center of the beam to the bottom edge), and I is the moment of inertia of the cross-section.
From Figure 1, we can see that the beam is rectangular with a width of 2 inches and a height of 4 inches. The moment of inertia of a rectangular cross-section is:
[tex]I = (b * h^3) / 12[/tex]
Where b is the width and h is the height. Plugging in the values for our beam, we get:
[tex]I = (2 * 4^3) / 12 = 10.67 in^4[/tex]
To find c, we need to determine the location of the neutral axis. For a rectangular cross-section, the neutral axis is located at the centroid, which is at the center of the cross-section. Since the height is 4 inches, the distance from the neutral axis to the bottom edge is 2 inches.
Now we can plug in our values into the formula for max stress:
[tex]max stress = (43 lb-ft * 2 in) / 10.67 in^4 = 8.07 psi[/tex]
Therefore, the maximum stress in the beam's cross-section is 8.07 psi.
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Tyson Gay's best time to run 100.0 meters was 9.69 seconds. What was his average speed during this run, in miles per hour? (3.281ft=1 m)(1 mile=5280 ft)
Report your answer to three significant figures (round your answer to one decimal place).
Tyson Gay's average speed during the 100-meter run was approximately 23.1 mph
To find Tyson Gay's average speed during the 100.0-meter run, we'll first convert meters to feet, then feet to miles, and finally seconds to hours.
1. Convert meters to feet: 100.0 meters * 3.281 ft/m = 328.1 feet
2. Convert feet to miles: 328.1 feet / 5280 ft/mile ≈ 0.0621 miles
3. Convert seconds to hours: 9.69 seconds * (1 hour / 3600 seconds) ≈ 0.00269 hours
Now we can calculate the average speed:
Average speed = distance/time = 0.0621 miles / 0.00269 hours ≈ 23.1 miles per hour
So, Tyson Gay's average speed during the 100-meter run was approximately 23.1 mph, reported to three significant figures and rounded to one decimal place.
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Traveling at a speed of 21 m/s, the driver of a car suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.72. How much time does it take for the car to come to stop? A) 1 sec B) 2 sec C) 3 sec D) 4 sec E) 5 sec
Option C is Correct. Traveling at a speed of 21 m/s, the driver of a car suddenly locks the wheels by slamming on the brakes so 3 sec need come to stop.
To solve this problem, we will use the concepts of kinetic friction and time. The formula to calculate the acceleration due to kinetic friction is:
a = μk × g
where a is the acceleration, μk is the coefficient of kinetic friction, and g is the acceleration due to gravity (approximately 9.81 m/s²).
1. Calculate the acceleration Speed due to kinetic friction:
a = 0.72 × 9.81 = 7.0632 m/s² (deceleration, since it's against the motion)
2. Next, we can use one of the equations of motion to find the time it takes for the car to stop. We'll use the following equation, where vf is the final velocity (0 m/s, as the car comes to a stop), vi is the initial velocity (21 m/s), a is the acceleration we calculated, and t is the time we want to find:
vf = vi + (a × t)
3. Solve for time, t:
0 = 21 + (-7.0632 × t)
7.0632 × t = 21
t = 21 / 7.0632 ≈ 2.97 sec
So, the answer is approximately 3 seconds.
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a) what magnitude point charge (in c) creates a 16,000 n/c electric field at a distance of 0.270 m? c (b) how large (in n/c) is the field at 10.0 m? n/c
To answer your question:
(a) The magnitude of the point charge that creates a 16,000 N/C electric field at a distance of 0.270 m can be calculated using the equation:
E = k*q/r^2
where E is the electric field, k is Coulomb's constant (9 x 10^9 Nm^2/C^2), q is the magnitude of the point charge, and r is the distance from the point charge.
Rearranging the equation to solve for q, we get:
q = Er^2/k
Substituting the given values, we have:
q = (16,000 N/C) x (0.270 m)^2 / (9 x 10^9 Nm^2/C^2)
q = 1.85 x 10^-8 C
Therefore, a magnitude point charge of 1.85 x 10^-8 C creates a 16,000 N/C electric field at a distance of 0.270 m.
(b) To find out how large the electric field is at a distance of 10.0 m, we can use the same equation:
E = k*q/r^2
But this time, we know the magnitude of the point charge (q) and the distance (r), and we need to solve for the electric field (E).
Substituting the values, we have:
E = (9 x 10^9 Nm^2/C^2) x (1.85 x 10^-8 C) / (10.0 m)^2
E = 3.7 x 10^-13 N/C
Therefore, the electric field at a distance of 10.0 m is 3.7 x 10^-13 N/C.
a) To find the magnitude of the point charge (in C) that creates a 16,000 N/C electric field at a distance of 0.270 m, you can use the formula for the electric field E:
E = k * |q| / r^2
where E is the electric field, k is Coulomb's constant (8.99 × 10^9 N m²/C²), |q| is the magnitude of the charge, and r is the distance.
16,000 N/C = (8.99 × 10^9 N m²/C²) * |q| / (0.270 m)^2
Solving for |q|, we get:
|q| ≈ 1.27 × 10^-6 C
b) To find the electric field (in N/C) at 10.0 m, we can use the same formula, with r = 10.0 m:
E = (8.99 × 10^9 N m²/C²) * (1.27 × 10^-6 C) / (10.0 m)^2
E ≈ 114 N/C
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a spring of force constant 245.0 n/m and unstretched length 0.280 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 22.0 n. How long will the spring now be, and how much work was required to stretch it that distance?
The spring constant is 245.0 N/m and the unstretched length is 0.280 m. The forces pulling on the spring increase to 22.0 N, pulling in opposite directions at opposite ends of the spring.
To find the new length of the spring, we can use Hooke's law, which states that the force applied to a spring is directly proportional to the amount of stretch or compression of the spring. The formula for Hooke's law is:
F = -kx
Where F is the force applied to the spring, k is the spring constant, and x is the amount of stretch or compression of the spring. The negative sign indicates that the force applied to the spring is in the opposite direction of the displacement of the spring.
We can rearrange this formula to solve for x:
x = -F/k
Plugging in the values we have:
x = -(22.0 N)/(245.0 N/m)
x = -0.0898 m
Therefore, the new length of the spring is:
L = Lo + x
L = 0.280 m - 0.0898 m
L = 0.1902 m
To find the work required to stretch the spring this distance, we can use the formula:
W = (1/2)kx^2
Plugging in the values we have:
W = (1/2)(245.0 N/m)(0.0898 m)^2
W = 0.975 J
Therefore, the work required to stretch the spring 0.0898 m is 0.975 J.
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Given the spring's force constant (k) is 245.0 N/m, and the force applied (F) is 22.0 N, we can use Hooke's Law to determine the change in the spring's length (Δx):
F = k * Δx
Rearranging the formula to find Δx:
Δx = F / k
Δx = 22.0 N / 245.0 N/m
Δx ≈ 0.0898 m
Now, to find the new length of the spring (L'), add the change in length (Δx) to the unstretched length (L):
L' = L + Δx
L' = 0.280 m + 0.0898 m
L' ≈ 0.3698 m
To calculate the work (W) done in stretching the spring, we use the formula:
W = 0.5 * k * Δx²
W = 0.5 * 245.0 N/m * (0.0898 m)²
W ≈ 1.003 J
So, the spring will now be approximately 0.3698 meters long, and 1.003 Joules of work was required to stretch it that distance.
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A good tutor will be correct in saying that velocity and acceleration are A) different concepts. B) the same concept, but expressed differently. C) expressions for changing speeds. D) rates of one another.
A good tutor will be correct in saying that velocity and acceleration are different concepts. The correct option is A.
Velocity and acceleration are distinct concepts in physics and describe different aspects of motion.
Velocity refers to the rate at which an object changes its position in a particular direction over time. It is a vector quantity, meaning it has both magnitude (speed) and direction. Velocity is calculated as the change in displacement divided by the change in time.
Acceleration, on the other hand, refers to the rate at which an object changes its velocity over time. It is also a vector quantity and is calculated as the change in velocity divided by the change in time.
While velocity and acceleration are related, they are not the same concept and are expressed differently. Velocity describes the speed and direction of motion, while acceleration describes how quickly the velocity changes.
They are both important in understanding the motion of objects and are fundamental concepts in physics.
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The gap in the outer and inner walls of a calorimeter (the device to measure heat) is filled with air because ita. allows only the heat to flow from inside to outside of the calorimeterb. restricts only the heat to flow from inside to outside of the calorimeterc. restricts the flow of heat from inside to outside and outside to inside of the calorimeter
The gap in the inner walls as well as outer walls of the calorimeter is filled with air because it restricts the flow of the heat from inside to outside as well as outside to inside of the calorimeter. Option C is correct.
The gap in the outer and inner walls of a calorimeter is typically filled with air to act as an insulator. Air is a poor conductor of heat, meaning that it restricts the flow of heat from inside to outside and outside to inside of the calorimeter. This helps to maintain the temperature inside the calorimeter and prevents heat from escaping or entering the calorimeter too quickly.
Hence, C. is the correct option.
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if an object has a moment of inertia 26 kg·m2 and rotates with an angular speed of 80 radians/s, what is its rotational kinetic energy?
The rotational kinetic energy of the object is 83,200 Joules.
The rotational kinetic energy of an object is the energy it possesses due to its rotation. It can be calculated using the formula:
[tex]K_rot = (1/2) * I * ω^2[/tex]
where K_rot is the rotational kinetic energy, I is the moment of inertia of the object, and ω is its angular velocity.
In this case, the object has a moment of inertia of [tex]26 kg·m^2[/tex]and is rotating with an angular speed of 80 radians/s. Substituting these values into the formula gives:
[tex]K_rot = (1/2) * 26 kg·m^2 * (80 radians/s)^2[/tex]
= 83,200 J
Therefore, the rotational kinetic energy of the object is 83,200 Joules.
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A section of a sphere is mirrored on both sides. If the magnification of an object is +3.70 when the section is used a concave mirror, what is the magnification of an object at the same distance in front of the convex side?
A section of a sphere is mirrored on both sides. If the magnification of an object is +3.70 the magnification is -4.70.
The magnification of an object at the same distance in front of the convex side of the mirrored section of a sphere can be found using the mirror equation:
1/f = 1/di + 1/do
where f is the focal length of the mirror, di is the distance of the object from the mirror, and do is the distance of the image from the mirror.
Since the section is mirrored on both sides, the focal length of the concave and convex sides will be the same. Therefore, we can use the magnification equation:
m = -di/do
where m is the magnification.
We know that when the section is used as a concave mirror, the magnification is +3.70. Therefore,
+3.70 = -di/do
Solving for do, we get
do = -di/3.70
Now, substituting this value of do into the mirror equation, we get
1/f = 1/di - 3.70/di
Simplifying this equation, we get
f = di/4.70
Therefore, the magnification of an object at the same distance in front of the convex side of the mirrored section will be
m = -di/(di/4.70)
m = -4.70
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) A rectangular bar is cut from AISI 1020 cold-drawn steel flat. The bar is 2.5in wide by 3/8in thick and has a 0.5-in-dia. Hole drilled through the center as depicted in Figure 1. The bar is concentrically loaded in push-pull fatigue by axial forces Fa, uniformly distributed across the width. Using a design factor of nd-2, estimate the largest force Fa that can be applied ignoring column action. 0.5 1020
The greatest force Fa that can be applied while ignoring column action is 12,000 lbf, while the question states that the bar's permissible stress is 20,000 psi.
What is force?A push or pull that causes a physical change in an item, such as a change in velocity, form, or size, is known as force. Forces may be physical or psychological. In contrast to non-contact forces like gravity, electricity, and magnetism, contact forces are generated by physical contact. A force's strength is often expressed in newtons (N).
The Goodman diagram can be used to calculate the bar's maximum load capacity.
The allowed stress and load factor are plotted on a chart called a Goodman diagram. The greatest load applied to the bar divided by the material's yield strength is known as the load factor.
The yield strength of cold-drawn flat steel AISI 1020 is about 40,000 psi. By dividing the yield strength by the design factor (nd-2), in this example 0.5, the allowed stress for the bar is calculated. Therefore, 20,000 psi is the maximum tension that the bar can withstand.
The Goodman diagram and 20,000 psi of permissible stress can be used to calculate the bar's maximum load capacity. The bar can support a maximum load of about 12,000 lbf. Therefore, 12,000 lbf is the maximum force Fa that can be applied while disregarding column action.
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What does resistances within a circuit have to do with the brightness of a light bulb within that same circuit?
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What do we call objects with high resistance and low resistance?
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Complete and full answer with explanation gets Brainliest
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People who misuse this and only use it to get points will be reported
Bats use echolocation to navigate. They can emit ultrasonic waves with frequencies as high as 1.0×105 Hz.
What is the wavelength of such a wave? The speed of sound in air is 340 m/s.
A) 3.4×103 m
B) 3.4×10−3 m
C) 3.4×105 m
D) 3.4×107 m
The wavelength of the ultrasonic wave emitted by bats is B) 3.4×10−3 m.
How to find wavelengthTo calculate the wavelength of the ultrasonic wave emitted by bats, we can use the formula:
Wavelength (λ) = Speed of sound (v) / Frequency (f)
Given that the frequency (f) is 1.0×10^5 Hz and the speed of sound (v) is 340 m/s, we can plug in the values:
λ = 340 m/s / 1.0×10^5 Hz
λ = 3.4×10^−3 m
So the correct answer is: B) 3.4×10^−3 m
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(a) Find the voltage drop in an extension cord having a 0.0600- Ω resistance and through which 5.00 A is flowing. (b) A cheaper cord utilizes thinner wire and has a resistance of 0.300 Ω . What is the voltage drop in it when 5.00 A flows? (c) Why is the voltage to whatever appliance is being used reduced by this amount? What is the effect on the appliance?
To find the voltage drop in the extension cord, we can use Ohm's Law, which states that V = IR, where V is the voltage drop, I is the current, and R is the resistance. Plugging in the given values, we get V = (5.00 A)(0.0600 Ω) = 0.3 V.
Using the same formula, we can find the voltage drop in the cheaper cord: V = (5.00 A)(0.300 Ω) = 1.5 V. The voltage drop occurs because the resistance of the cord causes some of the electrical energy to be converted into heat, rather than being delivered to the appliance. This reduces the voltage that reaches the appliance, which can affect its performance. For example, a motor might run more slowly, or a light bulb might be dimmer when the voltage is reduced. In some cases, the reduced voltage can also cause the appliance to draw more current, which can lead to further voltage drops and potentially damage the cord or the appliance.
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To find the voltage drop in the extension cord, we can use Ohm's Law, which states that V = IR, where V is the voltage drop, I is the current, and R is the resistance. Plugging in the given values, we get V = (5.00 A)(0.0600 Ω) = 0.3 V.
Using the same formula, we can find the voltage drop in the cheaper cord: V = (5.00 A)(0.300 Ω) = 1.5 V. The voltage drop occurs because the resistance of the cord causes some of the electrical energy to be converted into heat, rather than being delivered to the appliance. This reduces the voltage that reaches the appliance, which can affect its performance. For example, a motor might run more slowly, or a light bulb might be dimmer when the voltage is reduced. In some cases, the reduced voltage can also cause the appliance to draw more current, which can lead to further voltage drops and potentially damage the cord or the appliance.
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a 30 g ball of clay is thrown horizontally at 20 m/s toward a 1.2 kg block sitting at rest on a frictionless surface. the clay hits and sticks to the block.
During the impact, kinetic energy lost is KE lost = 5.8495 J as heat and sound.
Since there are no outside forces operating on the system and it is isolated, the overall momentum before and after the impact must be the same for frictionless surface. Therefore: The conservation of momentum principle, which asserts that the overall momentum of a system stays constant if no external forces impinge on it, must be used to address this issue.
Let's first calculate the initial momentum of the clay before it hits the block:
[tex]p_c = m_c* v_c\\p_c = 0.03 kg * 20 m/s\\p_c = 0.6 kg*m/s[/tex]
Since the block is at rest initially, its momentum is zero. After the clay hits and sticks to the block, the total momentum of the system is:
[tex]p_t = p_c + p_b\\\\p_t_b = p_t_av_f= 0.6 kg*m/s / (0.03 kg + 1.2 kg)\\\\v_f= 0.4878 m/s[/tex]
Therefore, the clay and block move together with a final velocity of 0.4878 m/s. To find the kinetic energy lost during the collision, we can calculate the initial and final kinetic energies of the clay:
[tex]KE_i = 0.5 * m_c * v_c^2\\KE_i = 0.5 * 0.03 kg * (20 m/s)^2\\KE_i = 6 J\\KE_f = 0.5 * (m_c + m_b) * v_f^2\\KE_f= 0.5 * 1.23 kg * (0.4878 m/s)^2\\KE_f = 0.1505 J[/tex]
Therefore, the kinetic energy lost during the collision is:
[tex]KE_l = KE_i - KE_f\\KE_l = 6 J - 0.1505 J\\KE_l = 5.8495 J[/tex]
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Correct Question:
A 30 g ball of clay is thrown horizontally at 20 m/s toward a 1.2 kg block sitting at rest on a frictionless surface. the clay hits and sticks to the block. Find the amount of kinetic energy lost.
In a polarization of light experiment an incandescent light source is used. The ratio polarized to unpolarized light intensity is (a) 25% (b) 50% (c) 75% (d) 100%
(b) 50%. In a polarization of light experiment an incandescent light source is used. The ratio polarized to unpolarized light intensity is 50%.
In an experiment to measure the polarization of light, an incandescent light source is used to emit light in all directions. However, the emitted light is unpolarized, meaning that the light waves vibrate in all possible planes perpendicular to the direction of propagation. To obtain polarized light, a polarizer is used to pass only the light waves that vibrate in a single plane. As a result, only half of the original light intensity can pass through the polarizer, and the other half is absorbed or blocked. Thus, the ratio of polarized to unpolarized light intensity is 1:1 or 50%. This result holds true for any polarizer that only allows light waves vibrating in a single plane to pass through.
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In the four trials of Exercise 1, one needs to accurately measure ... a. The length of four different radii and the corresponding time for 10 revolutions b. The weight hanger c. The mass of the stopperd. All of the above
In the four trials of Exercise 1, one needs to accurately measure the length of four different radii and the corresponding time for 10 revolutions in order to calculate the speed of the stopper in revolutions per minute.
Additionally, one needs to measure the mass of the stopper in order to calculate the centripetal force acting on it. Therefore, the correct answer is d. All of the above.
one needs to accurately measure all of the above options, which include:
a. The length of four different radii and the corresponding time for 10 revolutions
b. The weight hanger
c. The mass of the stopper
Accurately measuring these factors ensures that the results and conclusions drawn from the experiment are reliable and valid.
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calculate the wavelength given a frequency of 7.187x106 mhz (1 mhz = 106 hz)
the wavelength of the given frequency is approximately 4.18 x 10^-5 meters.
To calculate the wavelength given a frequency of 7.187x106 MHz, we need to use the equation:
wavelength = speed of light / frequency
The speed of light is approximately 3x108 meters per second.
First, we need to convert the frequency from MHz to Hz, since the speed of light is in meters per second and the frequency needs to be in hertz.
7.187x106 MHz = 7.187x106 x 106 Hz = 7.187x1012 Hz
Now we can plug in the values:
wavelength = 3x108 / 7.187x1012 = 4.17x10-5 meters
Therefore, the wavelength for a frequency of 7.187x106 MHz is approximately 4.17x10-5 meters.
To calculate the wavelength given a frequency, you can use the following formula:
wavelength = speed of light / frequency
Given the frequency of 7.187 x 10^6 MHz, first convert it to Hz:
7.187 x 10^6 MHz x 10^6 Hz/MHz = 7.187 x 10^12 Hz
Now, using the speed of light (c) which is approximately 3 x 10^8 m/s:
wavelength = (3 x 10^8 m/s) / (7.187 x 10^12 Hz)
wavelength ≈ 4.18 x 10^-5 m
So, the wavelength of the given frequency is approximately 4.18 x 10^-5 meters.
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according to phil, the only way we know how to get accurate stellar masses is group of answer choices when they have iron absorption lines when they are incredibly dim when they are incredibly bright when they have hydrogen absorption lines when they are in a binary system
According to Phil, the only way we know how to get accurate stellar masses is when they are in a binary system.
In a binary system, two stars orbit each other, and their gravitational interaction can be observed and measured. This interaction allows astronomers to determine the stars' masses using Kepler's laws and other astrophysical methods. Other methods, such as using iron or hydrogen absorption lines, can provide information about the stars' compositions and temperatures, but not their masses with the same accuracy as binary systems.
To obtain accurate stellar masses, it is essential to observe stars in a binary system, as their gravitational interaction provides the most reliable measurements.
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A healthy mind and body is important. The decisions you make can have a serious effect on your health and the health of those around you. Think about what you have learned in the module. Use the information from your milestone 1 and milestone 2 projects to help you complete your final project.
Your project should be about the diseases that can be caused by hereditary traits.
Choose a healthy behavior and develop a positive health media message.
TV commercial
Radio commercial
Magazine ad
Internet popup ad
Teacher approved idea
Examples of hereditary diseases include heart disease, Type 1 Diabetes, or many other diseases you may have learned about in your interviews.
You will present a message that tries to get people to make health choices that will help them reduce the impact of hereditary disease. Your message can be presented as a video or by using a Web 2.0 tool such as Weebly or Jing.
Your presentation will focus on:
Making people aware of hereditary traits.
Getting people to make good health choices to help lessen the risk of hereditary disease.
Answer the following questions to help you get started on your project:
What is the purpose of your media message? (to inform or persuade)
What type of medium is it? (commercial, advertisement, or teacher approved)
Who is your intended audience?
What is the message? (main idea or main points)
What information can be omitted?
What techniques are you going to use to get attention?
(Please put an actual answer please.)
Which one of the following does Kepler's Second Law indirectly describe?
- the masses of planets - the sizes of planets - the densities of the materials that planets are made of - the speeds with which planets travel in their orbits
Kepler's Second Law indirectly describes the speeds with which planets travel in their orbits.
This law, also known as the Law of Equal Areas, states that a line connecting a planet to the sun sweeps out equal areas in equal times, implying that planets move faster when closer to the sun and slower when farther away.
According to Kepler's Second Law, a line that connects a planet to the sun, known as the radius vector, sweeps out equal areas in equal times as the planet moves along its elliptical orbit.
This means that a planet covers the same amount of area in its orbit during equal time intervals, regardless of where it is in its orbit. This implies that a planet moves faster when it is closer to the sun and slower when it is farther away.
This observation has significant implications for our understanding of planetary motion. As a planet moves closer to the sun, it experiences a stronger gravitational pull, which accelerates its motion and causes it to move faster.
Conversely, as a planet moves farther away from the sun, the gravitational pull weakens, resulting in a slower motion. This is consistent with Kepler's Second Law, which states that planets move faster in the inner parts of their orbits (when closer to the sun) and slower in the outer parts (when farther away from the sun).
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a. Sisyphus is pushing a 95 kg flat stone up a 30 frictionless slope. How much force must he apply to push it up the slope at a constant speed of 22 cm/s? Hint: you might want to do part b first. force, including the normal force. You can use g 10 m/s. the ramp. If the stone has a constant acceleration downward of 2.6 m/s. What is a likely coefficient of b. Draw a fully labeled force diagram for the stone. Include all magnitudes for each c. Let's say the slope does have considerable friction, and Sisyphus lets the stone freely slide back down kinetic friction μ? μ
A 95 kg flat stone is being pushed by Sisyphus up a 30° frictionless slope. To move it up the hill at a steady pace of 22 cm/s, he needs exert 475 N of effort. The kinetic friction coefficient is 0.26.
Therefore Acceleration, 0.26 is likely to be the kinetic friction coefficient.
Here: Mass of stone, m = 95 kg
Speed, v = 22 cm/s
Slope, θ = 30°g = 10 m/s²(a)
The force required to push the stone up the slope at a constant speed can be found using the formula:
Force = Weight x Component of Weight along the slope
F = mgsinθF = 95 x 10 x sin30°F = 475 N
Therefore, the force required to push the stone up the slope at a constant speed is 475 N.
b. Let's say the slope does have considerable friction, and Sisyphus lets the stone freely slide back down the ramp. If the stone has a constant acceleration downward of 2.6 m/s², then the likely coefficient of kinetic friction μ can be found using the formula:
μ = a/gμ = 2.6/10μ = 0.26
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two lamps illuminate a screen equally. the first lamp has an intensity of 12.5 cd and is 3.0 m from the screen. the second lamp is 9.0 m from the screen. what is its intensity?
The intensity of the second lamp can be calculated using the inverse square law, which states that the intensity of light decreases with the square of the distance from the source. The equation for the inverse square law is:
I2 = I1 * (d1/d2)^2
where I1 is the intensity of the first lamp, d1 is the distance from the first lamp to the screen, I2 is the intensity of the second lamp, and d2 is the distance from the second lamp to the screen.
Substituting the given values, we get:
I2 = 12.5 cd * (3.0 m/9.0 m)^2
I2 = 12.5 cd * (1/3)^2
I2 = 12.5 cd * 0.111
I2 = 1.39 cd
Therefore, the intensity of the second lamp is 1.39 cd.
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A lighted candle is placed 38.0 cm in front of a diverging lens. The light passes through the diverging lens and on to a converging lens of focal length 14.0 cm that is 6.0 cm from the diverging lens. The final image is real, inverted, and 42.0 cm beyond the converging lens. Find the focal length of the diverging lens.
Let's use f1 to represent the diverging lens's focal length. The thin lens equation may be used to connect the problem's distances and focal lengths: Consequently, the divergent lens's focal length is 9.15 cm.
Now: 1/f1 = 1/d1 - 1/d2
1/f2 = 1/d2 - 1/d3
Here d1 is the distance from the candle to the diverging lens, d2 is the distance between the two lenses, and d3 is the distance from the converging lens to the final image. We can also use the magnification equation to relate the magnifications produced by the two lenses:
m = -(d2/f1)(f2/d3)
Here the negative sign indicates that the final image is inverted.
Values and solving the equations simultaneously, we get:
d1 = 38.0 cm
d2 = 6.0 cm
d3 = 42.0 cm
f2 = 14.0 cm
1/f1 = 1/d1 - 1/d2 = 1/38.0 cm - 1/6.0 cm = -0.0263 cm^-1
1/f2 = 1/d2 - 1/d3 = 1/6.0 cm - 1/42.0 cm = 0.1667 cm^-1
m = -(d2/f1)(f2/d3) = -(6.0 cm/(-0.0263 cm^-1))(0.1667 cm^-1/42.0 cm) = 1.53
Using the magnification equation, we can also relate the distances and focal lengths:
m = -f2/f1
Value of m and the given value of f2, we get:
1.53 = -14.0 cm/f1
f1 = -14.0 cm/1.53 = -9.15 cm
Since the focal length of a lens cannot be negative, we take the absolute value and get:
f1 = 9.15 cm
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The earth makes one complete revolution on its axis in 23 h 56 min. Knowing that the mean radius of the earth is 3960 mi, determine the linear velocity and acceleration of a point on the surface of the earth (a) at the equator, (b) at Philadelphia, latitude 40° north, (c) at the North Pole.
At the equator, Philadelphia, and the North Pole, a point's linear velocity and acceleration are determined as described above.
(a) At the equator:
The Earth's equator has a radius of 3960 miles. Therefore, the linear velocity of a point on the surface of the Earth at the equator is:
v = ωr = (2π / 23.9333 hours) * 3960 miles = 1037.564 mph
The acceleration of a point on the surface of the Earth at the equator can be calculated using the formula:
[tex]a = v^2 / ra = (1037.564 mph)^2 / 3960 miles = 0.273 g[/tex]
(b) At Philadelphia, latitude 40° north:
r = 3960 miles * cos(40°) = 3004.05 miles
The linear velocity of a point on the surface of the Earth at Philadelphia is:
v = ωr = (2π / 23.9333 hours) * 3004.05 miles = 784.166 mph
[tex]a = v^2 / r = (784.166 mph)^2 / 3004.05 miles = 0.154 g[/tex]
(c) At the North Pole:
r = 3960 miles * cos(90°) = 3960 miles
The linear velocity of a point on the surface of the Earth at the North Pole is:
v = ωr = (2π / 23.9333 hours) * 3960 miles = 1038.993 mph
The acceleration of a point on the surface of the Earth at the North Pole is:
[tex]a = v^2 / r = (1038.993 mph)^2 / 3960 miles[/tex] = 0.034 g
A key idea in physics is acceleration, which defines how rapidly an object's velocity alters over time. In other words, it is the rate at which the velocity of an object changes in relation to time. An item is considered to be accelerating when its speed changes, whether it is increasing or decreasing.
Depending on how the velocity changes, acceleration can be either positive or negative. For instance, an object's acceleration is positive while it is speeding up, whereas it is negative when it is going down. According to Newton's second equation of motion, acceleration is directly linked to the force pulling on an object.
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At the equator, Philadelphia, and the North Pole, a point's linear velocity and acceleration are determined as described above.
(a) At the equator:
The Earth's equator has a radius of 3960 miles. Therefore, the linear velocity of a point on the surface of the Earth at the equator is:
v = ωr = (2π / 23.9333 hours) * 3960 miles = 1037.564 mph
The acceleration of a point on the surface of the Earth at the equator can be calculated using the formula:
[tex]a = v^2 / ra = (1037.564 mph)^2 / 3960 miles = 0.273 g[/tex]
(b) At Philadelphia, latitude 40° north:
r = 3960 miles * cos(40°) = 3004.05 miles
The linear velocity of a point on the surface of the Earth at Philadelphia is:
v = ωr = (2π / 23.9333 hours) * 3004.05 miles = 784.166 mph
[tex]a = v^2 / r = (784.166 mph)^2 / 3004.05 miles = 0.154 g[/tex]
(c) At the North Pole:
r = 3960 miles * cos(90°) = 3960 miles
The linear velocity of a point on the surface of the Earth at the North Pole is:
v = ωr = (2π / 23.9333 hours) * 3960 miles = 1038.993 mph
The acceleration of a point on the surface of the Earth at the North Pole is:
[tex]a = v^2 / r = (1038.993 mph)^2 / 3960 miles[/tex] = 0.034 g
A key idea in physics is acceleration, which defines how rapidly an object's velocity alters over time. In other words, it is the rate at which the velocity of an object changes in relation to time. An item is considered to be accelerating when its speed changes, whether it is increasing or decreasing.
Depending on how the velocity changes, acceleration can be either positive or negative. For instance, an object's acceleration is positive while it is speeding up, whereas it is negative when it is going down. According to Newton's second equation of motion, acceleration is directly linked to the force pulling on an object.
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2. how do the results of this simulation exercise support the law of conservation of momentum? explain your answer.
This agreement between the simulation results and the Law of Conservation of Momentum serves as evidence that the law holds true in the simulated scenario.
The results of this simulation exercise support the Law of Conservation of Momentum by showing that the total momentum before an event (collision or separation) is equal to the total momentum after the event.
1. In the simulation exercise, you likely observed two objects interacting, such as colliding or separating.
2. Before the event, you can calculate the total momentum by adding the individual momenta of the objects (momentum = mass x velocity).
3. After the event, you can calculate the total momentum again by adding the individual momenta of the objects with their new velocities.
4. Comparing the total momentum before and after the event, you'll notice that they are equal or very close to equal, which demonstrates the Law of Conservation of Momentum in action.
Hence, the results of this simulation exercise support the Law of Conservation of Momentum.
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Listen Choose the items that help to fully describe VOLTAGE in a parallel circuit. 1) Directly related to resistance 2) Inversely proportional to current 3) Directly related to current 4) Used to slow the current 5) The resistance to the flow of current UND 6) Inversely proportional to resistance 7) Also known as Potential difference UN 8) Remains the same everywhere in a PARALLEL circuit UD 99 is provided by the battery 10) Is the flow of electricity 11) Directly related to voltage 12) Adds up to the total resistance
The items that help to fully describe VOLTAGE in a parallel circuit are 3) Directly related to current, 6) Inversely proportional to resistance, 7) Also known as Potential difference, and 11) Directly related to voltage.
Voltage can be used to slow the current, but it is not directly related to resistance in a parallel circuit.In a parallel circuit, the voltage remains the same everywhere, and it is provided by the battery. The total resistance in a parallel circuit is the sum of all the individual resistances. Current is the flow of electricity, and it is directly related to voltage and inversely proportional to resistance.To learn more about voltage, visit:
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A railroad car with a mass of 13000 kg collides and couple with a second car of mass 20,000kg that is initially at rest. the first car is moving with a speed of 3.5 m/s prior to the collision. a) what is the initial momentum of the first car ? b) if external forces can be ignored, what is the final velocity of the two railroad cars after they couple.
The initial momentum of the first car with the given data is:45500 kg*m/s
a) The initial momentum of the first car can be calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity. Thus, the initial momentum of the first car is:
p = 13000 kg * 3.5 m/s
p = 45500 kg*m/s
b) Since external forces can be ignored, we can use the law of conservation of momentum, which states that the total momentum of a system is conserved in the absence of external forces. Thus, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision:
Total momentum = p1 + p2
where p1 is the momentum of the first car and p2 is the momentum of the second car, which is initially zero.
Total momentum = 45500 kg*m/s + 0
Total momentum = 45500 kg*m/s
After the collision:
Total momentum = p1 + p2
where p1 and p2 are the final momenta of the two cars.
Since the two cars couple together after the collision, their final momentum is shared between them. We can assume that the final velocity of the two cars is v, which we want to find.
Thus, the final momentum of the two cars can be calculated using the formula p = (m1 + m2) * v, where m1 and m2 are the masses of the two cars.
Total momentum = (13000 kg + 20000 kg) * v
Total momentum = 33000 kg * v
Equating the total momentum before and after the collision, we get:
45500 kg*m/s = 33000 kg * v
Solving for v, we get:
v = 1.38 m/s
Therefore, the final velocity of the two railroad cars after they couple is 1.38 m/s.
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what is the energy required to accelerate a 1765 kg car from rest to 29 m/s?
The energy required to accelerate a 1765 kg car from rest to 29 m/s is approximately 373,128,250 Joules.
To calculate the energy required to accelerate a car from rest to 29 m/s, we can use the formula:
E = (1/2)mv^2
where E is the energy, m is the mass of the car, and v is the final velocity.
First, we need to convert the mass of the car from kilograms to grams:
m = 1765 kg = 1,765,000 g
Next, we can substitute the values into the formula:
E = (1/2)(1,765,000 g)(29 m/s)^2
Simplifying the equation:
E = (1/2)(1,765,000)(841) JE = 373,128,250 J
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a 0.142 kgkg baseball leaves a pitcher's hand at a speed of 28.5 m/sm/s. If air drag is negligible, how much work has the pitcher done on the ball by throwing it?
The pitcher has done 57.68 Joules of work on the 0.142 kg baseball by throwing it.
To calculate the work done on a 0.142 kg baseball by the pitcher, we need to consider the initial speed of the ball (28.5 m/s) and the terms "speed" and "work."
First, let's calculate the ball's kinetic energy (KE) using the formula: KE = 0.5 * mass * speed^2
KE = 0.5 * 0.142 kg * (28.5 m/s)^2
Now, solve for the kinetic energy:
KE = 0.071 * 812.25
KE = 57.68 J (Joules)
Since air drag is negligible, the work done by the pitcher on the ball is equal to the ball's kinetic energy. So, the pitcher has done 57.68 Joules of work on the 0.142 kg baseball by throwing it at a speed of 28.5 m/s.
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Suppose the number of turns in a rectangular coil of wire that is rotating in a magnetic field is tripled, what happens to the induced emf, assuming all the other variables remain the same?
A. It is reduced by a factor of 3
B. It is reduced by a factor of 9
C. It is increased by a factor of 3
D. It it reduced by a factor of 9
E. It remains the same
If the number of turns in a rectangular coil of wire that is rotating in a magnetic field is tripled, the induced emf is increased by a factor of 3. (C)
This is because the emf is directly proportional to the number of turns in the coil. So, if the number of turns is tripled, the induced emf will also be tripled. It is important to note that this assumes all other variables, such as the magnetic field strength and the angular velocity of the coil, remain constant.
In summary, increasing the number of turns in a rotating rectangular coil of wire will increase the induced emf, while decreasing the number of turns will decrease the induced emf. This principle is used in many electrical devices, such as generators and motors, to control the amount of electrical energy produced or consumed.
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A section of a sphere is mirrored on both sides. If the magnification of an object is +4.10 when the section is used a concave mirror, what is the magnification of an object at the same distance in front of the convex side?
_______________
Magnification is the relationship between the size of an image and the size of the item that created it in optics. The ratio of the image length to the object length, as measured in planes perpendicular to the optical axis, is referred to as linear magnification, also known as lateral or transverse magnification.
Since the section of the sphere is mirrored on both sides, the focal length of the concave mirror and the convex mirror will be the same. Therefore, we can use the mirror formula:
1/f = 1/u + 1/v
Where f is the focal length, u is the distance of the object from the mirror, and v is the distance of the image from the mirror.
When the section is used as a concave mirror, the magnification is given by:
m = -v/u = +4.10
Since the magnification is positive, the image is upright.
Now, when the same object is placed in front of the convex side at the same distance u, the image will be virtual and erect. The magnification is given by:
m = v/u
To find v, we need to first find f. We know that:
m = -v/u = +4.10
Therefore, v = -4.10u
Now, using the mirror formula, we can find f:
1/f = 1/u + 1/v
1/f = 1/u - 1/4.10u
1/f = (4.10 - 1)/4.10u
f = 4.10u/3.10
f = 1.32u
Now that we know the focal length, we can find the image distance v:
1/f = 1/u + 1/v
1/1.32u = 1/u + 1/v
v = -0.32u
Therefore, the magnification is: m = v/u = -0.32
So, the at the same distance in front of the convex side is -0.32.
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