Answer:
[tex]500 \: \mathrm{kg} \cdot \mathrm{m/s}[/tex]
Explanation:
The momentum of an object is given as [tex]p=mv[/tex]. Since Amy has a mass of 50 kg and is travelling 10 m/s, her momentum is [tex]p=mv=50\cdot 10 =\fbox{$500\: \mathrm{kg\cdot m/s}$}[/tex].
Answer:
500
Explanation:
This illustration shows two opposing forces pulling on a wagon. Which description best describes how the wagon will move?
10N
30N
The wagon will slow down
The wagon will move to the left.
The wagon will not move because the forces are opposite
The wagon will move to the right.
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Answer:
The wagon will move to the right.
Explanation:
From the question given above, the following data were obtained:
Force applied to the left (Fₗ) = 10 N
Force applied to the right (Fᵣ) = 30 N
Direction of the wagon =.?
To determine the direction in which the wagon will move, we shall determine the net force acting on the wagon. This can be obtained as follow:
Force applied to the left (Fₗ) = 10 N
Force applied to the right (Fᵣ) = 30 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 30 – 10
Fₙ = 20 N to the right
From the calculations made above, the net force acting on the wagon is 20 N to the right. Hence the wagon will move to the right.
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 32.7 m/s2 with a beam of length 5.29 m , what rotation frequency is required
Answer:
The rotation frequency required is 23.78 RPM
Explanation:
Given;
radial acceleration, a = 32.7 m/s²
length of the beam, r = 5.29 m
The linear velocity is calculated as;
[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{ar}[/tex]
where;
v is linear velocity
The angular velocity is calculated as;
[tex]\omega = \frac{v}{r} \\\\Recall, v = \sqrt{ar} \\\\Then, \omega = \frac{\sqrt{ar}}{r}} \\\\ \omega = \frac{\sqrt{32.7 \times5.29}}{5.29}\\\\\omega = 2.49 \ rad/s\\\\Angular \ frequency \ is \ calculated \ as;\\\\\omega = 2\pi f\\\\f = \frac{\omega}{2\pi} \\\\f = \frac{2.49}{2\pi} \\\\f = 0.396 \ rev/s\\\\f = 23.78 \ rev/min[/tex]
Therefore, the rotation frequency required is 23.78 RPM
A toy car has an initial acceleration of 2m/s^2 across a horizontal surface after it is released from rest. After the car travels for a time t consisting of only the car an open system or a closed system, and why?
a. Open system because the acceleration of the car is not constant.
b. Open system, because an external force is applied to the car that causes it to accelerate.
c. Closed system, because the speed of the car is as expected in the case where an object has uniform acceleration for a timet
d. Closed system because mechanical energy was not removed from the system as a result of a net force
Answer:
Open system, because an external force is applied to the car that causes it to accelerate.
Explanation:
In this case, the system under study is the toy car. A system may be open or closed.
An open system is acted upon by external forces but a closed system is not acted upon by any external force.
Since an external force acted on the toy car causing it to accelerate, then it is an open system.
In the figure, given ∆x=30cm, k=200N/cm, g=10m/s². Find the mass of the object
Answer:
600 Kg
Explanation:
From the question given above, the following data were obtained:
Extention (∆x) = 30 cm
Spring constant (K) = 200 N/cm Acceleration due to gravity (g) = 10 m/s²
Mass (m) of object =?
Next, we shall determine force exerted. This can be obtained as follow:
Extention (∆x) = 30 cm
Spring constant (K) = 200 N/cm
Force (F) =?
F = K∆x
F = 200 × 30
F = 6000 N
Finally, we shall determine the mass of the object. This can be obtained as follow:
Acceleration due to gravity (g) = 10 m/s²
Force (F) = 6000 N
Mass (m) of object =?
F = mg
6000 = m × 10
Divide both side by 10
m = 6000 / 10
m = 600 Kg
Thus, the mass of the object is 600 Kg
If the force of gravity suddenly stopped acting on planets, they would
A.) spiral slowly towards the sun
B.) continue to orbit the sun
C.) move in straight lines tangent to thier orbits
D.) spiral slowly away from the sun
E.) fly straight away from the sun
PLEASE HELP!!!! WILL GIVE BRAINLIEST!!! WORTH 20 PTS!!!!!
Please write the processes and tectonic plate interactions that are causing the geological event to occur. Include an explanation of convection currents.
Answer:
Explanation:
These tectonic plates rest upon the convecting mantle, which causes them to move. The movements of these plates can account for noticeable geologic events such as earthquakes, volcanic eruptions, and more subtle yet sublime events, like the building of mountains.
Answer:
Tectonic plates recurring over and over again cause a rupture in the Earth's crust. This agitates the magma in volcanoes, causing the volcano to erupt.
Explanation:
3. A car is traveling up a 3% grade, with the speed of 85mph, on a road that has good, wet pavement. A deer jumps out onto the road and the driver applies the brakes 290-ft from it. The driver hits the deer at a speed of 20mph.If the driver did not have antilock brakes, and the wheels were locked the entire distance, would a deer-impact speed of 20mph be possible
Answer:
Explanation:
From the given information;
Let assume that:
the wheel radius = 15 inches
the driveline slippage = 3%; &
the gear reduction ratio (overall) = 2.5 to 1
So; using the equation:
[tex]v_1= \dfrac{2 \pi r n_o (1 -i)}{\varepsilon_o}[/tex]
[tex]v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (1 -0.03)}{2.5}[/tex]
[tex]v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (0.97)}{2.5}[/tex]
[tex]v_1 = 126.92 \ fp^3[/tex]
[tex]frl = 0.01 ( 1+ \dfrac{v}{147}) \ if \ v \ is \ ft/sec[/tex]
[tex]frl = 0.01 \Bigg( 1+ \dfrac{\dfrac{126.92 +(20)1.47 }{2} }{147}\Bigg)[/tex]
[tex]S = \dfrac{v_b ( v_1^r-v_2^r)}{2g(n_b \mu + frl \pm sin \ y}[/tex]
where;
[tex]\mu = 0.6[/tex]
[tex]291 = \dfrac{1.64( 126.92^2-29.9^2)}{64.4(n_b \times 0.6 +0.01532 +0.03}[/tex]
[tex]n_b = 1.33 \to which \ is \ not \ possible[/tex]
However;
[tex]n_b \mu = 1.33(0.6) = 0.80[/tex]
[tex]\mu = 0.9 \to[/tex] if the car's anti-clock breaking system did not fail
Thus;
[tex]n_b (0.9) = 0.80[/tex]
[tex]n_b =\dfrac{ 0.80}{(0.9) }[/tex]
[tex]n_b = 0.89[/tex]
Hence, the distance is possible if the anti-clock breaking system did not fail.
A jogger takes 30 s to run a total distance of 200 m by running 150 m [E] and then 50 m [W]. (a)
Calculate the jogger’s average speed (b) Calculate the jogger’s average velocity
Answer:
(a) 6.67 m/s (b) 3.33 m/s
Explanation:
if i understood your question correctly, the jogger goes 150m east before going in the opposite direction west for 50m. (see diagram)
in avg speed, we use the total distance (150m + 50m = 200m) and total time (30s)
v = d/t =200/30 =6.67 m/s
in avg velocity, we use the total displacement (150m - 50m = 100m) and total time (30s)
v= d/t = 100/30 = 3.33 m/s
A pingpong ball has 2 kg/s of momentum when
thrown 8 m/s. Find the mass of the ball.
Answer:
0.25 kg
Explanation:
p = mv
2 = m(8)
2/8 = m(8)/8 *cancels
m = 1/4 OR 0.25 kg
Problem Solving: Show your solution:
Calculate the wavelengths of the following electromagnetic waves.
a) Microwave in an oven with a frequency of 2.45 GHz.
b) Radio waves used by submarines with a frequency of 24 kHz
Answer:
(a) λ = 0.1224 m
(b) λ = 12500 m = 12.5 km
Explanation:
The wavelength can be calculated with the help of the frequency of the waves. The formula utilized for this purpose is given as follows:
[tex]\lambda = \frac{c}{f}[/tex]
where,
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of the wave
f = frequency of the wave
(a)
f = 2.45 GHz = 2.45 x 10⁹ Hz
λ = ?
Therefore,
[tex]\lambda = \frac{3\ x\ 10^{8}\ m/s}{2.45\ x\ 10^{9}\ Hz}\\[/tex]
λ = 0.1224 m
(b)
f = 24 KHz = 2.45 x 10³ Hz
λ = ?
Therefore,
[tex]\lambda = \frac{3\ x\ 10^{8}\ m/s}{24\ x\ 10^{3}\ Hz}\\[/tex]
λ = 12500 m = 12.5 km
A ball bearing is projected vertically upwards from the ground with a velocity of 15ms. Calculate the time taken by the ball to return to the ground (g=10ms^2)
Answer:
t = 3 [s]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}-g*t[/tex]
where:
Vf = final velocity [m/s]
Vo = initial velocity = 15 [m/s]
g = gravity acceleration = 10 [m/s²]
t = time [s]
Now replacing we have:
[tex]0 = 15 -10*t\\10*t=15\\t= 1.5[s][/tex]
Note: In the equation above the gravity acceleration is negative, because the movement of the ball bearing is pointing againts the gravity acceleration.
The time calculated is only when the ball bearing reaches the highest elevation, and it will take the same time for descending, therefore the total time is:
t = 1.5 + 1.5 = 3 [s]
An object is released from rest at a height H near and above the surface of Earth. As the object falls toward the surface, Earth’s atmosphere exerts a resistive force on the object such that it reaches a terminal velocity before it reaches the ground. Which of the following claims is true? Select two answers.
The system consisting of only the object is an open system.
The system consisting of only the object is an open system.
A
Earth’s atmosphere does negative work on the object as it falls toward the surface.
Earth’s atmosphere does negative work on the object as it falls toward the surface.
B
The change in the object’s kinetic energy from the instant it is released from rest, to the instant it reaches terminal velocity, is zero.
The change in the object’s kinetic energy from the instant it is released from rest, to the instant it reaches terminal velocity, is zero.
C
The total mechanical energy of the object-Earth system remains constant at all times in which the object is in motion.
Answer:
Second and Last Option Are Correct
Explanation:
A bug starts out at rest, 3 m to the right of the origin. It then starts moving on a trip. After 1 s, the bug is seen at 9 m to the right of the origin, travelling at 5 m/s to the right. After 7 s (from the start of the trip), the bug is seen at 2 m to the left of the origin, travelling at 8 m/s to the right. a.) What is the position vector of the bug 1 s after the start of the trip
Answer:
d = 6 i^ m
Explanation:
This is an exercise in kinematics, where the position vector is the displacement of the body from one point to another.
In this case we are told that for t = 0 the worm is at x₀ = 3m and with velocity starts from zero, after t = 1 s it is at x₁ = 9m and with a velocity of
v₁ = 5 m / s
They ask what is the displacement for the time of 1 s
d = x₁ - x₀
d = 9 -3 m
d = 6 m
Bold indicates vector, displacement vector is
d = 6 i ^ m
Learning task 2: Using the information you gathered from Learning Task 1, make a concept web of the contributions of the following scientist in the DEVELOPMENT OF MAGNETIC THEORY
A. Andre- Marie Ampere
B. Michael Faraday
C. Heinrich Herts
D. James Clerk Maxwell
E. Hans Christian Oersted
Answer:
The contributions of the following scientist in the DEVELOPMENT OF MAGNETIC THEORY
James Clerk Maxwell Hans Christian OerstedExplanation:
George Green was the first personality to formulate a mathematical principle of magnetism and electricity and his system created the framework for the work of different scientists such as William Thomson, James Clerk Maxwell, and others. Magnetism is the power exercised by magnets when they drag or deflect each other. Magnetism is produced by the movement of electric charges.
The contributions of James Clerk Maxwell and Hans Christian Oersted, et al in the DEVELOPMENT OF MAGNETIC THEORY are as follows:
They discovered that the speed at which electromagnetic waves traveled was similar to that of lightThey proved that there was a proportional connection between electricity and magnetismAccording to the given question, we are asked to show the contributions which the aforementioned scientists had in the development of the magnetic theory.
As a result of this, we can see that James Maxwell first developed this theory in the nineteenth century and the theory was modified by other scientists who made the framework for the electrical system and magnetism.
Read more here:
https://brainly.com/question/17913237
John and Tom were given one mirror each by their teacher. Tom found his image to be erect and of the same size whereas John found her image erect and smaller in size. This means that the mirrors of John and tom are, respectively
(a) plane mirror and concave mirror.
(b) concave mirror and convex mirror.
(c) plane mirror and convex mirror.
(d) convex mirror and plane mirror
Answer:
(d)
Explanation:
John- convex mirror
Tom - plane mirror
Answer:
(d) convex mirror and plane mirror
Explanation:
A plane mirror forms an image that is : virtual { behind the mirror } , image and object are at the same distance from the flat mirror, image is upright and image size is the same as object size. Tom's image.
A concave mirror form both real and virtual images. When a concave mirror is very near to an object , the image is virtual and magnified. When the distance between object and mirror is increased, a real image is formed and the size is reduced.
In a convex mirror, the image formed is smaller than the object, it is upright and is located behind the mirror. The image is virtual. John's image.
Which is the best explanation for why Toms technique works ?
Read the scenario and solve these two problems.
When traveling at top speed, a roller coaster train with a mass of 12,000 kg has a velocity of 30 m/s. The kinetic energy of the train at top speed is
J.
Given this kinetic energy, what is the tallest hill this roller coaster train can reach the top of?
The train can climb a hill that is
m high.
Answer:
45.9m
Explanation:
Given parameters:
Mass of roller coaster train = 12000kg
Velocity = 30m/s
Unknown:
Tallest hill the roller coaster train can reach = ?
Solution:
Kinetic energy is the energy due to the motion of body.
Potential energy is the energy due to the position of a body.
So;
[tex]\frac{1}{2}[/tex] m v² = mgh
m is the mass, g is the acceleration due to gravity, h is the height and v is the velocity
m v² = 2mgh
v² = 2gh
h = [tex]\frac{v^{2} }{2g}[/tex]
h = [tex]\frac{30^{2} }{2 x 9.8}[/tex] = 45.9m
True or False. Facts are based on observations. *
True
False
Answer:
TRUE
Explanation:
IM SMART
Answer:
Stay Safe! ,God bless you . The answer is false ,
Explanation:
A weightlifter curls a 32 kg bar, raising it each time a distance of 0.50 m. How many times must he repeat this exercise to burn off the energy in one slice of pizza? Assume 25% efficiency.
Answer:
Explanation:
Average energy contained by a slice of pizza is 860 J .
energy used in lifting 32 kg bar by .50 m = mgh
= 32 x 9.8 x .5 = 156.8 J
efficiency is 25 % , so energy used up = .75 x 156.8 = 117.6 J
So number of times exercise to be repeated to burn off energy of a slice of pizza
= 860 / 117.6
= 7.3 or 7 times .
An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-quency will the car and passenger vibrate onthe springs? Answer in units of Hz. The acceleration of gravity is 9.81 m/s^2.
Answer:
The required frequency = 0.442 Hz
Explanation:
Frequency [tex]f = ( \dfrac{1}{2 \pi}) \omega[/tex]
where;
[tex]\omega = \sqrt{\dfrac{k}{m} }[/tex]
Then;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )[/tex]
However;
[tex]k = \dfrac{F}{x}[/tex] and;
mass [tex]m = m_{car } + m_{person}[/tex]
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )[/tex]
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )[/tex]
where;
[tex]F = m_{person}g[/tex]
Then;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )[/tex]
replacing the values;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )[/tex]
[tex]\mathbf{f = 0.442 \ Hz}[/tex]
A 12-kg object is moving rightward with a constant velocity of 4 m/s. How much net force is required to keep the object moving with
the same speed and in the same direction?
Calculate the extension of a 20cm spring that has a spring contrast of 45000N/m and 1500N of elastic potential energy.
Answer:
See the answer below
Explanation:
The elastic potential energy can be calculated by means of the following equation:
[tex]E_{el}=\frac{1}{2} *k*x^{2}[/tex]
where:
Eel = elastic energy = 1500 [J]
k = spring constant = 45000 [N/m]
x = extension [m]
[tex]1500=\frac{1}{2} *45000*x^{2} \\3000=45000*x^{2} \\x=\sqrt{0.06666}\\x=0.258[m] = 25.8 [cm][/tex]
How do sound waves travel?
What is nature/nurture debate and why is it important in psychology
Water will expand more than ___
A. door
B. juice
C. air
Answer:
Its A. door because liquids expand better than solids
WHat does that mean?
true or false A person's speed around the Earth is faster at the poles than it is at the equator.
Answer:False
Explanation:The Earth rotates faster at the equator than at the poles.
Where do the rain forest plants get their nutrients?
from plants that fall to the forest floor
from animals that die on the forest floor
from plants and animals that die and fall to forest floor
from sunlight
Answer:
From plants and animals that die and fall to forests floor
Explanation:
Hope it helps
Mark brainliest
at what speed does the kg ball move ?
Answer: Choice A) 2 meters per second
=======================================================
Explanation:
The smaller ball has momentum of
p = m*v
p = (1 kg)*(4 m/s)
p = 4 kg*m/s
All of this momentum transfers into the larger ball because the smaller ball comes to a complete stop.
For the larger ball, we have p = 4 and m = 2. Let's find v.
p = m*v
4 = 2*v
4/2 = v
2 = v
v = 2 m/s which is why the answer is choice A
The larger ball moves at a speed of 2 meters per second. The speed is cut in half compared to the smaller ball because the larger ball has more inertia (aka more mass), and therefore it takes more energy to move it. If you apply the same energy to each, then the smaller object moves faster.
A 2.6 kg ball is accelerated at 4.5 m/s2.
Calculate the force needed to achieve this feat.
Show all work including formula and units!
Answer:
[tex]12\:\mathrm{N}[/tex]
Explanation:
Force is given by the equation [tex]F=ma[/tex].
Plugging in given values, we have:
[tex]F=ma=2.6\cdot 4.5=11.7=\fbox{$12\:\mathrm{N}$}[/tex] (two significant figures).