Answer:
Explanation:
The force occurring on a spring F = mg
If x should be the expansion in the spring.
Then F = mg = -kx
where;
[tex]x = \dfrac{mg}{k}[/tex]
[tex]x = \dfrac{2.5 \ kg*9.81 \ m/s^2 }{58 \ N/m}[/tex]
x = 0.423 m
x = 42.3 cm
The final reading of the spring when the end of the spring lines up with the ruler mark is = 42.3 cm + 17 cm
= 59.3 cm
≅ 60 cm
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 34.5 N is applied tangent to the rim of the disk.
a) What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through .200 revolution?
b) What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through .200 evolution?
Answer:
a) v = 1.01 m/s
b) a = 5.6 m/s²
Explanation:
a)
If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:[tex]\tau = I * \alpha (1)[/tex]
Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:τ = F*r (2)For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).Replacing (2) and (3) in (1), we can solve for α, as follows:[tex]\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)[/tex]
Since the angular acceleration is constant, we can use the following kinematic equation:[tex]\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)[/tex]
Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:[tex]0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)[/tex]
Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:[tex]\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)[/tex]
Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:[tex]v = \omega * r (8)[/tex]
where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.Replacing this value and (7) in (8), we get:[tex]v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)[/tex]
b)
There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:[tex]a_{t} = \alpha * r (9)[/tex]
where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.Replacing this value and (4), in (9), we get:[tex]a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)[/tex]
Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.The radial acceleration is just the centripetal acceleration, that can be expressed as follows:[tex]a_{c} = \omega^{2} * r (11)[/tex]
Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.Replacing this value and (7) in (11) we get:[tex]a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)[/tex]
The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:[tex]a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)[/tex]
The speed limit on some segments of interstate 4 is 70 mph. What is this in km/h?
Answer:
112.63km/hr
Explanation:
The given dimension is :
70mph
We are to convert this to km/hr
1 mile = 1.609km
so;
70mph x 1.609 = 112.63km/hr
So,
The solution is 112.63km/hr
What happens to Average Speed if distance decreased & time stayed the same?
Answer:
I think you are trying to ask 'what happens to the average speed (of humans or animals.) if the distance decreased and time stayed the same (the time such as what hour it is or what second?)' If that is the case then the average speed will be more since the distance decreased. And the time staying the same might affect it I am not fully sure so I will not say anything about that. I hope this helps!
need help pleaseee !
Answer:
08 and 09 with be 67
Explanation:
i know this because it was the vase major
Block X and block Y travel toward each other along a horizontal surface with block X traveling in the positive direction. Block X has a mass of 4kg and a speed of 2ms. Block Y has a mass of 1kg and a speed of 1 ms. A completely inelastic collision occurs in which momentum is conserved. What is the approximate speed of block X after the collision
Answer:
v = 1.4 m / s
Explanation:
To solve this problem we use the law of conservation of momentum, for this we define a system formed by the two blocks in such a way that the force during the collision have been internal
initial instant. Just before the crash
p₀ = M v₁ - m v₂
final instant. Right after the crash
p_f = (M + m) v
the moment is preserved
p₀ = p_f
M v₁ - m v₂ = (M + m) v
v = [tex]\frac{1}{M+m}[/tex] (M v₁ - mv₂)
let's calculate
v = [tex]\frac{1}{4+1}[/tex] (4 2 - 1 1)
v = 1.4 m / s
in the same direction as the largest block (M)
The approximate speed of block X after the collision is 1.4 m/s.
Given information:
Block X and block Y travel toward each other along a horizontal surface with block X traveling in the positive direction.
Block X has a mass of [tex]m_1=[/tex] 4 kg and a speed of [tex]u_1=[/tex] 2 m/s.
Block Y has a mass of [tex]m_2=[/tex] 1 kg and a speed of [tex]u_2=[/tex] -1 m/s.
The collision between them is perfectly inelastic. So, the final velocity of both the objects will be the same.
The momentum of the system will be conserved. Let v be the final velocity of both the blocks.
The final velocity v can be calculated as,
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\4\times 2-1\times1=(4+1)v\\5v=7\\v=1.4\rm\;m/s[/tex]
Therefore, the approximate speed of block X after the collision is 1.4 m/s.
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When measuring wellness you must consider
A all components of health
B your physical fitness being in the top 10% of the population
C being free of diseases
D both physical and mental health
Answer:
The answer is A) all components of health
Explanation:
Got it right on edge
Can someone help me out please I got it wrong
Answer:
3 maybe since protons=atomic
a spring stretches from an initial height of 5 cm to a final stretch of 10 cm. the spring constant is 800 n/m.How much work was done on the spring?
what is the final force on the spring when it is at its 10 cm stretch?
explain why it is not appropriate to use the equation w=f//d when considering springs.
Answer:
.
Explanation:
F = kx so k = 800/((10-5)/100) = 16000 N/m
W = 1/2 kx^2 = 1/2 * 16000 * .05^2 = 20 J.
(sorry if it's wrong)
Why are models of atoms and molecules useful explain
Answer:
The models are larger versions of atoms/molecules. It is a visualization so we can understand the super small particles.
Explanation:
Atoms and molecules are very, very, small. They average 100 picometers across.
They definitely cannot be seen with the naked human eye. We can only see them with special microscopes (electron, optical, etc). that magnify hundreds of thousands of times.
So, we use models of atoms and molecules. It allows us to study, visualize, understand, and "see" the particles on a much larger scale.
To review the solution to a similar problem, consult Interactive Solution 1.43. The magnitude of a force vector is 87.4 newtons (N). The x component of this vector is directed along the x axis and has a magnitude of 72.1 N. The y component points along the y axis. (a) Find the angle between and the x axis. (b) Find the component of along the y axis.
Answer:
(a) 34.4°
(b) 49.4 N
Explanation:
(a) From the diagram,
Amgle between the x axis can be calculated as,
cosΘ = adjacent/hypoteneous
cosΘ = 72.1/87.4
cosΘ = 0.8249
Θ = cos⁻¹(0.8249)
Θ = 34.4°.
Hence the angle between the x axis is 34.4°
(b) To find the component along the y axis we make use of pythagoras theorem.
a² = b²+c²................... Equation 1
Where a = 87.4 N, b = 72.1 N, c = y.
Substitute these values into equation 1
87.4² = 72.1² + y²
y² = 87.4²-72.1²
y² = 2440.35
y = √(2440.35)
y = 49.4 N
I will give brainly
Defend Democritus' work on the atom and its contribution to the modern atomic model.
What is the strength of an electric field that will put a force of
1.28 x 10-15 N on a proton?
Answer: E = 7,490.6 N/C
Explanation:
If we have a field E, and a particle with a charge q, the force that the particle experiences is:
F = E*q
In this case, we know that the force is:
F = 1.2*10^(-15) N
And we know that the particle is a proton, where the charge of a proton is:
q = 1.602*10^(-19) C
Then we can replace these two values in the equation to get:
1.2*10^(-15) N = E*1.602*10^(-19) C
We just need to isolate E.
(1.2*10^(-15) N)/(1.602*10^(-19) C) = E
7,490.6 N/C = E
That is the strength of the electric field.
In which part of a lab report would be the following sentence most likely occur? “Since the data showed that the
Answer:
most likely be included in the analysis section of a lab report
Explanation:
Plz help this is so confusing
The correct answer is 5 km/h
Explanation:
The speed at which the duck travels can be found by using the equation that is given (speed= distance/ time). The first step to do this is to replace distance and time using the values given. Here is the process:
speed = distance / time
speed = 10 km / 2 hours
Now, solve this equation
speed = 10 km/ 2 hours
10 / 2 or 10 divided 2 = 5
Finally, use the units, in this case, the correct units are km/h
Artificial satellites in space can help you find locations on
Earth. True or false?
Please answer this question I don't know how to do it.
A 2 kg toy car moves at a speed of 5 m/s. If a child applies a 5N force for 2 m in the same direction the car is already moving, what is the change in kinetic energy of the car?
Answer:
[tex]10\: \mathrm{J}[/tex]
Explanation:
The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.
The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].
After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.
To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]
We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].
To find acceleration:
[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].
Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].
Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].
Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).
The change in the kinetic energy of the car is 10 J.
The given parameters;
mass of the car, m = 2 kginitial velocity of the car, u = 5 m/sforce applied by the child, F = 5 Ndistance traveled, s = 2 mThe acceleration of the car is calculated as follows;
[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]
The final velocity of the car is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]
The change in the kinetic energy of the car is calculated as follows;
[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]
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A 62 kg student, starting from rest, slide down an 10.6 m high water slide. How fast is he going at the bottom of the slide? Use g = 10 m/s2
Answer:
14.6m/s
Explanation:
Given parameters:
Mass of the student = 62kg
Initial velocity = 0m/s
Height of slide = 10.6m
g = 10m/s²
Unknown:
Speed at the bottom of the slide = ?
Solution:
The speed at the bottom of the slide is the final velocity;
v ² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 0² + 2x 10 x 10.6
v² = 212
v = 14.6m/s
In a parallel circuit, if one connection is broken
A. all of the connections stop working. B. B only broken one will stop working
C. Nothing happens to circuit.
Explain:
A safety plug is designed to melt when the pressure inside a metal tank becomes too high. A gas
at 51.0 atm and a temperature of 23.0°C is contained in the tank, but the plug melts when the
pressure reaches 75.0 atm. What temperature did the gas reach?
Question 12 of 25
What type of energy is thermal energy?
O A. Chemical energy
O B. Nuclear energy
O C. Kinetic energy
O D. Gravitational potential energy
It’s C kinetic energy
Small household electrical devices, such as vacuum cleaners, televisions, and floor lamps, each draw a different amount of current, but all require 120 volts to operate. Are the outlets in of a power-strip, then, wired in series or parallel
Answer:
the selected configuration is PARALLEL
Explanation:
There are two ways to connect electrical circuits.
In series in this type of circuit the equipment is connected to the same line, in this case the current of the circuit circulates through all the inputs is the same, but the voltage in each one is a part of the total voltage, this circuit has the problem that if an appliance is disconnected or damages the power and the entire circuit is interrupted
V_total = ∑ V_i
In a parallel circuit the equipment is connected in two two cables, for which the current is divided in each branch, the voltage in all the branches is the same, this type of circuit has the advantage that if one equipment is damaged or disconnects the others they can continue working
V_total = V
Therefore, in consequence of the above, the selected configuration is PARALLEL
In the picture shown below A represents a characteristic of only geocentric model, B represents a characteristic common to both geocentric and heliocentric models, C represents a characteristic of only heliocentric model, and D represents a characteristic which the geocentric and heliocentric models do not have.
Under which label will the characteristic, "The sun and planets revolve around a central moon in the solar system" fall?
A
B
C
D
how much power does it take to do 104 J of work in 8 sec?
Answer:
Given:
W=104J
t=8sec
Solve:
P=?
Equation:
P=W/t or Fd/t
P= 104J'8sec
P=13W
How does the geosphere interact with the biosphere during a fire?
Answer: A forest fire could affect the geosphere by ash and the heat from the fire, but the fire would affect the biosphere by less oxygen in the air and would make it hard for human to breath affecting climate change. Also many homes to wildlife that live in the forest would lose their homes. This would result in lose of money from no hunting and that affect the economy.
an incline projection can achive maximium horizontal horizontal range at an angle of projection,ፀ=45 why?
Answer:
Explanation:
For an incline projection made at angle of θ , expression of horizontal range R can be given by the following equation .
R = u² sin 2θ / g
u is initial velocity of throw .
For range R to be maximum , the value of sin 2θ must be maximum . The maximum value of sin of an angle is 1 , so
For maximum R ,
sin2θ = 1 = sin90
2θ = 90 .
θ = 45⁰ .
So for projection made at 45⁰ , horizontal range is maximum .
the diameter of the wheels on your car ( including the tires) is 25 inches. you are going to drive 250 miles today. each of your wheels is goingnto turn by an angle of
Let's say you were using a single-pulley system to lift a flag up a flagpole. If you wanted to raise the flag 10 meters into the air, how far would you have to move the rope?
a. 5 meters
b. 10 meters
c. 20 meters
d. 50 meters
Answer:
b. 10 meters
Explanation:
In a single pulley , which is a lever , velocity ratio is one . In other words , velocity of load is equal to velocity of effort . The load shifts by the same distance as the force or effort moves . The only advantage of this pulley is that direction of force is reversed .
Hence to raise a flag by 10 m , rope would have to move by 10 m .
Some giant ocean waves have a wavelength of 25 m and travel at 6.5 m/s with a frequency of 0.26 HZ. What is the period of such a wave ?
Answer:
3.85s
Explanation:
Given parameters:
Wavelength = 25m
Velocity = 6.5m/s
Frequency = 0.26Hz
Unknown:
Period of the wave = ?
Solution:
The period of a wave is the inverse of the frequency of the wave.
Period = [tex]\frac{1}{frequency}[/tex]
Period = [tex]\frac{1}{0.26}[/tex] = 3.85s
A north magnetic pole is facing another north magnetic pole with a distance x. If the distance between the poles becomes 12x, what happens to the magnitude of the field energy between them?
The answer is "The field energy will increase"
What will happen to the magnitude of the field energy between them is that
The field energy will increaseWhenever there is a stored energy in an object at rest, the object will move in the direction that causes the energy stored to decrease, which will be replaced by movement energy.
Therefore, if two magnets with like poles are pointing together, then the energy stored will decrease as they move apart. Hence, as the distance between poles decreases, field energy increases.
What are magnetic poles?Magnetic pole are region at each end of a magnet where the external magnetic field is strongest.
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