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tino4ka555 [31]
3 years ago
5

A student conducts an experiment in which an object travels across a horizontal surface while for 2 s a net force is applied to

a 2 kg object that initially travels with a speed of 0.5 m/s. Data collected from the experiment are used to create the graph of the magnitude of the applied force exerted on the object as a function of time is shown. All frictional forces are considered to be negligible. Can the student use the graph and the known data to determine the momentum of the object after the force has been applied?
Yes, because the average force exerted on the object can be used with the time interval of 2 s to determine the change in momentum of the object by using Δp⃗ =F⃗ Δt.


Yes, because the area bound by the best-fit line and the horizontal axis from 0 s to 2 s can be used to determine the change in momentum of the object.


No, because the data is too scattered to determine the average force that is exerted on the object.


No, because the student needs to know the direction that the force is applied to the object because the applied force will be in the same direction as the change in momentum of the object.
Physics
1 answer:
erma4kov [3.2K]3 years ago
3 0

Answer:

  D.  No, because the student needs to know the direction that the force is applied

Explanation:

The change in momentum depends on the direction of the force as well as its magnitude. Since the graph only supplies force magnitude information, it is insufficient to allow the student to calculate the change in momentum.

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A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it
daser333 [38]

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

T = FR = 18 * 2.4 = 43.2 Nm

According to Newton 2nd law, the angular acceleration would be

\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2

It starts from rest, then after 15s it would achieve a speed of

\omega = \alpha t = 0.021 * 15 = 0.31 rad/s

(b) The distance angle swept by it is:

\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad

Hence the work by the child

W = T\theta = 43.2 *2.314  \approx 100 J

c) Average power to work per time unit

P = \frac{W}{t} = \frac{100}{15} = 6.67 W

7 0
3 years ago
A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving downward at 30 cm/s just before l dith t
viktelen [127]

Answer:

The speed of the cart and clay after the collision is 50 cm/s .

Explanation:

Given :

Mass of lump , m = 500 g = 0.5 kg .

Velocity of lump , v = 30 cm/s .

Mass of cart , M = 1 kg .

Velocity of cart , V = 60 cm/s .

We know by conservation of momentum :

mv+MV=(m+M)v'

Here , v' is the speed of the cart and clay after the collision .

Putting all value in above equation .

We get :

0.5\times 30+1\times 60=(0.5+1)\times v'\\\\v'=\dfrac{15+60}{1.5}\\\\v'=50\ cm/s

Hence , this is the required solution .

3 0
3 years ago
Identify each picture as either an inelastic collision or elastic collision
Ivan

Answer:

<u>Inelastic collision:</u>

A collision in which there is a loss of Kinetic Energy due to internal friction of the bodies colliding.

<u>Characteristics of an inelastic collision:</u>

  • <em>the momentum of the system is conserved</em>
  • <em>the momentum of the system is conservedloss of kinetic energy</em><u> </u>

<em>I</em><em>n</em><em> </em><em>a perfectly elastic collision</em><em>, the two bodies </em><em>that</em><em> </em><em>collide with each other stick together.</em>

<u>Elastic </u><u>collision</u><u>:</u>

A collision in which the kinetic energy of the two bodies, before and after the collision, remains the same.

<u>Characteristic</u><u>s</u><u> </u><u>of</u><u> </u><u>elastic</u><u> </u><u>collision</u><u>:</u>

  • <em>the</em><em> </em><em>momentum</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>system</em><em> </em><em>is</em><em> </em><em>conserved</em>
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In everyday life, no collision is perfectly elastic.

__________________

ANSWER:

<u>Given examples:</u>

  • Two cars colliding with each other form an example of inelastic collision.

<u>Reason:</u>

<em>(</em><em>T</em><em>hey</em><em> </em><em>lose</em><em> </em><em>kinetic</em><em> </em><em>energy</em><em> </em><em>and</em><em> </em><em>come</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>stop</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>collision</em><em>.</em><em>)</em>

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<em>(a very less amount of kinetic energy is lost)</em>

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v = v₀ + at

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Given values:

v₀ = 0m/s (starts from rest), a = 9.81m/s², t = 3s

Plug in and solve for v:

v = 0 + 9.81(3)

v = 29.4m/s

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