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Neko [114]
2 years ago
10

A 3.0-kg cart moving to the right with a speed of 1.0 m/s has a head-on collision with a 5.0-kg cart that is initially moving to

the left with a speed of 2.0 m/s. After the collision, the 3.0-kg cart is moving to the left with a speed of 1.0 m/s. What is the final velocity of the 5.0-kg cart?
Physics
1 answer:
Maurinko [17]2 years ago
3 0

Answer:

v = 0.8 m/s towards left

Explanation:

As we know that there is no external force on the system of two cart so total momentum of the system is conserved

so we will say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now plug in all data into the above equation

3(1) + 5(-2) = 3(-1) + 5 v

here we assumed that left direction of motion is negative while right direction is positive

so we can solve it for speed v now

3 - 10 = - 3 + 5 v

5 v = -4

v = -0.8 m/s

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To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

PE= \frac{GMm}{d}

The kinetic energy can be written as,

KE= \frac{1}{2} mv^2

Where,

G = 6.67*10^{-11}m^3/kg\cdot s^2Gravitational Universal Constant

m = 5.972*10^{24}kg Mass of Earth

h = 56*10^6m  Height

r = 6.378*10^6m Radius of Earth

From the conservation of energy:

\frac{1}{2} mv^2 = \frac{GMm}{d}

Rearranging to find the velocity,

v = \sqrt{\frac{2Gm}{d}} \rightarrow  Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

d = r+h

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Replacing the values we have that

v = \frac{2(6.67*10^{-11})(5.972*10^{24})}{6.378*10^6+56*10^6}

v = 3.6km/s

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3 0
2 years ago
A small smooth object slides from rest down a smooth inclined plane inclined at 30degrees horizontal.What is the acceleration do
asambeis [7]
The acceleration is given as:

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7 0
2 years ago
a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
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Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
2 years ago
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Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

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Explanation: Please see the attachments below

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010) Identify the true statement. Group of answer choices The height of waves is determined by wind strength and fetch. Wave bas
AlekseyPX

Answer:

The height of the wave is determined by the wind strength and fetch.

Explanation:

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The more the strength and the more the fetch size the more will be the height of the wave.

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7 0
2 years ago
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