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Natali5045456 [20]
3 years ago
9

Debbie places two shopping carts in a cart Corral. she pushes the first cart, which then pushes a second cart. what force is bei

ng exerted?
Physics
1 answer:
Bingel [31]3 years ago
6 0

When Debbie pushes the first cart she is using an applied force. An applied force is created when someone or something pushes another thing using, of course, an applied force. Now, when the second cart is being pushed by the first cart, this is also an applied force. You can tell because the first cart is being pushed using forced and this causes the second cart to be pushed using some of the force that is being transmitted to the first cart.


Debbie exerts applied force on the first cart. The first cart exert applied force on the second cart.



- Marlon Nunez

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A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
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Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

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Mass of object = 5 kg

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Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

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