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Whitepunk [10]
4 years ago
15

Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its conten

ts, of 5.0 kg, and the other has total mass of 3.0 kg. If the 3.0 kg cart is moving at 3.0 m/s after the spring is allowed to push the carts apart, what is the velocity of the 5.0 kg cart after release? The system is closed.A) -1.8 m/s B) -5.0 m/sC) -0.56 m/sD) -0.0 m/s
Physics
1 answer:
ladessa [460]4 years ago
3 0

Answer:

A) - 1.8 m/s

Explanation:

As we know that whole system is initially at rest and there is no external force on this system

So total momentum of the system must be conserved

so we will have

m_1v_1 + m_2v_2 = 0

now plug in all data into above equation

5(v) + 3(3)

5v = -9

v = -1.8 m/s

so correct answer is

A) - 1.8 m/s

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D. If a dog has a mass of 12 kg, what is its weight on Neptune?<br> 11.7N/kg
steposvetlana [31]

Answer:

133.8 N

Explanation:

Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2

Therefore, the weight of the dog on this planet would be:

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3 years ago
Explain the difference between si base units and derived units. give an example of each
34kurt
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6 0
3 years ago
A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor
Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force F_{1}=6\ N

Second force F_{2}=10\ N

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

\tau=Force\times lever\ arm

So, Here,

\tau_{2}=5 \tau_{1}

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

F_{2}\times d_{2}=5\times F_{1}\times r_{1}

r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}

Where, F_{1}=First force

F_{1}=First force

F_{2}=Second force

r_{1}= distance

Put the value into the formula

r_{2}=\dfrac{5\times6\times0.2}{10}

r_{2}=0.6\ m

Hence, The moment arm is 0.6 m

6 0
4 years ago
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