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dimaraw [331]
2 years ago
15

A 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot

al momentum of the two vehicles (don’t forget to assign a NEGATIVE velocity to one of the vehicles, because they are travelling in opposite directions)
Please answer ASAP
Physics
1 answer:
Molodets [167]2 years ago
3 0

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

M' = 22750 kgm/s.

Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

Hence the total momentum of the two vehicles is 16250 kgm/s due south

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7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

7 0
3 years ago
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