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Ratling [72]
3 years ago
13

Which of the following is not part of the forces that work together to move the tectonic plates?

Physics
1 answer:
SSSSS [86.1K]3 years ago
8 0

creo que soy yo pero lo veo en ingles

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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
Two energy transfers take place when a book hits the ground, Which type of energy transfers are those
bazaltina [42]

Answer:

I think that when a book hits the ground its potential energy converts into kinetic energy and then kinetic energy is transformed into sound and heat energy.

Explanation:

6 0
3 years ago
The speed of an electric locomotive is 90kmh .express this speed in m/s with method​
Maksim231197 [3]

Answer:

24.3 m/s

Explanation:

1 kmh = 0.27 m/s, that makes a conversion ratio of 0.27/1kmh

\frac{90 kmh}{1} x \frac{0.27 m/s}{1 kmh}

The "kmh" n the top and bottom cancel out. And  then you just multiply the top 90 x 0.27 and the bottom 1 x 1 to get

\frac{24.3 m/s}{1}

and since its over 1 its just 24.3 m/s

8 0
3 years ago
An ambulance is traveling east at 61.9 m/s. Behind it there is a car traveling along the same direction at 28.5 m/s. The ambulan
inessss [21]

Answer: 0.4 m

Explanation:

Given

Speed of ambulance, vs = 61.9 m/s

Speed of car = 28.5 m/s

Frequency of ambulance siren, f = 694 Hz

Velocity of sound in air, v = 343 m/s

With speed of ambulance being (61.9 m/s) -> We solve using

fd = f(v + vr) / (v - vs), where vr = 0

fd = 694 * (343 + 0) / (343 - 61.9)

fd = 694 * (343 / 281.1)

fd = 694 * 1.22

fd = 847 Hz

Recall,

λ = v/f

λ = 343/847

λ = 0.4 m

Therefore, the wavelength of the sound of the ambulance’s siren if you are standing at the position of the car is 0.4 m

7 0
3 years ago
Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is r
devlian [24]

Answer:

F_{net} = 31.88 N

Explanation:

When top block is just or about to slide on the lower block then we can say that the frictional force on it will be maximum static friction

So we will have

F_{net} = ma

F_s = ma

\mu_s mg = ma

a = \mu_s g

a = (0.50)(9.81)

a = 4.905 m/s^2

now for the Net force on two blocks to move together

F_{net} = (m_1 + m_2) a

F_{net} = (2.3 + 4.2)(4.905)

F_{net} = 31.88 N

4 0
3 years ago
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