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Svetllana [295]
2 years ago
8

Dave and Alex push on opposite ends of a car that has a mass of 875 kg. Dave pushes the car to the right with a force of 250 N,

and Alex pushes to the
left with a force of 315 N. Assume there is no friction.
What is the net force on the car in the x-direction? Help meee please
Physics
2 answers:
White raven [17]2 years ago
8 0

The net force acting on the car is 65 N to the left

The net force acting on an object is simply defined as the resultant force acting on the object.

From the question given, we obtained the following data:

  • Force applied to the right (Fᵣ) = 250 N
  • Force applied to the left (Fₗ) = 315 N
  • Net force (Fₙ) =?

The net force acting on the car can be obtained as follow:

Fₙ = Fₗ – Fᵣ

Fₙ = 315 – 250

<h3>Fₙ = 65 N to the left </h3>

Therefore, the net force acting on the car is 65 N to the left

Learn more on net force: brainly.com/question/19549734

zvonat [6]2 years ago
5 0

Since there is no friction, the net force on the car in the x - direction will be 65 Newtons

Given that Dave and Alex push on opposite ends of a car that has a mass of 875 kg.

Dave pushes the car to the right with a force of 250 N, which is positive direction and Alex pushes to the left with a force of 315 N which is negative direction. Since there is no friction, only the horizontal forces will be considered.

To calculate the net force on the car in the x-direction, we will find the difference between the two forces because of their directions are opposite to each other.

Net force = 315 - 250

Net force = 65 Newtons

Therefore, the net force on the car in the x - direction will be 65 Newtons

Learn more about resolution of forces here: brainly.com/question/11556949

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Explanation:

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v=\frac{d}{t}

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For the jet in this problem, we have:

d = 1293 mi is the distance covered

t = 2.1 h is the time interval

Therefore, the average  speed is:

v=\frac{1293}{2.1}=615.7 mph

We can also convert into SI units (m/s), keeping in mind that:

1 mi = 1609 m\\1 h = 3600 s

And so

v=615.7 \frac{mi}{h}\cdot \frac{1609 m/mi}{3600 s/h}=275.2 m/s

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A ball with a mass of 3kg is dropped from the top of a building this is 20m high. what is the velocity of the ball when it is 10
anzhelika [568]

Answer:

Velocity=14[m/s]

Explanation:

We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy.

In the attached image we can see the illustration of the ball falling from the height of 20 meters, at this time the potential energy will have the following value.

Ep=m*g*h\\where:\\m=3[kg]\\h=20[m]\\

Ep=3*9.81*20\\Ep=588.6[J]

When the ball passes through half of the distance (10m) its potential energy will have decreased by half as shown below.

Ep=3*9.81*10\\Ep=294.3[m]

If we know that potential energy is transformed into kinetic energy, we can find the value of speed.

Ek=\frac{1}{2} *m*v^{2} \\therefore\\v=\sqrt{\frac{Ek*2}{m} } \\v=\sqrt{\frac{294.3*2}{3} } \\\\v=14[m/s]

7 0
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Tatiana [17]

Answer:

75 m/s

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We can apply motion equations here

V = U + a * t

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V = 0+ 3 * 25

V = 75 m/s

After  25 seconds , subjected to the given acceleration velocity is increased from 0 to 75 m/s

3 0
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