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vichka [17]
3 years ago
7

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

ind the acceleration of the sled? Check all that apply.
The force of gravity must be broken into its parallel and perpendicular components.

Acceleration can be found by dividing the net force by mass. Acceleration can be found by multiplying the net force by mass.

The magnitude of the force components must be multiplied by gravity.

Trigonometry can be used to solve for the magnitude of the force components.

Net force must be found before acceleration can be found.
Physics
2 answers:
olga55 [171]3 years ago
8 0

Answer:

i. The force of gravity must be broken into its parallel and perpendicular components.

ii. Acceleration can be found by dividing the net force by mass.

iv. Trigonometry can be used to solve for the magnitude of the force components.

vi. Net force must be found before acceleration can be found.

Explanation:

Since the force of gravity is a vector quantity, it has both magnitude and direction, then it has both horizontal and vertical components. But it acts always in the vertical direction so that the vertical component holds. While the horizontal component is zero because it does not act in the horizontal direction.

Therefore, the vertical component of the force of gravity must be broken into its parallel and perpendicular components.

Form Newton's second law;

             F = ma

Thus,      

                  a = \frac{F}{m}

So, acceleration can be found by dividing the net force by mass. And the net force must be found before acceleration can be found.

Llana [10]3 years ago
3 0
-The force of gravity must be broken into its parallel and perpendicular components.-Acceleration can be found by dividing the net force by mass.-Trigonometry can be used to solve for the magnitude of the force components.<span>-Net force must be found before acceleration can be found.</span>
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A) Traveling with an initial speed of 70 km/h, a car accelerates at 6000km/h^2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?

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B) At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?

Bullet A is fired upwards with velocity 450 m/s

Bullet B is fired upwards with velocity 600 m/s too

Using the equations of motion, we can obtain a relation for when vertical distance covered by the bullets and time since they were fired.

y = ut + ½at²

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u = initial velocity = 450 m/s

a = acceleration due to gravity = -9.8 m/s²

y = 450t - 4.9t² (eqn 1)

For the bullet B, fired 3 seconds later,

u = initial velocity = 600 m/s

a = acceleration due to gravity = -9.8 m/s²

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y = 600T - 4.9T²

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But, note that, since T starts reading, 3 seconds after t started reading,

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y = 450t - 4.9t²

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