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sergij07 [2.7K]

3 years ago

11

A roller coaster begins at the top of a hillIf it accelerates at the rate of 2m/s ^2 And has a mass of 2000kg, what net force is

acting on it? (F=ma)

1 answer:

Luden [163]3 years ago

5 0

The correct answer to the question is : 4000 N

EXPLANATION:

As per the question, the mass of the roller coaster is given as m = 2000 kg

The acceleration of the roller coaster is given as a = 2m/s2

We are asked to calculate the net force acting on the roller coaster.

From Newton's second laws of motion,we know that net external force acting on a particle is equal to the product of mass with acceleration of the particle.

Mathematically it is written as-

F = ma

Hence,net force acting on the roller coaster is calculated as-

F = 2000Kg×2m/s2

= 4000 N.

Here, Newton (N) is the unit of force.

Hence,net force acting on the roller coaster is 4000 N.

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A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

Likurg_2 [28]

Answer:

Part a)

$T = 0.81 s$

Part b)

$v_x = 3.33 m/s$

Part c)

$v_y = 3.91 m/s$

Part d)

$\theta = 49.55 degree$

Part e)

$T = 1.11 s$

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

$\Delta y = 1.80 - 1.02$

$\Delta y = 0.78 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(0.78)$

$v_y = 3.91 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 3.91 - 9.81 t_1$

$t_1 = 0.39 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(1.80 - 0.95)$

$v_y = 4.08 m/s$

time taken by it to reach this height

$4.08 = v_i + at$

$4.08 = 0 + 9.81 t_2$

$t_2 = 0.42 s$

$T = t_1 + t_2$

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Part b)

Horizontal velocity

$v_x = \frac{x}{t}$

$v_x = \frac{2.70}{0.81}$

$v_x = 3.33 m/s$

Part c)

vertical velocity is the intial y direction velocity

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Part d)

Take off angle is given as

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Part e)

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$\Delta y = 1.30 m$

$v_f^2 - v_y^2 = 2a \Delta y$

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$v_f^2 - 0 = 2(9.81)(2.50 - 0.72)$

$v_y = 5.9 m/s$

time taken by it to reach this height

$5.9 = v_i + at$

$5.9 = 0 + 9.81 t_2$

$t_2 = 0.60 s$

$T = t_1 + t_2$

$T = 1.11 s$

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