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2 years ago

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A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin

g opposite the car's motion. How far does the car move during 2.5 s? *

1 answer:

Dennis_Churaev [7]2 years ago

5 0

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

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3 years ago

A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of

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Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

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3 years ago

The roof of a two-story house makes an angle of 29° with the horizontal. A ball rolling down the roof rolls off the edge at a sp

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Answer:

$t = 0.93 s$

Part b)

$d = 3.98 m$

Part c)

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Explanation:

The two components of the velocity of the ball is given as

$v_x = 4.9 cos29$

$v_x = 4.28 m/s$

$v_y = 4.9 sin29$

$v_y = 2.37 m/s$

Part a)

now we know that the displacement in y direction is given as

$\Delta y = 6.4 m$

so we have

$\Delta y = v_y t + \frac{1}{2}gt^2$

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$t = 0.93 s$

Part b)

Distance of the ball in x direction of the motion is given as

$d = v_x t$

$d = 4.28 \times 0.93$

$d = 3.98 m$

Part c)

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$v_x = 4.28 m/s$

while in Y direction we can use kinematics

$v_y = v_{oy} + at$

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$v_y = -11.49 m/s$

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3 years ago

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3 years ago