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3 years ago

If you go twice as fast, your kinetic energy becomes.

Physics

2 answers:

olga55 [171]3 years ago

7 0

C. Two times bigger

xenn [34]3 years ago

5 0

Answer:

2 times bigger

Explanation:

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66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st

kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t Equation 2

y: The vertical distance the ball moves at time t

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t

Known information

We know the following data:

Vi=15 m / s

$g =9.8 \frac{m}{s^{2} }$

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)

a) t=1s

$y = 15*1 - ½ 9.8*1^{2}$ =15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

$y = 15*1.5 - ½ 9.8*1.5^{2}$ =22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

$y = 15*2 - ½ 9.8*2^{2}$ = 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

3 0

3 years ago

The pulse site located at the point where the upper leg bends is called the

VLD [36.1K]

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3 years ago

____ waves are often used in heat lamps fill in the blank

Ludmilka [50]

Heat your answer is Heat.

Hoped I helped.

8 0

3 years ago

Read 2 more answers

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antiseptic1488 [7]

Answer:

that's fusion

Explanation:

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3 years ago

Read 2 more answers

A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle.

chubhunter [2.5K]

Answer:

v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

p₀ = m v₁₀

final. After the crash

$p_{f}$ = m $v_{1f}$ + m v_{2f}

Recall that velocities are a vector so it has x and y components

p₀ = p_{f}

we write this equation for each axis

X axis

m v₁₀ = m v_{1fx} + m v_{2fx}

Y Axis

0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

sin 23.3 = v_{2fy} / v_{2f}

cos 23.3 = v_{2fx} / v_{2f}

v_{2fy} = v_{2f} sin 23.3

v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

1.83 = v_{1f} cos θ + 1.15 cos 23.3

0 = - v_{1f} sin θ + 1.15 sin 23.3

1.83 = v_{1f} cos θ + 1.0562

0 = - v_{1f} sin θ + 0.4549

v_{1f} sin θ = 0.4549

v_{1f} cos θ = -0.7738

we divide these two equations

tan θ = - 0.5878

θ = tan-1 (-0.5878)

θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

v_{1f} cos (-30.45) = - 0.7738

v_{1f} = -0.7738 / cos 30.45

v_{1f} = -0.8976 m / s

the component and this speed is

v_{1fy} = v1f sin θ

v_{1fy} = 0.8976 sin (30.45)

v_{1fy} = - 0.4549 m / s

8 0

3 years ago