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Papessa [141]
2 years ago
15

A 0.5-kilogram ball is thrown vertically upward with an initial kinetic energy of 25 joules. Ap-proximately how high will the ba

ll rise?
Physics
1 answer:
tatuchka [14]2 years ago
4 0
It's gravitational potential energy at the top will roughly equal it's kinetic energy when it was released (a little is lost to air resistance).  Note this will assume the release point is zero potential energy.  (we are free to define it that way, just letting you know).  Gravitational potential energy is mgh.
mgh=25J
h=25J/(0.5kg x 9.81m/s^2) = 5.097m
So it goes about 5.1 meters above the point where it was released
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A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular a
stepan [7]

Answer:

a) α = 1.875 \frac{rad}{s^{2} }

b) t = 8 s

Explanation:

Given:

ω1 = 0 \frac{rad}{s}

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*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*

a) α = ?

(ω2)^2 = (ω1)^2 + 2α(theta)

15^{2} = 0^{2} + 2(α)(60)

225 = 120α

α = 1.875 \frac{rad}{s^{2} }

b)

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True or false velocity is the slope of the acceleration versus time graph
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A sports car accelerates from rest for 5 seconds reaching a velocity of 23.0 m/s.
denis-greek [22]

Answer:

<h2>4.6 m/s²</h2>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

<h3>a =  \frac{v - u}{t}  \\</h3>

where

v is the final velocity

u is the initial velocity

t is the time taken

a is the acceleration

Since the body is from rest u = 0

From the question we have

a =  \frac{23 - 0}{5}  =  \frac{23}{5}  \\

We have the final answer as

<h3>4.6 m/s²</h3>

Hope this helps you

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