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aleksandrvk [35]
3 years ago
5

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

ed of 16 m/s at an angle 47 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
What is the maximum height the ball goes above the ground?
Physics
1 answer:
Oxana [17]3 years ago
7 0
V^2 = u^2 + 2as

0 = (16 sin 47)^2 + 2 (-9.81)s

s = 6.98m

= 7.0m (2s . f)

Hence, maximum height is 8.5 m

hope this helps
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About 4.7e25 kilograms
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A weightlifter curls a 30 kg bar, raising it each time a distance of 0.60 m. How many times must he repeat this exercise to burn
Sholpan [36]

Answe

given,

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distance of rise, h = 0.60 m

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( energy from pizza ) x (efficiency) = n m g h

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( 1260 x 10³) x (0.25) = n x 30 x 9.8 x 0.6

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3 years ago
Estimate the power you produce in running up a flight of stairs. Give your answer in horsepower (1 hp = 746 W). Suppose you clim
GarryVolchara [31]
The first thing you should do is calculate the work done when climbing the stairs. This work by definition will be given by:
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3 0
3 years ago
Read 2 more answers
A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t
mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 \frac{m}{s^{2}}

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

K=U

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}

part 3

Acceleration can be find using Newton's second law:

F=ma

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

-kx=ma

a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s

So between 0 and 4.5 cm we have half a period:

t=\frac{T}{2}=0.66s

7 0
2 years ago
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RSB [31]
Hey there

Crying is a signal <span>of distress to babies.</span>
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