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n200080 [17]
3 years ago
12

A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the

stone hits the water 2.00 seconds later, what is the height of the bridge above the water?
Remember to identify all your data, write the equation, and show your work.
Physics
1 answer:
hodyreva [135]3 years ago
6 0
Vi = 15 m/s
t = 2 s
a = 9.8 m/s^2
y = ?

The kinematic equation that has all of our variables is d = Vi*t + 0.5*a*t^2
y = 15*2 + 0.5*9.8*2^2 = 49.6 m
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The magnetising current in a transformer is rich in ______________ harmonic?
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•<br> What happens to similar charged<br> particles and differently charged<br> particles
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Like charges repel

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When particles of similar charges are brought together, they repel each other and increase the distance of separation. Repulsion occurs because both two electrons have negative electrical charge forcing their lines of force to repel. However, when particles of opposite charges are brought nearer to each other, they attract each other and reduce the distance of separation.

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What change does this cause concerning weather?
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More extreme weather.

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The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
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Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

The acceleration is same, a=-10.0 m/s^2

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





7 0
3 years ago
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