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Ostrovityanka [42]
3 years ago
10

a stone is thrown down off a bridge with a veloctiy of 5.6m/s. What is the velocity afer 3 seconds have passed?

Physics
2 answers:
Flura [38]3 years ago
6 0
I think it's 16.8 because you multiply 5.6x3=16.8
nexus9112 [7]3 years ago
5 0

Answer:

Final speed of the stone, v = 35 m/s

Explanation:

It is given that,

Initial speed of the stone, u = 5.6 m/s

Let v is the velocity of the stone after 3 seconds have passed. It is moving down under the action of gravity. So, using the first equation of motion as :

v=u+at

v=u+gt

v=5.6+9.8\times 3

v = 35 m/s

So, the final speed of the stone is 35 m/s. Hence, this is the required solution.

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<h3>What is induced voltage?</h3>

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3 0
2 years ago
Which is more dense, cold freshwater or warm seawater?
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4 0
3 years ago
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A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
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