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Viefleur [7K]
3 years ago
8

A 0.250 kg block on a vertical spring with a spring constant of 5.00 x 103 N/m is pushed downward compressing the spring 0.100 m

. When released, the block leaves the spring and travels upward vertically. How high does it rise above its point of release
Physics
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer:

The height at point of release is 10.20 m

Explanation:

Given:

Spring constant k = 5 \times 10^{3} \frac{N}{m}

Compression x = 0.10 m

Mass of block m = 0.250 kg

Here spring potential energy converted into potential energy,

   mgh = \frac{1}{2} kx^{2}

For finding at what height it rise,

  0.250 \times 9.8 \times h = \frac{1}{2}  \times 5 \times 10^{3} \times (0.10) ^{2}                ( ∵ g = 9.8 \frac{m}{s^{2} } )

  h = 10.20 m

Therefore, the height at point of release is 10.20 m

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A hockey puck sliding along frictionless ice with speed v to the right collides with a horizontal spring and compresses it by 1.
patriot [66]

Answer:

The spring's maximum compression will be 2.0 cm

Explanation:

There are two energies in this problem, kinetic energy K= \frac{mv^{2}}{2} and elastic potential energy U= \frac{kx^{2}}{2} (with m the mass, v the velocity, x the compression and k the spring constant. ) so the total mechanical energy at every moment is the sum of the two energies:

E=K+U

Here we have a situation where the total mechanical energy of the system is conserved because there are no dissipative forces (there's no friction), so:

E_{i}=E_{f}

K_{i}+U_{i}=K_{f}+U_{f}

Note that at the initial moment where the hockey puck has not compressed the spring all the energy of the system is kinetic energy, but for a momentary stop all the energy of the system is potential elastic energy, so we have:

K_{i} = U_{f}

\frac{mv^{2}}{2}=\frac{kx^{2}}{2} (1)

Due conservation of energy the equality (1) has to be maintained, so if we let k and m constant x has to increase the same as v to maintain the equality. Therefore, if we increase velocity to 2v we have to increase compression to 2x to conserve the equality. This is  2(1.0) = 2.0 cm

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3 years ago
Which use of a simple machine changes the direction of a force? A. using a ramp to load boxes onto a truck B. using an axe to sp
kiruha [24]
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3 years ago
If a fixed length simple pendulum is found to have three times the period on an unknown planet’s surface (compared to Earth), wh
balu736 [363]

Answer:

g/9

Explanation:

length of the pendulum = L

time period on the earth = T

Time period on the planet = 3T

Let the acceleration due to gravity on the earth is g and on the planet is g'.

Use the formula for the time period of a simple pendulum for the time period on earth

T=2\pi \sqrt{\frac{L}{g}}     .... (1)

Time period on the surface of planet is

3T=2\pi \sqrt{\frac{L}{g'}}      .... (2)

Divide equation (2) by equation (1)

\frac{3T}{T}= \sqrt{\frac{g}{g'}}

g' = g/9

Thus, the acceleration due to gravity on the planet is g /9

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3 years ago
You are on the roof of the physics building of your school, 46.0 m above the ground. Your physics professor, who is 1.80 m tall,
Elena L [17]

Answer:

The professor should be 3.60 meters away from the place where the egg would fall.

Explanation:

To calculate this first we need to calculate the time it would take for the egg to fall the distance from the roof to the professor's head (46m-1.8m=44.2m).

The formula for free fall is:

y=1/2*g*t^{2}

solving for t it would be

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replacing

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Since he is traveling at a constant speed the formula is

x=v*t

replacing

1.20m/s * 3.00 s = x = 3.60 m

8 0
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F being force
That would make the equation
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5 0
4 years ago
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