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Vera_Pavlovna [14]

3 years ago

15

You pull the plunger back on a pinball machine. The spring is pulled back 25 cm from its rest position and has a spring constant

of 78 N/m. What speed will a 50.0 g ball in front of the spring reach when the spring has returned to its rest position?

1 answer:

NemiM [27]3 years ago

5 0

Answer:

9.9 m/s

Explanation:

from the question we are given the following

length (L) = 25cm = 25 / 100 = 0.25m

spring constant (K) = 78 N/m

mass of the ball (m) = 50g = 50 / 1000 = 0.05 kg

At maximum compression all energy is potential energy and at maximum velocity all energy is kinetic energy, and from the conservation of energy the two energy must be he same. Therefore

potential energy at max compression = kinetic energy at max velocity

(1/2) x K x L^2 = (1/2) x m x V^2

(1/2) x 78 x 0.25^2 = (1/2) x 0.05 x V^2

V^2 = 97.6

V = 9.9 m/s

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Why must the Moon travel more than a full orbit around the Earth for the full moon to be complete?

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Answer:

The difference between the sidereal and synodic months occurs becuase as our moon moves around the earth, the earth also moves around our sun. Our moon must travel a little farther in its path to make up for the added distance and complete the phase cycle.

Explanation:

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2 years ago

an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s.where will the potential energy of the object b

Kay [80]

Answer:

333.3 m

Explanation:

Given

$m =100g\ =\ 0.1kg\\v = 100 m/s\\g = 10 m/s ^2$

Potential energy = $\frac{2}{3}\ of\ Kinetic\ energy$ ......Equation(1)

We know that

Potential energy=mgh

Kinetic energy = $\frac{1}{2} mv^{2}$

Now From the Equation(1)

$mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\ gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\ m$

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2 years ago

The average lifetime of μ-mesons with a speed of 0.95c is measured to be 6 x 10^6 s. Find the average lifetime of μ-mesons in a

Mnenie [13.5K]

Answer:

19.2*10^6 s

Explanation:

The equation for time dilation is:

$t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}$

Then, if it is observed to have a life of 6*10^6 s, and it travels at 0.95 c:

$t = \frac{6*10^6}{\sqrt{1-\frac{(0.95c)^2}{c^2}}} = 19.2*10^6 s$

It has a lifetime of 19.2*10^6 s when observed from a frame of reference in which the particle is at rest.

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3 years ago

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

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The Milky Way galaxy is made up of billions of stars, as well as gas and dust, which are all drawn to one another by gravitational pull, as well as a significant amount of dark matter. Our galaxy is approximately 100,000 light years [e1] across.

To know more about milky way galaxy you may visit the link :

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10 months ago