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MAXImum [283]
2 years ago
14

A relaxed spring of length 0. 13 m stands vertically on the floor; its stiffness is 1180 N/m. You release a block of mass 0. 5 k

g from rest, with the bottom of the block 0. 7 m above the floor and straight above the spring. How long is the spring when the block comes momentarily to rest on the compressed spring?.
Physics
1 answer:
Daniel [21]2 years ago
4 0

The length of spring when the block comes momentarily to rest on the compressed spring will be 0.054 m.The length of the spring by the letter x.

<h3>What is the potential energy of the spring?</h3>

The energy is stored in the spring when it is stretched or compressed by some length. It is the product of mass, gravity and distance compressed or stretched. Mathrmatically it is given by;

PE=mgh

The given data in the problem is ;

m is the mass of the block is 0.5 kg height,

h is the height is released is 0.7 m

x initial length of the spring = 0.13 m.

K is the force constant of the spring = 1180 N/m.

By the law of conservation of energy,

The potential energy of the spring gets converted

\rm mgh=\frac{1}{2}KL^2 \\\\ \rm L= \sqrt{\frac{2mgh}{K} } \\\\ \rm L= \sqrt{\frac{2\times 0.5\times9.81 \times 0.7}{1180} \\\\  \\\\

\rm L= 0.054m

Hence the length of spring when the block comes momentarily to rest on the compressed spring will be 0.054 m

To learn more about the potential energy of the spring refer to the link;

brainly.com/question/2730954

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A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3
Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,  

\text { Electric power }(p)=\frac{V^{2}}{R}

\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}

\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

For calculating number of turns

\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

Given that,

80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }

\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}

\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .

We need to find the number of turns in the secondary winding \left(N_{S}\right) to run the bulb at 120W \left(P_{2}\right)

Firstly find the secondary voltage in the transformer use, \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}

V_{2}^{2}=\frac{120^{2} \times 120}{80}

V_{2}^{2}=\frac{1728000}{80}

V_{2}^{2}=21600

V_{2}=\sqrt{21600}

V_{2}=146.9 \mathrm{V}=V_{S}

Now, finding the number of turns in secondary coil. Use, \frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

\frac{30}{N_{S}}=\frac{65}{146.9}

N_{S}=\frac{30 \times 146.9}{65}

N_{S}=\frac{4407}{65}N_{S}=67.8

The number of turns in the secondary winding are 67.8 turns.

6 0
3 years ago
How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s
DIA [1.3K]

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

a = 3 m/s^2

Formula

a = (vf - vi) /t              Substitute the givens into this formuls

Solution

3 = (40 - 10) / t          Multiply both sides by t

3*t = t(40 - 10)/t        Combine. Cancel t's on the right

3*t = 30                     Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

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calculate the distance in km that alex can run in 2.0 hours if she maintains the constant average speed of 8.00 km/h
Tema [17]
16 kilometers is the answer i came up with. hope this helps.
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How do you rationalize the tension being used in Tennis Racket strings using the concept of impulse and momentum?
zheka24 [161]

Answer:

The momentum, ΔP, and therefore, kinetic energy given to the ball in a serve is the result of the product of the tension force, 'F', in the string and the time of contact, Δt, between the ball and the string

ΔP = F × Δt

Explanation:

The impulse, ΔP, is the produce of the force, 'F', applied to a body for a given period of time, Δt', that gives motion to the body, and it is equal to the change of momentum of the body

ΔP = F × Δt

The momentum, 'P', of a body is the product of the mass, 'm', of the body and its velocity, 'v'

P = m × v

Tension is the axial pulling force of a string

T = Axial Force, F_{axial}

The tension used in Tennis Racket strings is between 40 to 65 lbs.

When high tension is used in the string, the string is taut, and the contact duration between the Racket string and the ball is minimal, and the player needs to use more force to obtain a high momentum, and therefore, energy in the ball, which reduces control, and increase stress, as force is more emphasized

When low tension is used in the string, the Tennis Racket strings are more elastic. During a serve, the ball pushes the strings further back into the racket, such that the ball spends more time in contact with the string, (Δt is larger), and therefore, the impulse, F·Δt = ΔP, given to the ball is larger, therefore, the ball has a larger change in momentum, and therefore more energy in the intended direction.

However, a very slackened string will increase the increase area and time (large Δt) of contact of the ball and the racket such that the force given to the ball, F = ΔP/(large Δt) is reduced and therefore reduce the likelihood of gaining points from a serve against an opponent with a much forceful return of a serve.

3 0
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Jacob is traveling at 5.00 m/s North. Jacob throws a ball with a velocity of 5.00 m/s South. Jacob throws the ball from a height
Nataly [62]

Answer:t=0.54 s

Explanation:

Given

Jacob is traveling 5 m/s in North direction

Jacob throw a ball with a in south direction with a velocity of 5 m/s

Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0

Time taken by ball to reach ground

s=ut+\frac{gt^2}{2}

1.45=0+\frac{9.81\times t^2}{2}

t^2=frac{2\times 1.45}{9.81}

t=0.54 s

Motion of ball will be straight line

3 0
3 years ago
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