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Vladimir79 [104]

3 years ago

11

A vertical spring with stiffness k originally is at rest with no mass attached. Then, a mass M is attached, and the spring rocks

back and forth until it is at rest at a new equilibrium position. What is this position in terms of g, M, and k?

1 answer:

Alina [70]3 years ago

8 0

Answer:

The position of the spring in terms of g, m & k is $x = \frac{m g}{k}$

Explanation:

Stiffness of the spring = k

Mass = m

When a mass m is attached with the spring then spring stretched. in that case the force exerted on the spring is equal to weight of the mass attached.

⇒ Force exerted on the spring F = k x

⇒ m g = k x

⇒ $x = \frac{m g}{k}$

This is the position of the spring in terms of g, m & k.

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A jogger accelerates at a constant rate as she travels 5.0

VladimirAG [237]

Answer:

2.0s

Explanation:

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3 0

3 years ago

Hey Tori's are the rim of a canyon yells hello toward the opposite side she has an echo hello four seconds later if the speed of

krek1111 [17]

Answer:

The wall is 680 meter away from the person.

Explanation:

Given data

Speed of sound = 340 $\frac{m}{s}$

Given that Persons said hello toward the opposite side she has an echo hello 4 seconds later means it takes 2 seconds for the sound to reach the wall & again 2 seconds to reach the persons ear.

Therefore the distance between the person & wall is

D = speed × Time

D = 340 × 2

D = 680 meter

Therefore the wall is 680 meter away from the person.

6 0

3 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

3 years ago

17. A distance-time graph indicates that an object moves 120 meters in 5 seconds and then remains at rest for 5 seconds. What is

irina1246 [14]

Answer:

12 m/s

Explanation:

Speed is distance moved per unit time and expressed as

S=d/t

S is speed, d is distance and t is time.

When at rest for five minutes, distance moved is zero hence speed is 0/5=0 m/s

When having moved 120 m for 5 s then speed is

S=120/5=24 m/s

Average speed is the average of these two speeda

Average speed=(24+0)÷2=24/2=12 m/s

Therefore, average speed is 12 m/s

3 0

3 years ago

Describe the motion of an automobile on an east-west

Tanya [424]

Answer:

1. The automobile is traveling due east and is speeding up.

2. The car is traveling due east and is is slowing down.

3. The automobile is traveling due east at a constant speed.

4. The car is traveling due west and is slowing down.

5. The automobile is traveling due west and is speeding up.

6. The automobile is traveling due west at a constant speed.

7. The automobile is accelerating due east from rest.

8. The automobile is accelerating due west from rest.

Explanation:

The key to understanding this is:

When the acceleration and initial velocity of the automobile have the same sign (positive or negative) then the automobile is speeding up. Explained further, if acceleration and the initial velocity are both positive or they are both negative the automobile is speeding up but whenever they have opposite signs (that is acceleration is positive and initial velocity is negative or vice versa) the automobile is slowing down. When the acceleration is zero the automobile is maintaining a unform motion at a constant speed (the speed is not changing with time). The + or - sign indicates the direction of travel. In this case east is + and west is -. It is my pleasure answering this question. I hope you find it helpful. Thank you.

4 0

3 years ago