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natulia [17]
2 years ago
14

If you were in a spaceship watching a ball hover at rest (inside the spaceship) in mid-air, and the spaceship suddenly began rap

idly accelerating, what would you see happen to the ball? Why?
Physics
1 answer:
USPshnik [31]2 years ago
4 0

Answer:

It will still hover until the spaceship "hits" or exerts a force on it.

Explanation:

Remember, if there is no net force, there is no acceleration or movement.

In this case, our ball is hovering in the spaceship, and in space, we can assume there is no F_g, and we can assume there is no F_N, nor no forces acting against it.

So, the ball would not move.

However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.

This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.

Think about it this way.

Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only  F_g is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.

If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.

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evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

5 0
3 years ago
Read 2 more answers
A lamp hangs from the ceiling at a height of 2.6 m. The lamp has a mass of 3.8 kg. The screws holding the lamp break, and it fal
iVinArrow [24]

Answer:

Explanation:

Given height of lamp from the ceiling = 2.6m

mass of the lamp = 3.8kg

acceleration due to gravity = 9.81m/s²

As the body falls to the ground, it falls under the influence of gravity.

Gravitational potential energy = mass*acc due to gravity * height

Gravitational potential energy  = 3.8*2.6*9.81

Gravitational potential energy  = 96.923 Joules

b) Kinetic energy = 1/2 mv²

m = mass of the body (in kg)

v = velocity of the body (in m/s²)

To get the velocity v, we will use the equation of motion v^{2} = u^{2}+2gh

v^{2} = 0^{2}+2(9.81)(2.6) \\v^{2} = 51.012\\v =\sqrt{51.012}\\ v = 7.14m/s

Since mass = 3.8kg

K.E =  1/2 * 3.8 *7.14^{2}\\ K.E = 96.86Joules

c) To know how fast the lamp is moving when it hits the ground, we will use the formula. When the body hits the ground, the height covered will be 0m. this means that the body is not moving once it hits the ground. It stays in one position. The energy possessed by the body at this point is potential energy. The correct answer is therefore 0 m/s

4 0
3 years ago
A boy uses a force to keep a rock from falling. (He is carrying the rock.) Some forces, like this one, act only when two objects
MArishka [77]
Gravity is a non contact force , jump from a little high place (don't do that :P) , you wont just get stuck in the air you will fall down , this is gravity you dont need any contact 
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35-20 = 15m/s difference
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If the energy bar was his only fuel, how far could a 68 kg person walk at 5 km/h?
goldfiish [28.3K]

An energy bar contains of 20grams of carbohydrates and 1gram is equal to 17 KJ. If the energy bar was his only fuel the total energy available is equal to (17,000 x 20) = 340,000j. Estimate KJ per hour is equal to 1539KJ, it has Time equal to (340,000/1539,000) is equal to 0.2209 hour and his Distance is equal to (5,000 x .2209) = 1.1045km. 

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3 years ago
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