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FromTheMoon [43]
3 years ago
5

A battery is connected to a small torch bulb so that a current flows. The traditional direction of the flow of current and the f

low of the electrons is ... a. The same from positive to negative. b. The same from negative to positive. c. The traditional flow of current is positive to negative and the flow of electrons is the opposite. d. The traditional flow of current is negative to positive and the flow of electrons is the same.
Physics
2 answers:
EastWind [94]3 years ago
8 0

Answer:

c. The traditional flow of current is positive to negative and the flow of electrons is the opposite.

Explanation:

By convention, the electron carries negative charges, so it would flow from the negative side to the positive side. The current, by convention, would have the opposite direction, from positive to negative.

ICE Princess25 [194]3 years ago
7 0

Answer

C.The traditional flow of current is positive to negative and the flow of electrons is the opposite

Explanation:

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Luis is trying to push a box of new soccer balls across the floor. In the illustration, the arrow on the box is a vector represe
solong [7]

Answer:

Complete question

Luis is trying to push a box of new soccer balls across the floor.

If the box is not moving, which of the following must be true?

A. The box is exerting a larger force on Luis than he is exerting on the box

B. There is another force acting on the box that balances Luis's force.

C. Luis is applying a force that acts at a distance.

D. There is no force of friction acting on the box.

Explanation:

Using newton second law of motion

ΣF = ma

Now, for a body not to move when a force is acting on it means that the body is in equilibrium and acceleration a = 0

Therefore,

Fnet = 0

So, if he is applying a force F to push the box and the box is not moving then, there is an external force that is pushing the force back opposite the direction he his pushing and this force counterbalance is own force.

F—F' = 0

F' = F

So, F' is the counter balance force and it is equal to the force applied by Luis

Or it might be frictional force, because if the static friction is not overcome, then, the body will not leave it's state of rest. So if the fictional force is very high, then the box will not leave it rest position and we also know that frictional force opposes motion,

F—Fr=0

F = Fr

So using this explanation,.

The answer is B

B. There is another force acting on the box that balances Luis's force.

3 0
3 years ago
1 point
Svetllana [295]

Answer:

Beta 17,000K, bc warmer is more blue

5 0
3 years ago
Acar goes from 20m/s to 30m/s in 10 seconds what is its acceleration
prohojiy [21]

Acceleration = change in velocity/change in time

                     = (30 - 20) / 10 - 0

                      = 10 / 10

Acceleration = 1 m/s²

8 0
4 years ago
The principal limitation of wind power is unpredictability. Please select the best answer from the choices provided
Lady_Fox [76]
<span>The correct answer is A: True. Although the ability to forecast weather has increased over the years, the ability to predict when and how much wind will occur at any given time and place is difficult. Weather conditions are subject to constant change and while some areas are known to have more wind than others, there is no guarantee about how much wind there will be as weather patterns vary in consistency.</span>
6 0
4 years ago
Read 2 more answers
A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.
victus00 [196]

Answer:

Explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A)  = N₀(B)

N = N₀ e^{-\lambda t}

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ e^{-\lambda_1 t}

For B

N(B) = N(B)₀ e^{-\lambda_2 t}

N(A) = N(B) , for t = 2 h

N(A)₀ e^{-\lambda_1 t} = N(B)₀ e^{-\lambda_2 t}

N(A)₀ e^{-\lambda_1 t} = 5 x N₀(A)  e^{-\lambda_2 t}

e^{-\lambda_1 t} = 5  e^{-\lambda_2 t}

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

half life = .693 / λ

For A

.77 =  .693 / λ₁

λ₁ = .9 h⁻¹

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

Putting t = 2 h , λ₁ = .9 h⁻¹

e^{\lambda_2\times  2} = 5  e^{.9\times  2}

e^{\lambda_2\times  2} = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .

6 0
3 years ago
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