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Oliga [24]
2 years ago
9

Suppose the position of a particle moving along the x-axis given by the equation X=(1m/s²)t²-(5m/s)t +0.5m

Physics
1 answer:
Rama09 [41]2 years ago
5 0

Answer:

sorry i dont know

Explanation:

i dont know

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A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov
Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

5 0
3 years ago
Two scientists interpret the same set of data differently. How would the scientific community deal with this problem?
alexdok [17]
The correct answer would be D. A new experiment would be needed to be done in order to test the conclusions. In science there is no authority, data is the only thing that matters. So if we have two different conclusions from the same date the only solution is to perform more tests and more experiments to see what is correct.
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3 years ago
Nanotechnology is currently used in electronics for
VMariaS [17]

Answer:

data storage, OLEDS, and wires

Explanation:

3 0
2 years ago
This. please help me.
kramer
The transfer of heat from the Sun is in the form of thermal radiation (also called infrared radiation)
The transfer of heat at the person’s feet is by thermal conduction. During the day the sand will be hot so if you step on it with bare feet there will be conduction of heat to the feet.
3 0
2 years ago
1. If a model train car with a momentum of 12 kg•m/s collides with a 2 kg model train car that is not moving, what is the total
wariber [46]

Answer:

P_{f} = 12 \ kg.m/s

Explanation:

Given data:

Momentum of moving model train, P_{1} = 12 \ kg.m/s

Mass of the stationary model train, m = 2 \ kg

Initial speed of the stationary model train, v = 0

Assume there is no external force is acting on the given train system.

In this case, the total linear momentum of the trains would be conserved.

Let the final linear momentum of the trains be P_{f}.

Thus,

P_{i} = P_{f}

P_{1} + P_{2} = P_{f}

P_{1} + mv = P_{f}

12 + 2 \times 0 = P_{f}

\Rightarrow \ P_{f} = 12 \ kg.m/s.

7 0
3 years ago
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