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Daniel [21]
3 years ago
12

The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif

ference between the two plates either a) decrease b) does not change c) increase?
Physics
1 answer:
IRISSAK [1]3 years ago
7 0

Since the electric field between the plates is constant

so we can say the relation between potential difference and electric field is given by

\Delta V = E. d

here from above relation we can say that is electric field is kept constant then potential difference is directly proportional to the distance between the plates

so here we will say that if the distance between the plates is decreased then the potential difference between the plates will also decreases

So correct answer will be

<u>a) decrease </u>

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1. 1500j of work was done to move a box 20m. What force was applied to the box ?
Fantom [35]

Answer:

1. 75N

2. 67,983 J (=67.98 kJ)

Explanation:

1. Work = Force x Distance

we are given that Work = 1,500J and Distance = 20m

hence,

Work = Force x Distance

1,500 = Force x 20

Force = 1,500 ÷ 20 = 75N

2. Potential Energy, PE = mass x gravity x change in height

we are given that mass = 165 kg and change in height = 42m

assuming that gravity, g = 9.81 m/s²

Potential Energy, PE = mass x gravity x change in height

Potential Energy, PE = 165 x 9.81 x 42 = 67,983 J (=67.98 kJ)

4 0
3 years ago
A rabbit runs 4.4 m across a lawn, stops, then runs 2.2 m back in the opposite direction. What is the rabbit’s displacement from
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Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

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3 years ago
When did ernest rutherford make his discovery
oee [108]

Answer:

1911

Explanation:

"In 1911, he was the first to discover that atoms have a small charged nucleus surrounded by largely empty space, and are circled by tiny electrons, which became known as the Rutherford model (or planetary model) of the atom."

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Explain two scenarios where a large truck can have the same momentum as a small car.
KengaRu [80]

The momentum, p, of any object having mass m and the velocity v is

p=mv\cdots(i)

Let M_L and M_S be the masses of the large truck and the car respectively, and V_L and V_S be the velocities of the large truck and the car respectively.

So, by using equation (i),

the momentum of the large truck = M_LV_L

and the momentum of the small car = M_SV_S.

If the large truck has the same momentum as a small car, then the condition is

M_LV_L=M_SV_S\cdots(ii)

The equation (ii) can be rearranged as

\frac {M_L}{M_S}=\frac {V_S}{V_L} \; or \; \frac{M_L}{V_S}=\frac{M_S}{V_L}

So, the first scenario:

\frac {M_L}{M_S}=\frac {V_S}{V_L}

\Rhghtarrow M_L:M_S=V_S:V_L

So, to have the same momentum, the ratio of mass of truck to the mass of the car must be equal to the ratio of velocity of the car to the velocity of the truck.

The other scenario:

\frac{M_L}{V_S}=\frac{M_S}{V_L}

\Rhghtarrow M_L:V_S= M_S:V_L

So, to have the same momentum, the ratio of mass of truck to the velocity of the car must be equal to the ratio of mass of the car to the velocity of the truck.

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How would you decide what numeric scale to use for each axis?
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