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almond37 [142]
4 years ago
9

Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true abo

ut them as they move apart? (There could be more than one correct choice.)a. Their electrical potential energy keeps decreasing.b. Their acceleration keeps decreasing.c. Their kinetic energy keeps increasing.d. Their kinetic energy keeps decreasing.e. Their electric potential energy keeps increasing.
Physics
1 answer:
Colt1911 [192]4 years ago
7 0

Answer:

Explanation:

correct options

a ) Their electrical potential energy keeps decreasing

Actually as they move apart , their electrical potential energy decreases due to increase of distance between them and kinetic energy increases

so a ) option is correct

b ) Their acceleration keeps decreasing

As they move apart , their mutual force of repulsion decreases due to increase of distance between them so the acceleration decreases .

c ) c. Their kinetic energy keeps increasing

Their kinetic energy increases because their electrical potential energy decreases . Conservation of energy law will apply .

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natta225 [31]

Answer:

Total resistance = 0.92Ω

Explanation:

For parallel connected resistors we have effective resistance

                 \frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+.....

Here parallel circuit made up of resistances of 2Ω, 3Ω, and 4Ω.

That is

             R₁ = 2Ω

             R₂ = 3Ω

             R₃ = 4Ω

Substituting

             \frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\\\frac{1}{R_{eff}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\\\\frac{1}{R_{eff}}=\frac{6+4+3}{12}\\\\R_{eff}=\frac{12}{13}=0.92ohm

Total resistance = 0.92Ω

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3 years ago
Examine the weather map and locate this symbol.
DochEvi [55]
The right answer is A
7 0
3 years ago
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As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

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v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

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We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

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3 years ago
Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each
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Answer with Explanation:

We are given that

I_1=29 A

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