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4 years ago

Two particles are separated by 0.38 m and have charges of -6.25 x 10-°C

Physics

2 answers:

Studentka2010 [4]4 years ago

5 0

Answer:

did u get it?

Explanation:

frfr

Evgesh-ka [11]4 years ago

4 0

Answer:

-4.53 X 10-6N

Explanation:

apex

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Electronic flash units for camera contain a capacitor for storing the energy used to produce the flash. In one such unit, the fl

nadezda [96]

Answer:

420J

Explanation:

Power is the time rate of change in energy. Power is the ratio of energy to time. The S.I unit of power is in watts.

Given that the flash lasts for 1/675 s, power output is 2.7 * 10⁵ W. Hence:

Power = Energy / time

Substituting:

2.7 * 10⁵ W = Energy / (1/675)

Energy = 2.7 * 10⁵ W * 1/675 = 400J

Therefore the energy emitted as light is 400J.

Since the conversion of electric energy to light is 95% efficient, hence the energy stored as electrical energy is:

Energy(capacitor) = 5% of 400J + 400J = 0.05*400 + 400

Energy(capacitor) = 420J

8 0

3 years ago

How many ways can motion change

meriva

Answer:

There are four main ways of doing that :-

Velocity
Acceleration
Momentum
Kinetic energy

Hope it helps!

7 0

4 years ago

Read 2 more answers

HELP ME PLEASE! MOTION CONCEPT MAP!

Zinaida [17]

a. speed b. direction c. Magnitude d. Direction

8 0

3 years ago

Why do the birds and butterflies keep flying into my window? (4 butterflies, 2 birds). I think there after my blood.

Zina [86]

Answer:

Is that an actual question for school?

6 0

3 years ago

Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta

Sergeu [11.5K]

Answer:

The heat flows into the gas during this two-step process is 120 cal.

Explanation:

Given that,

Number of moles = 3

Heat capacity at constant volume = 4.9 cal/mol.K

Heat capacity at constant pressure = 6.9 cal/mol.K

Initial temperature = 300 K

Final temperature = 320 K

We need to calculate the heat flow in to gas at constant pressure

Using formula of heat

$\Delta H_{1}=nC_{p}\times\Delta T$

Put the value into the formula

$\Delta H_{1}=3\times6.9\times(320-300)$

$\Delta H_{1}=414\ cal$

We need to calculate the heat flow in to gas at constant volume

Using formula of heat

$\Delta H_{1}=nC_{v}\times\Delta T$

Put the value into the formula

$\Delta H_{1}=3\times4.9\times(300-320)$

$\Delta H_{1}=-294\ cal$

We need to calculate the heat flows into the gas during two steps

Using formula of total heat

$\Delta H_{T}=\Delta H_{1}+\Delta H_{2}$

$\Delta H_{T}=414-294$

$\Delta H_{T}=120\ cal$

Hence, The heat flows into the gas during this two-step process is 120 cal.

7 0

4 years ago