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Elza [17]
3 years ago
8

Back 8 8. 9. 10. 11. 12. 13. 13

Physics
1 answer:
deff fn [24]3 years ago
3 0

Answer:

i dont know the answer for this question sorry for the misunderstanding

Explanation:

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A 50-ω resistor is connected to a 9.0 V battery. How much thermal energy is produced in 7.5 minutes?1.2 * 10^2 J1.3 * 10^3 J3.0
emmasim [6.3K]

In order to calculate the thermal energy, first let's calculate the power, using the formula:

P=\frac{V^2}{R}

For a voltage V = 9 Volts and a resistance R = 50 ohms, we have:

\begin{gathered} P=\frac{9^2}{50} \\ P=\frac{81}{50}=1.62\text{ W} \end{gathered}

Now, multiplying the power by the time (in seconds), we can find the energy:

\begin{gathered} E=P\cdot t \\ E=1.62\cdot7.5\cdot60 \\ E=729\text{ J} \end{gathered}

In scientific notation, we have an energy of 7.3 * 10^2 J, therefore the correct option is the fourth one.

4 0
1 year ago
Explain why a ball thrown in space could keep moving forever, while a ball thrown here on Earth will come to a stop.
Dmitriy789 [7]

Answer:

In space there is no air resistance. on earth there is

in space there is no opposite forces acting on stopping the ball, so if you throw it it will go on forever.

on earth there is air resistance and gravity, this will pull the ball towards the ground and slow it down.

4 0
2 years ago
Which of the following statements is true about static electricity?
natta225 [31]
Static electricity can be an alternating current.
3 0
2 years ago
Read 2 more answers
The diagram shows the scales used for recording
padilas [110]

Answer: 212

Explanation:

3 0
3 years ago
Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both fil
MatroZZZ [7]

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

<u>FOR BULB 1:</u>

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴   ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

<u>FOR BULB 2:</u>

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴   ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

<u>A₁/A₂ = 0.44</u>

6 0
3 years ago
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