Ask question

Ask question

All categories

stealth61 [152]

3 years ago

10

If 100 J of electrical energy enter the bulb and 5 J of light energy leave the bulb, how many joules of heat energy leave the bu

lb?

1 answer:

Novay_Z [31]3 years ago

6 0

As per energy conservation we know that

Energy enter into the bulb = Light energy + Thermal energy

so now we have

energy enter into the bulb = 100 J

Light energy = 5 J

now from above equation we have

$100 = 5 + heat$

$Heat = (100 - 5) J$

$Heat = 95 J$

You might be interested in

This is heat transfer due to differences in density

Gala2k [10]

Answer:

convection

Explanation:

Heat transfer by convection is caused by differences of temperature and density within a fluid.

8 0

3 years ago

Read 2 more answers

Copper exists in nature as two isotopes. The atomic masses and relative abundance of these isotopes is given in the table. What

Alona [7]

<span>The correct answer should be B) 63.55. That's because the most precise number is 63.546, but you would write 55 because 46 is rounded that way in the equation. The others are a bit higher, while E is a completely different element, Iodine. This isn't the most precise piece of data because in reality there would be a slight differentiation of +- 0,003u</span>

7 0

3 years ago

An apple falls from a tree and hits the ground 9.98 m below. with what speed will it hit the ground

cupoosta [38]

The answer to this question is <span>13,537</span>

8 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

1. A man travels 45 m in 35 seconds. How fast is that?<br> What’s the answer

Natalija [7]

Answer:

1.28

Explanation:

If you want to find the m/s you would divide distance by time, so

45 divided by 35 would equal 1.28571429 and so on.

you can just write the three first numbers.

8 0

3 years ago