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4 years ago

12

A certain 100W light bulb has an efficiency of 95%. How much thermal energy will this light bulb add to the inside of a room in

2.5h if the bulb is on for the whole time?

1 answer:

Usimov [2.4K]4 years ago

7 0

Since the bulb consumes 100 watts of power and its efficiency is 95%,
it generates 95 watts of light energy and 5 watts of heat energy whenever
it's turned on.

5 watts means 5 joules of energy per second.

(2.5 hours) x (3,600 seconds/hour) = 9,000 seconds

(9,000 seconds) x (5 joules/second) = 45,000 joules of heat in 2.5 hours

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Helppppo please help help I have no time quick I’m in the exam

Sunny_sXe [5.5K]

Answer:

D

Explanation:

calculate effective resistance of the resistors

= 80 + 120

= 200

overall resistance = 48 ohms

so overall current flowing through circuit

I = V / R

I = 12 / 200

I = 0.06 A

since circuit is in series, current is same at every point so...

potential difference across 80 = R x I

= 80 x 0.06

answer is potential difference = 4.8 V

hope this helps

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5 0

3 years ago

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular accelera

nalin [4]

Answer:

5.74s

Explanation:

We can first solve for the initial angular velocity using the following formula

$\omega^2 - \omega_0^2 = 2\alpha\theta$

Where $\omega = -22.4rad/s$ is the final angular velocity, $\alpha = -22.4 rad/s^2$ is the angular acceleration and $\theta = 0$ is the angular displacement

$22.4^2 - \omega_0^2 = 2*(-7.8)*0$

$\omega_0^2 = 22.4^2$

$\omega_0 = 22.4rad/s$

So for the wheel to get from 22.4 to -22.4 with angular acceleration of -7.8 then the time it takes must be

$t = \frac{\Delta \omega}{\alpha} = \frac{-22.4 - 22.4}{-7.8} = 5.74s$

8 0

3 years ago

A liquid of density 1270 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the f

FinnZ [79.3K]

Answer:

${P_2}-P_1=49.99\ KPa$

Explanation:

$v_1=9.43\ m/s$

$d_1=11.7\ cm/s$

$d_2=17.5\ cm/s$

From continuity equation

$A_1v_1=A_2v_2$

$v_1d_1^2=v_2d_2^2$

$v_2=\dfrac{9.43\times 11.7^2}{17.5^2}$

$v_2=4.21\ m/s$

$\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2$

${P_1}+\rho\dfrac{v_1^2}{2}+\rho y_1g={P_2}+\rho\dfrac{v_2^2}{2}+\rho y_2g$

$\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1$

$1270\times \dfrac{9.43^2}{2}+1270\times 0\times 10 -1270\times\dfrac{4.21^2}{2}-1270\times 0.175\times 10={P_2}-P_1$

${P_2}-P_1=49.99\ KPa$

4 0

3 years ago

What is the value of work done on an object when a 10–newton force moves it 30 meters and the angle between the force and the di

Tom [10]

The component of force which is parallel to the path of the motion is 10N * cos25 = 9.06 N <span>
This component is the only "effective" force, and it is the one doing work.

Work = force * distance </span>

Work = 9.06 N * 30 m

<span>Work = 2.7×10^2 J</span>

8 0

3 years ago

Read 2 more answers

A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release

iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= $2pie\frac sqrt {m}{k}$

$\frac {{f2}/times {fo}}$ =1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒ $\frac{f2} {f1}$ = $\frac sqrt{m1}[m2}$

⇒ $\frac{f2}{fo}$ = 1:3

⇒f2= $\frac{fo} {3}$

6 0

3 years ago