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QveST [7]
3 years ago
14

A lightbulb has a power of 100 W and is used for 4 hours. A microwave has a power of 1200 W and is used for 5 minutes. How much

energy is used by the lightbulb? How much energy is used by the microwave? The lightbulb has an efficiency of 1.8%. How much heat energy does the lightbulb produce in 4 hours?
Physics
1 answer:
ladessa [460]3 years ago
4 0

Answer:

a. E=1440KJ

b. E=360KJ

c. E=1.8 J

Explanation:

I have the power (Watts) is expressed as Energy (Joules) / Time (seconds), also I have to1hour * \frac{60min}{1hour}*\frac{60s}{1min}\\1h=3600s

so, for the lightbulb

100W= \frac{Energy}{3600s*4}\\Energy=100W*14400s\\ E=1440000J =1440KJ

Analogously, for the microwave

1200W= \frac{Energy}{60s*5}\\Energy=1200W*300s\\ E=360000J =360KJ

Now, I have to express the efficiency as Heat energy / power * 100 so1.8= \frac{Heat Energy}{100W}*100\\Heat Energy=1.8J

Done

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<h2>Answer:V=\sqrt{G\frac{M}{R}}  </h2>

The velocity of a satellite describing a circular orbit is <u>constant</u> and defined by the following expression:

V=\sqrt{G\frac{M}{R}}     (1)

Where:

G is the gravity constant

M the mass of the massive body around which the satellite is orbiting

R the radius of the orbit (measured from the center of the planet to the satellite).

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. I<u>t depends on the mass of the massive body.</u>

In addition, this orbital speed is constant because at all times <u>both the kinetic energy and the potential remain constant</u> in a circular (closed) orbit.

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
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Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

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