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rosijanka [135]
3 years ago
14

A car radio draws 0.18 A of current in the autos 12-V electrical system. (a) How much electric power does the radio use?

Physics
1 answer:
Lera25 [3.4K]3 years ago
8 0

Answer:

Explanation:

a) Electric Power = V X I ,  where V is potential difference  and I is current

= 12 x .18 =2.16 Watt .

b) Resistance = volt / current ( ohm's Law )

Resistance = 12/0.18 = 66.67 ohm.

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A 30 kg child rides a 20 kg bicycle. Together, the child and the bicycle have a momentum of 110 kg-m/s. What is the velocity of
love history [14]

Answer:

The velocity of the boy and the bicycle is 2.2 m/s.

Explanation:

We have,

Mass of child is 30 kg and the mass of bicycle is 20 kg. The combined momentum of the child and the bicycle is 110 kg-m/s.

It is required to find the velocity of the boy and the bicycle. The momentum of an object is given in terms of mass and its velocity. So,

p=Mv

M is combined mass of child and bicycle

v=\dfrac{p}{M}\\\\v=\dfrac{110}{30+20}\\\\v=2.2\ m/s

So, the velocity of the boy and the bicycle is 2.2 m/s.

8 0
3 years ago
Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou
spin [16.1K]

Answer:

(a) 28690 m/s (b) 2.46x10^{33}J

Explanation:

The orbital speed is define as:

v = \frac{2 \pi r}{T}   (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

v = \frac{d}{t}   (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

d = vt   (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

8min . \frac{60s}{1min} ⇒  480s

t = 480s

Replacing all those values in equation 3 it is gotten:

d = (3.0x10^{8}m/s)(480s)

d = 1.44x10^{11}m

Kepler’s third law is defined as:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

T = \sqrt{r^{3}}

T = \sqrt{(1.44x10^{11}m)^{3}}

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

T = \sqrt{1AU}

T = 1AU

That can be expressed in units of years.

T = 1AU . \frac{1year}{1AU}

T = 1year

But there are 31536000 seconds in one year:

T = 1year . \frac{31536000s}{1year}

T = 31536000s

Finally, equation  1 can be used:

v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}

v = 28690 m/s

<u>So Earth orbital speed around the Sun is 28690 m/s.</u>

<em>b) What is its kinetic energy?</em>

The kinetic energy is defined as:

E = \frac{1}{2}mv^{2}  (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

F = \frac{GMm}{r^{2}}

ma = \frac{GMm}{r^{2}}  (5)

M will be isolated in equation 5:

M = \frac{r^{2}a}{G}

Where r is the radius of the Earth (6.38x10^{6}m)

M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})

M = 5.98x10^{24} Kg

E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}

E = 2.46x10^{33}Kg.m^{2}/s^{2}

E = 2.46x10^{33}J

<u>Hence, the kinetic energy of Earth is 2.46x10^{33}J.</u>

8 0
3 years ago
A physics student swings a pail of water in a vertical circle 1.0 m in radius at a constant speed. If the water is NOT to spill
love history [14]

Answer:

(A) 3.1 m/s

(B) 2.0 s

Explanation:

At the minimum speed, the force of gravity equals the centripetal force.

mg = m v² / r

v = √(gr)

v = √(9.8 m/s² × 1.0 m)

v = 3.1 m/s

The time is the circumference divided by the speed.

t = (2π × 1.0 m) / (3.1 m/s)

t = 2.0 s

7 0
3 years ago
LAST ONE! ASAP PLEASE
Doss [256]

Answer:

There are so many questions which one you don't know

4 0
3 years ago
Read 2 more answers
A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0
3 years ago
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