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Travka [436]
3 years ago
5

Which electrical device makes it possible to transmit electrical energy efficiently from a power plant to users?

Physics
2 answers:
Debora [2.8K]3 years ago
6 0
Solar pannels are the energy sorce that lets you transfer energy from a plant source to the use or the consumer 
Aneli [31]3 years ago
3 0
Hey, Name's Jessy. I hope, I answer your question.

A Transformer. It current carrying wire.
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Which statement corresponds to emission spectra?
monitta
B
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7 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

acceleration=\frac{30}{5}=6m/s^{2}

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0
3 years ago
A toy car moves 8 min 4 s at the constant velocity. What is the car's velocity?
Schach [20]

Explanation:

<u>Formula:</u>

velocity = (d \div t)

<u>d = distance given</u>

<u>t</u><u> </u><u>=</u><u> </u><u>the amount of time </u><u>given</u>

<u>Substitute the given values into the formula for velocity</u><u>:</u>

v = 8 \div 4

velocity is shortened for v.

8 (distance) divided by 4 (time) equals the velocity.

<u>Solve:</u>

2 = 8 \div 4

The velocity of the toy car equals: B. 2 m/s.

6 0
2 years ago
A Roller Derby Exhibition recently came to town. They packed the gym for two
arlik [135]

Answer:

14.36m/s

Explanation:

From the law of conservation of linear momentum

m1u1 + m2u2 = v(m1 + m2)

68×17 + 76×12= v(68+76)

1156+912 = 144v

2068 = 144v

v = 2068/144

=14.36 m/s

7 0
3 years ago
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