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wariber [46]
2 years ago
13

Suppose a current flows through a copper wire. Which two things occur?

Physics
1 answer:
Bingel [31]2 years ago
4 0

Answer:

a

Explanation:

as the copper wire is very dangerous so now if these two thing happens then it would easily help the current flows through it so it might be a little bit easy for the current to flow through it

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DedPeter [7]
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8 0
2 years ago
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Question 33
pantera1 [17]

Answer:

A. The waves in the water travel faster and at a higher frequency than they travel on land.

Explanation:

The main reason why human ears can hear dolphins' vocalizations while under the water but cannot hear them well on land is because water is denser than air and air particles travel faster in denser particles.

Denser particles also ensures that the frequency of the waves move faster which in turn produces a faster and louder result.

3 0
2 years ago
A 3250-kg aircraft takes 12.5 min to achieve its cruising
scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

n = \frac{P_0}{P}

P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}

kinetic energy= \frac{1}{2} mv^2  =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2

kinetic energy = 90590389.66 kg m^2/s^2

gravitational energy = mgh = 3250*9.8*10000 = 315500000.00  kg m^2/s^2

total energy = 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2

P_o =\frac{409091242.28}{750} = 545454.99 j/s

effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49

effeciency n = = 49%

8 0
2 years ago
What is the centripetal force that would be required to keep a 4.0 kg mass moving in a horizontal circle with a radius of 0.80 m
KIM [24]

Answer:

D. 1.8 × 102 newtons radially inward

Explanation:

The magnitude of the centripetal force is given by:

F=m\frac{v^2}{r}

where

m is the mass of the object

v is the tangential speed

r is the radius of the circular trajector

In this problem, we have m = 4.0 kg, v = 6.0 m/s and r = 0.80 m, therefore substituting into the equation we get

F=(4.0 kg)\frac{(6.0 m/s)^2}{0.80 m}=180 N

The centripetal force is the force that keeps the object in a circular trajectory, so it is a force that is always directed inward (towards the centre of the circular path) and radially. Therefore, the correct answer is

D. 1.8 × 102 newtons radially inward

4 0
3 years ago
Read 2 more answers
During a solar eclipse, the Moon is positioned directly between Earth and the Sun. Find the magnitude of the net gravitational f
dezoksy [38]

Answer:

F= 2.3733 x10^{20} N

Explanation:

Let's define the variables to proceed with the operations,

So,

The masses

M_ {sun} = 1.99 * 10^{30} Kg

M_ {Earth} = 5.98 * 10 ^ {24} Kg

M_ {Moon} = 7.36 * 10 ^{22} Kg

Average distances

\bar {x} _ {Sun \rightarrow Earth} = 1.5 * 10 ^ {11} m

\bar {x} _ {Earth \rightarrow Moon} = 3.84 * 10 ^ 8m

Gravitational constant

G = 6.67 * 10^ {-11} \frac {Nm ^ 2} {kg ^ 2}

The formula of the Gravitational Force between the Moon and the Earth would be,

F = \frac {GM_ {Earth} M_ {Moon}} {(\bar {x} _ {Earth \rightarrow Moon}) ^ 2}

F= \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(5.98*10^{24}Kg)(7.36*10^{22}Kg)}{(3.84*10^8m)^2}

F = 1.9908 * 10 ^{20} N

This force is in the direction of the earth.

We perform the same process but now between the Sun and the Moon, like this,

F_2 = \frac {GM_ {Sun} M_ {Moon}} {(\bar {x} _ {Sun \rightarrow Earth} - \bar {x} _ {Earth \rightarrow Moon}) ^ 2}

F_2 = \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(1.99*10^{30}Kg)(7.36*10^{22}Kg)}{(1.5*10^{11}m-3.84*10^8m)^2}

F_2 = 4.3641*10^-{ 20} N

This force is in the direction of the Sun

The net force must be

F_ {net} = F_2-F

F_ {net} = 2.3733*10^{20} N

This in the direction of the Sun.

8 0
3 years ago
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