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3 years ago

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The source of power in a nuclear power plant comes from

1 answer:

alukav5142 [94]3 years ago

6 0

The source of power in a nuclear power plant is nuclear reactors. These reactors split uranium atoms, releasing energy in the form of heat, in a process called fission. The heat converts water into steam which powers a turbine generator to produce electricity.

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A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire

soldi70 [24.7K]

Answer:

a) $v=4.0m/s$

b) $B=2.958T$

Explanation:

From the question we are told that:

Wire Length $l=10cm=>0.10m$

Resistance $R=0.35$

Force $F=1.0N$

Power $P=4.0W$

a)

Generally the equation for Power is mathematically given by

$P=Fv$

Therefore

$v=\frac{P}{F}$

$v=\frac{4.0}{1.0}$

$v=4.0m/s$

b)

Generally the equation for Magnetic Field is mathematically given by

$B=\frac{\sqrt{PR}}{vl}$

$B=\frac{\sqrt{4*0.35}}{4*0.10}$

$B=2.958T$

5 0

2 years ago

A water droplet falling in the atmosphere is spherical. Assume that as the droplet passes through a cloud, it acquires mass at a

ArbitrLikvidat [17]

Answer:

it b

Explanation:

bc A water droplet falling in the atmosphere is spherical

4 0

2 years ago

A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th

Evgen [1.6K]

Answer:

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Explanation:

Given that

$\dfrac{dx}{dt}= -1\ m/s$

From the diagram

$tan\theta=\dfrac{21}{x}$

By differentiating with time t

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

When x= 10 m

$tan\theta=\dfrac{21}{10}$

θ = 64.53°

Now by putting the value in equation

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

$sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)$

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Therefore rate of change in the angle is 0.038\ rad/s

8 0

3 years ago

What does the object below model?

harina [27]

Answer: c

Explanation: it is c because i used my brain to answer it

3 0

2 years ago

By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 3?

s344n2d4d5 [400]

Answer:

Amplitude is decreased by a factor of $\sqrt3$ if intensity is decreased by a factor of 3.

Explanation:

Intensity of a sound wave is directly proportional to the square of its amplitude.

Therefore, if intensity is $I$ and amplitude is $A$ , then

$I=kA^2$ , where, $k$ is constant of proportionality.

Now, if intensity of sound wave is decreased by a factor of 3. So,

New intensity is, $I_{new}=\frac{I}{3}$

$I_{new}=kA_{new}^2\\\frac{I}{3}=kA_{new}^2$

Plug in $kA^2$ for $I$ . This gives,

$\frac{kA^2}{3}=kA_{new}^2\\A_{new}^2=\frac{A^2}{3}\\A_{new}=\sqrt{\frac{A^2}{3}}=\frac{A}{\sqrt{3}}$

Therefore, amplitude is decreased by a factor of $\sqrt3$ .

4 0

3 years ago