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3 years ago

Which feature or process is unique to nuclear power plants when compared to conventional coal-burning power plants?

Physics

2 answers:

Mazyrski [523]3 years ago

5 0

Answer:

Control rods

Explanation:

Control rods is unique to nuclear power plants as they are mainly used in nuclear reactors to stabilize the rate of fission of plutonium or uranium. They are usually composed of chemical elements such as cadmium, silver, boron, and indium.

Maru [420]3 years ago

4 0

Answer:

Control rods

Explanation:

Control rods are used to control the rate of nuclear chain reaction in nuclear reactors. The reactivity of the nuclear reactor is greatly affected by the distance of control rods inserted and number of control rods inserted. Control rods control the rate of fission of uranium, etc., in a nuclear reactor

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A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?

jeka57 [31]

Answer:

<h3>The answer is 45 J</h3>

Explanation:

The work done by an object can be found by using the formula

<h3>workdone = force × distance</h3>

From the question

distance = 3 meters

force = 15 newtons

We have

workdone = 15 × 3

We have the final answer as

Hope this helps you

7 0

3 years ago

2. What do you understand by balanced and unbalanced force

labwork [276]

Answer:

forces that are equal in size and opposite in direction. Balanced forces do not result in any change in motion. unbalanced. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.

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hope helpful ~

8 0

2 years ago

A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What

kramer

Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let $p_{1}$ = momentum of the 1st ball

$p_{2}$ = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

$(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}$

$(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}$

Since we are dealing with identical balls, all the m terms cancel out so we are left with

$(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} + (v_{2})_{f}\cos \theta_{2}$

Putting in the numbers, we get

$1.33 = (0.750) \cos(33.30) + (v_{2})_{f} \cos \theta_{2}$

$= > (v_{2})_{f} \cos \theta_{2} = 0.703$

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

$0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}$

$= > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30) = 0.412$

Taking the ratio of the sine equation to the cosine equation, we get

$\frac{ \sin \theta _{2}}{ \cos \theta_{2} } = \tan \theta_{2} = \frac{0.412}{0.703} = 0.586$

$\theta_{2} = { \tan}^{ - 1} (0.586) = 30.4$

Solving now for $(v_{2})_{f}$ ,

$(v_{2})_{f} = \frac{0.412}{ \sin(30.4) } = 0.815 \: \frac{m}{s}$

3 0

3 years ago

Can someone help ? I need help please

MAXImum [283]

Answer:

Explanation:

to see how fast she is going per second, you would have to divide the distance traveled by the seconds it took to travel the distance

5 0

2 years ago

Read 2 more answers

When jumping, a flea accelerates at an astounding 1000 m/s2 but over the very short distance of 0.50 mm. If a flea jumps straigh

Nadusha1986 [10]

Answer:

The flea reaches a height of 51 mm.

Explanation:

Hi there!

The equations of height and velocity of the flea are the following:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

While in free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the flea at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration of the flea due to the jump.

v = velocity of the flea at time t.

g = acceleration due to gravity.

First, let's calculate how much time it takes the flea to reach a height of 0.0005 m. With that time, we can calculate the speed reached by the flea during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference on the ground, then, h0 = 0. Since the flea is initially at rest, v0 = 0. Then:

h = 1/2 · a · t²

We have to find the value of t for which h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Now, let's find the velocity reached in that time:

v = v0 + a · t (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

When the flea is at a height of 0.50 mm, its velocity is 1.00 m/s. This initial velocity will start to decrease due to the downward acceleration of gravity. When the velocity is zero, the flea will be at the maximum height. Using the equation of velocity, let's find the time at which the flea is at the maximum height (v = 0):

v = v0 + g · t

At the maximum height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, let's find the height reached by the flea in that time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

The flea reaches a height of 51 mm.

5 0

3 years ago