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galina1969 [7]
3 years ago
5

In a nuclear power plant, the temperature of the water in the reactor is above 100°C because of what?

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

Answer:

The temperature of the water increases because the nuclear reactor heats it producing steam

Explanation:

The nuclear power plants are usually defined as those thermal plants where the nuclear reactors are used in order to generate heat that eventually leads to the rotating of the turbines and produces electricity. Here the nuclear reactor heats the water, and it increases above a temperature of 100°C, where this heat energy plays a key role in the entire process. It is an efficient method as it does not lead to the emission of any green house gases that are harmful to the environment.

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If you travel 1.7 km north from your house at noon, and at 6:00 PM you travel 5.4 km south, what is your displacement? 3.7 km no
Yakvenalex [24]

<u>Answer</u>

3.7 Km south


<u>Explanation</u>

The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.

Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).

Now lets add the two values.

(+1.7) + (-5.4) = 1.7 - 5.4

                    = - 3.7 Km      But negative was towards south.

∴ Answer = 3.7 Km south.


6 0
3 years ago
A machinist is required to manufacture a circular metal disk with area 1900 cm2. (a) What radius produces such a disk? (Round yo
alexgriva [62]

Answer:

24.5987 cm

Explanation:

A = 1900 cm^2

Let r be the radius of disc.

The area of disc is given by

A = π r²

Where, π = 31.4

1900 = 3.14 x r²

r² = 605.095

r = 24.5987 cm

6 0
3 years ago
A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energ
lesantik [10]

Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC  at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

The formula for the potential energy between the two charges is given by

U=\frac{Kq_{1}q_{2}}{r}

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

U = U_{1,2}+U_{2,3}+U_{3,1}

U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}

U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}

U = - 2.7 x 10^-6 J

3 0
3 years ago
How to find the density of air?
nikitadnepr [17]
You have to divide the pressure exerted by the air into two partial pressures: of the dry air and of the water vapor. Combining these two values gives you the parameter.
8 0
2 years ago
[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
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