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3 years ago

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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Name two effects of using coal as energy

german

Answer:

Here is some useful background facts...

Explanation:

They include mercury, lead, sulfur dioxide, nitrogen oxides, particulates, and various other heavy metals. Health impacts can range from asthma and breathing difficulties, to brain damage, heart problems, cancer, neurological disorders, and premature death

7 0

3 years ago

Read 2 more answers

A certain wave has a wavelength of 45.0 meters and a frequency of 9.00 Hz. What is the speed of the wave? 405 m/s 5.00 m/s 0.200

kirill115 [55]

Wave speed = (wavelength) x (frequency)

= (45 meters) x (9 per second)

= 405 meters per second .

5 0

3 years ago

Read 2 more answers

Object A has mass 83.0 g and hangs from an insulated thread. When object B, which has a charge of +140 nC, is held nearby, A is

erastova [34]

Answer:

a) -238 nC

b) 0.889 N

Explanation:

Concepts and Principles

Particle in Equilibrium: If a particle maintains a constant velocity (so that a = 0), which could include a velocity of zero, the forces on the particle balance and Newton's second law reduces to:

∑F = 0 (1)

Coulomb's Law: the magnitude of the electrostatic force exerted by a point charge q1 on a second point charge q2 separated by a distance r is directly proportional to the product of the two charges and is inversely proportional to the square of the distance between them:

F_12 = k*| q1 |*| q2 |/r^2 (2)

where k = 8.99 x 10^9 N m^2/C^2 is Coulomb constant.

Given Data

mA (mass object A) = (83 g)*(1/1000g)=0.09 kg

qB (charge of object B) = (140 nC)*(1/10^9 nC) = 130 x 10^-9 C

Object A is attracted to object B.

Ф(angle made by object A with the vertical) = 7.2°

( r (distance between the two objects) = (5 cm) * (1 m/ 100 cm) =0.05 m

Object A is in equilibrium.

Required Data

In part (a), we are asked to determine the charge qA of object A.

In part (b), we are asked to determine the tension T in the thread.

(a) The FBD in Figure 1 shows the forms acting on object A; Fe is the electric force exerted on object A by object B, T is the tension force exerted on the thread, and m_a*g is the gravitational force exerted on object A.

Model object A as a particle in equilibrium in the horizontal and vertical direction and apply Equation (1) to it:

∑F_x = F_e-Tsin = 0 F_e=TsinФ (3)

∑F_y = TcosФ - m_a*g= 0 m_a*g=TsinФ (4)

Divide Equation (3) by Equation (4) to eliminate T:

F_e/m_a*g=tanФ

F_e=m_a*g*tanФ

Substitute for F_e by using Coulomb's law from Equation (2):

k*| q_A |*| q_B |/r = m_a*g*tanФ

Solve for q_A :

| q_A | = m_a*g*tanФ_r/k*| q_B |

Substitute numerical values from given data:

| q_A | = 238 nC

Because object A is attracted to object B. it has an opposite negative charge. Therefore, the charge on object A is | q_A | = -238 nC

(b)

Solve Equation (4) for T:

T = m_a*g/cosФ

Substitute numerical values from given data:

T = (0.09 kg)(9.8 m/s^2) /cos 7.2°

= 0.889 N

4 0

3 years ago

Two identical projectiles were projected at 200 km/h the first one is at an angle 60 to the horizontal while the second one is a

Setler79 [48]

Answer:

The angle that will have the most range (Goes further) will be the one that is shot with 60 ° with the horizontal. 

Since the one that shoots with respect to the vertical, it will only reach higher. And it won't go as far as the horizontal.

Greetings.

Danna.

3 0

2 years ago

Please help!!

xz_007 [3.2K]

Displacement is the area under the velocity/time graph. So for example this object's displacement in the first 3 seconds is (1/2)(3sec)(12.5 m/s)= 18.75m. (and then it starts backing up, displacement decreasing, after 3sec when velocity is negative).

But This object is never speeding up. Its velocity is smoothly decreasing at (25/6) m/s^2 (the slope of the graph). So the answer to the question is actually zero.

3 0

3 years ago