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3 years ago

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At an oceanside nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water th

at is discharged back into the ocean. The amount that the water temperature is raised has a uniform distribution over the interval from 10° to 25° C. Suppose that a temperature increase of more than 18° C is considered to be potentially dangerous to the environment. What is the probability that at any point in time, the temperature increase is potentially dangerous?

1 answer:

Mice21 [21]3 years ago

8 0

Answer is 30. just took it

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What is the melting point of substance A?

Misha Larkins [42]

Answer:

Solids are easily recognized by their ability to retain a fixed shape and definite volume. Particles making

up a solid are held together in a rigid form. They are not free to move about or slide past one another and

the solid does not have the ability to flow. (Although the particles of a solid do not move position to position, they do have motion in that they are constantly vibrating.

To change the temperature of a solid, heat energy must be added. The amount of heat energy that changes

the temperature of 1.0 g of a solid by 1.0°C is called its specific heat (c). Each substance has its own

specific heat. The specific heat of ice is 2.1 Joules/g°C. In other words we must supply 1.0 gram of ice

with 2.1 Joules of heat energy to raise its temperature by 1.0 °C.

The general equation for calculating heat energy to change the temperature of a solid is:

Heat = Mass x Specific Heat (solid) x Temperature Change

Q = m c DT

10 g 10 g 10 g 10 g 10 g 10 g

Calculate the heat necessary to change 10 g of ice(s) at -20 °C to 10 g of ice(s) at 0°C. (A-B)

Q = mc∆T = (10 g) (2.1 J/g°C) (20°C) = 420 J

If you continue to add heat energy once the temperature of the ice reaches 0°C , the heat absorbed is called

the heat of fusion (Lf). This heat is used to cause a change of phase (from a solid to a liquid). This heat is

increasing the potential energy of the molecules of the solid. No temperature change takes place. Each

substance has its own heat of fusion. The heat of fusion for ice is 340 Joules/g. Exactly the same amount

of heat is given up when 1.0 g of water is changed to ice. This heat is called the heat of crystallization.

The general equation for calculating heat energy to change a solid to a liquid is:

Heat = Mass x Heat of Fusion

Q = m Lf

Calculate the heat necessary to change 10 g of ice(s) at 0°C to 10 g of water(l) at 0°C.(B-C)

Explanation:

Q = mLf = (10 g)( 340 J/g) = 3400 J

3 0

3 years ago

The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of t

oee [108]

The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

By definition the gravitational potential energy is given by,

$PE=\frac{GMm}{r}$

Where,

m = Mass of Moon

G = Gravitational Universal Constant

M = Mass of Ocean

r = Radius

First we calculate the mass through the ratio given by density.

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

Then the potential energy at this point would be,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

At the nearest point we perform the same as the previous process, we calculate the radius

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

The we calculate the Potential gravitational energy,

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

7 0

3 years ago

If a car has a momentum of 2.04 x 104 kgm/s and a velocity of 18 m/s, what is its mass?

muminat

Answer:

Mass = 1133.33 kg (Approx.)

Explanation:

Given:

Momentum = 2.04 x 10⁴ kg[m/s]

Velocity = 18 m/s

Find:

Mass

Computation:

Mass = Momentum / Velocity

Mass = [2.04 x 10⁴] / 18

Mass = 1133.33 kg (Approx.)

5 0

3 years ago

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

<u>Time taken to descent the total height:</u>

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

<u>Now the total time taken by the ball to hit the ground:</u>

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

Two people are standing on rollerskates. One is more massive than the other. They push against each other and move away. How do

Gala2k [10]

If they push away, C. The more massive person moves slower.

6 0

3 years ago