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3 years ago

Typical Pressurized Water Reactors can produce 1100 to 1500

Physics

1 answer:

lozanna [386]3 years ago

3 0

Answer: about 1,100,000,000 to 1,500,000,000 Joules/second

Explanation:

1 MW (megawatt) = 1,000,000.00 J/s (joules per second)

1100(1,000,000) = 1,100,000,000

1500(1,000,000) = 1,500,000,000

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A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

$T_{mi}$ = initial temperature of metal = 100 °C

$m_{a}$ = mass of aluminum cup = 100 g

$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$

7 0

4 years ago

How would you design an experiment to condition a rabbit to salivate to the ringing of a cell phone?

Klio2033 [76]

Food causes the response (salivation). Pair the food each time with a cell phone ringing. Continue this process with repetition.
• Eventually, the rabbit will learn to salivate at the ringing

5 0

3 years ago

1. A truck with a mass of 8, 000 kg is traveling at 26.8 m/s when it hits the brakes. A.)What is the momentum of the truck befor

NikAS [45]

Answer:

1. A.) The moment of the truck before it hits the brakes is 214,400 kg·m/s

B.) The force it takes to stop the truck is approximately 17,290.4 N

Explanation:

1. A.) The given parameters are;

The mass of the truck, m = 8,000 kg

The velocity of the truck when it hits the brakes, u = 26.8 m/s

Momentum = Mass × Velocity

The moment of the truck = The mass of the truck × The velocity of the truck

Therefore;

The moment of the truck before it hits the brakes = 8,000 kg × 26.8 m/s = 214,400 kg·m/s

B.) The amount of momentum lost when the truck comes to a stop = The initial momentum of the truck

The time it takes the truck to come to a complete stop, t = 12.4 s

The deceleration, "a" of the truck is given by the following kinematic equation of motion

v = u - a·t

Where;

v = The final velocity of the truck = 0 m/s

u = The initial velocity = 26.8 m/s

a = the deceleration of the truck

t = The time of deceleration of the truck = 12.4 s

Substituting the known values gives;

0 = 26.8 - a × 12.4

Therefore;

26.8 = a × 12.4

a = 26.8/12.4 ≈ 2.1613

The deceleration (negative acceleration) of the truck, a ≈ 2.1613 m/s²

Force = Mass × Acceleration

The force required to stop the truck = The mass pf the truck × The deceleration (negative acceleration) given to the truck

∴ The force it takes to stop the truck = 8,000 kg × 2.1613 m/s² ≈ 17,290.4 N.

8 0

3 years ago

According to Coulomb's law, if the separation between two particles of the same charge increases four times, the potential energ

Lera25 [3.4K]

Answer:

c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R

V2 / V1 = (R1 / R2) = 1/4

V2 = V1 / 4

8 0

3 years ago

An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car

Rudik [331]

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)= $3.2 m/s^2$

Speed of car after 4.5 s

using equation of motion

v=u+at

$v=4.92+3.2\times 4.5=4.92+14.4$

v=19.32 m/s

Displacement of the car after 4.5 s

$v^2-u^2=2as$

$19.32^2-4.92^2=2\times 3.2\times s$

$349.05=2\times 3.2\times s$

s=54.54 m

4 0

3 years ago