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3 years ago

How does a nuclear power plant produce energy

1 answer:

8 0

The steam then turns turbines to produce electricity. The difference is that nuclear plants do not burn anything. Instead, they use uranium fuel, consisting of solid ceramic pellets, to produce electricity through a process called fission. best i can do hope it helps

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Wil-E-Coyote drops a bowling ball off a cliff to try to catch the Roadrunner. The cliff is

PtichkaEL [24]

Answer:

t = 5.19 s

Explanation:

We have,

Height of the cliff is 132 m

It is required to find the time taken by the ball to fall to the ground. Let t is the time taken. So, using equation of kinematics as :

$y=ut+\dfrac{1}{2}gt^2\\\\\text{since}\ u=0\\\\y=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2y}{g}}\\\\t=\sqrt{\dfrac{2\times 132}{9.8}}\\\\t=5.19\ s$

So, it will take 5.19 seconds to fall to the ground.

8 0

2 years ago

how can you tell, as you walk close to a parked car, if it had been running recently? describe your reasoning in terms of energy

Blababa [14]

This question is probably referring to heat energy transferring from the car to its surroundings.

4 0

3 years ago

Which is a rule of boundary plates colliding or sliding past each other?

marissa [1.9K]

Explanation:

C...earthquakes occur

6 0

2 years ago

Read 2 more answers

The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le

lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = $\sqrt{2gh}$

V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

= 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

= - 0.329 J

Hence, kinetic energy is not conserved.

8 0

2 years ago

A sound from a source has an intensity of 270 dB when it is 1 m from the source.

Rufina [12.5K]

Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: $\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
where
$I_{2}$ is the intensity at distance 2
$I_{1}$ is the intensity at distance 1
$d_{2}$ is distance 2
$d_{1}$ is distance 1

We can infer for our problem that $I_{1}=270$ , $d_{1}=1$ , and $d_{2}=3$ . Lets replace those values in our formula to find $I_{2}$ :
$\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
$\frac{I_{2} }{270} =( \frac{1}{3} )^2$
$\frac{I_{2} }{270} = \frac{1}{9}$
$I_{2}= \frac{270}{9}$
$I_{2}=30$ dB

We can conclude that the intensity of the sound when is 3 m from the source is 30 dB.

8 0

3 years ago