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3 years ago

What is the sequence of energy transformations that occur in a nuclear reactor?

Physics

2 answers:

Pavel [41]3 years ago

7 0

Answer:

B. nuclear to thermal to mechanical to electrical

Explanation:

Nuclear reactions release energy used to heat steam which spins a turbine to generate electrical energy.

svet-max [94.6K]3 years ago

6 0

Nuclear reactions release energy used to heat steam which spins a turbine to generate electrical energy. Thus,
B. nuclear to thermal to mechanical to electrical

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An electron is released a short distance above earth's surface. a second electron directly below it exerts an electrostatic forc

slamgirl [31]

The mass of an electron is 9.109 x 10⁻³¹ kg

The weight of the electron is (mass) x (g) = 8.926 x 10⁻³⁰ Newton

The charge on an electron is -1.602 x 10⁻¹⁹ Coulomb

The repelling force between the two electrons is (K · q₁ · q₂ / r²) =

(8.98755 x 10⁹ N-m²/C²) x (1.602 x 10⁻¹⁹ C)² / D²

In order for the bottom one to just exactly hold the top one up at a distance 'D', the repelling force has to be exactly equal to the weight of the upper electron.

8.926 x 10⁻³⁰ N = (8.98755 x 10⁹ N-m²/C²)·(1.602 x 10⁻¹⁹ C)² / D²

We have to solve THAT ugly mess for ' D '.

Clean up the units first:

Cancel the C² on the right side, then divide each side by Newton:

8.926 x 10⁻³⁰ = (8.98755 x 10⁹ m²) x (1.602 x 10⁻¹⁹)² / D²

Now, let's multiply both sides by (D² x 10²⁹) :

D² x 8.926 x 10⁻¹ = (8.98755 m²) x (1.602)²

Divide each side by (0.8926):

D² = (8.98755 x 1.602²) / (0.8926) meter²

D² = 25.84 m²

Take the square root of each side:

<em>D = 5.08 meters</em>

I am shocked, impressed, and amazed !

Are you shocked, impressed, or amazed ?

3 0

4 years ago

A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel

anzhelika [568]

Answer:

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck $a_{t$ = 1 m/ $s^{2}$ (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

$a_{A}$ = 0.5 m/ $s^{2}$ (upward) , $a_{B}$ = 0.5 m/ $s^{2}$ (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( $m_{A}$ + $m_{B}$ )g = ( $m_{A}$ + $m_{B}$ ) $a_{B}$

= 2T- (400 + 50)*(9.81 m/ $s^{2}$ ) = (400 + 50)*(0.5 m/ $s^{2}$ )

T = 2320N

Truck: $M_{R}$ = ∑( $M_{R}$ )eff: = $-N_{f}$ (3.4m) + $m_{T}$ (2.0m) - T (0.6m)= $m_{T} a_{T}$ (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/ $s^{2}$ )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/ $s^{2}$ ) = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: $N_{f}$ + $N_{R}$ - $m_{T}$ g = 0

10544 + $N_{R}$ - (2000kg)(9.81 m/ $s^{2}$ ) = 0

$N_{R}$ = 9076N

∑fx (to the left) = ∑(fx)eff: $F_{R}$ - T = $m_{T} a_{T}$

$F_{R}$ = 2320N + (2000kg)(9.81 m/ $s^{2}$ ) = 4320N

(a) reaction at each front wheel:

1/2 $N_{f}$ (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: $N_{B}$ + $M_{B}$ g - $m_{B}$ $a_{B}$

$N_{B}$ = (400kg)(9.81 m/ $s^{2}$ ) + (400kg)(0.5 m/ $s^{2}$ ) = 4124N (compression)

3 0

4 years ago

Read 2 more answers

Determine Force. 589J of work was accomplished over a distance of 1100m, how much force was exerted? Include the correct units.

LiRa [457]

Answer:

Explanation:

Given:

A = 589 J

D = 1 100 m

____________

F - ?

A = F·D

F = A / D = 589 / 1100 ≈ 0.54 N

5 0

1 year ago

(4%) Problem 9: A mass is connected to a spring and is allowed to move horizontally. The mass is at a position L when the spring

skad [1K]

Answer:

a) Acceleration is zero , c) Speed is cero

Explanation:

a) the equation that governs the simple harmonic motion is

x = A cos (wt +φφ)

Where A is the amplitude of the movement, w is the angular velocity and φ the initial phase determined by the initial condition

Body acceleration is

a = d²x / dt²

Let's look for the derivatives

dx / dt = - A w sin (wt + φ)

a = d²x / dt² = - A w² cos (wt + φ)

In the instant when it is not stretched x = 0

As the spring is released at maximum elongation, φ = 0

0 = A cos wt

Cos wt = 0 wt = π / 2

Acceleration is valid for this angle

a = -A w² cos π/2 = 0

Acceleration is zero

c) When the spring is compressed x = A

Speed is

v = dx / dt

v = - A w sin wt

We look for time

A = A cos wt

cos wt = 1 wt = 0, π

For this time the speedy vouchers

v = -A w sin 0 = 0

Speed is cero

5 0

4 years ago

Explain sound waves.sources of sound waves and the experiment to demonstrate sound waves

kipiarov [429]

The sound wave is series of rarefaction and compressions traveling through substance . and the waves travel through medium like solid , liquid , and gases

8 0

4 years ago