Answer:
change in mass = 2.41*10^{8}kg
Explanation:
The change in the mass can be computed by using the relation
(1)
That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

Hence, by replacing in the equation (1) you have (c=3*10^{8}m/s)

HOPE THIS HELPS!!
Explanation:
To find the average of these numbers, we just have to add the three numbers together and divide by 3.
- 2.07 + 0. 74 + 1.33 = 4.14. 4.14 / 3 = 1.38
- 1.09 + 1.40 + 0.31 = 2.8. 2.8 / 3 ≈ 9.3333333/ 9 1/3
- 0.95 + 1.61 + 0.56 = 3.12 / 3 = 1.04
- 0.81 + 1.89 + 1.08 = 3.78 / 3 = 1.26
a = 3.09 m/s²
<h3>Explanation</h3>
This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the <em>timeless </em>suvat equation is likely what you need for this question.
In the <em>timeless</em> suvat equation,

where
is the acceleration of the car;
is the <em>final</em> velocity of the car;
is the <em>initial</em> velocity of the car; and
is the displacement of the car.
Note that <em>v</em> and <em>u</em> are velocities. Make sure that you include their signs in the calculation.
In this question,
Apply the <em>timeless</em> suvat equation:
.
The value of
is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.
Answer:
v= 103.5 V; energy =1.65 x 10^-17
Explanation:
the deflected energy eV sin θ
Complete Question
Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?
a. 3/8q
b. 1/4q
c. 3/4q
d. q
e. 5/8q
f. 1/3q
g.1/2q
h. 0
Answer:
The correct option is b
Explanation:
From the question we are told that
The charge carried by A is q C
The charge carried by B is 0 C
The charge carried by C is 0 C
When A and B are brought close and then separated the charge carried by A and B is mathematically evaluated as

When C and B are brought close and then separated the charge carried by C and B is mathematically evaluated as

When C and A are brought close and then separated the charge carried by C and A is mathematically evaluated as

Looking at these calculation we can see that the charge carried by B is
