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!!
<span>F x L = W x X whereW=weight is total load = 80, L is length from fulcrum which is the unknown and what we are solving for. x= length we know. and F equals 50 force we know. So (W*X)/F=LL equals 64</span>
Answer:
Explanation:
By conservation of the linear momentum:
m1*V1 = (m1+m2)*V2
Solving for V2:
The kinetic energies are:
Simplifying:
The ratio will be:
Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
Jogger moves in three displacements
d1 = 10 blocks East
d2 = 5 blocks South
d3 = 2 blocks East
now we can say
total displacement towards East direction will be
Total displacement towards South
now to find the net displacement we can use vector addition
<em>so magnitude of net displacement will be equal to 13 blocks</em>
