The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as
Here,
= Change in height
m = mass of super heroine
g = Acceleration due to gravity
The change in height will be,
The final position of the heroin is below the ground level,
The initial height will be the zero point of our system of reference,
Replacing all this values we have,
Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J
Answer:
Your answer should be A. 0.25 J and moving to the right
Explanation:
Answer:
a)Q=71.4 μ C
b)ΔV' = 10.2 V
Explanation:
Given that
C ₁= 8.7 μF
C₂ = 8.2 μF
C₃ = 4.1 μF
The potential difference of the battery, ΔV= 34 V
When connected in series
1/C = 1/C ₁ + 1/C₂ + 1/C₃
1/ C= 1/8.4 +1 / 8.4 + 1/4.2
C=2.1 μF
As we know that when capacitor are connected in series then they have same charge,Q
Q= C ΔV
Q= 2.1 x 34 μ C
Q=71.4 μ C
b)
As we know that when capacitor are connected in parallel then they have same voltage difference.
Q'= C' ΔV'
C'= C ₁+C₂+C₃ (For parallel connection)
C'= 8.4 + 8.4 + 4.2 μF
C'=21 μF
Q'= C' ΔV'
Q'=3 Q
3 x 71.4= 21 ΔV'
ΔV' = 10.2 V
Answer:
total momentum = 8.42 kgm/s
velocity of the first cart is 3.660 m/s
Explanation:
Given data
mass m1 = 2.3 kg
mass m2 = 1.5 kg
final velocity V2 = 4.9 m/s
final velocity V3 = - 1.9 m/s
to find out
total momentum and velocity of the first cart
solution
we know mass and final velocty
and initial velocity of second cart V1 = 0
so now we can calculate total momentum that is m1 v2 + m2 v2
total momentum = 2.3 ×4.9 + 1.5 ×(-1.9)
total momentum = 8.42 kgm/s
and
conservation of momentum is
m1 V + m2 v1 = m1 v2 + m2 v3
put all value and find V
2.3 V + 1.5 ( 0) = 2.3 ( 4.9 ) + 1.5 ( -1.9)
V = 8.42 / 2.3
V = 3.660 m/s
so velocity of the first cart is 3.660 m/s
Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration, 
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
