charge stored in the capacitor=3.29 x 10⁻⁴ C
Explanation:
we use the formula
Q= C V
Q= charge
C= capacitor=25.3 μF= 25.3 x 10⁻⁶ F
V= voltage= 13 V
Q=(25.3 x 10⁻⁶ ) (13)
Q= 3.29 x 10⁻⁴ C
The work done on the car is -20 J.
Work done on the car is negative, meaning that the car actually does work on the external system.
<h3>Energy and law of conservation of energy</h3>
- Energy is the ability to do work
- the law of conservation of energy states that the total energy in a system is conserved
From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.
- Initial energy = 100 J
- Initial energy = Final energy - work done on car
- Final Energy = Work done on car + initial energy
80J = Work done on car + 100 J
Work done on car = 80 - 100J
Work done on car = -20 J
Hence, the work done on the car is -20 J
Work done on car is negative.
Since work done on the car is negative, it means that the car actually does work on the external system. Hence, the decrease in the energy of the car.
Learn more about energy and work at: brainly.com/question/13387946
Answer:
Explanation:
mass of the fellow ( m ) = 66kg
acceleration of fellow a
v = u + at
4.5 = 0 + a x 2
a = 4.5 /2
= 2.25 m / s²
Net force acting on fellow in upward direction by the surface of elevator
R - mg where R is reaction force of the surface of the elevator
Applying Newton's law of motion
R - mg = ma
R = m (g +a )
= 66 x ( 9.8 + 2 )
= 778.8 N
This will be the scale reading .
Answer:
The correct option is D
Explanation:
In trying to achieve what the student wanted to see, which is to see the relationship between the weight the cord can hold and how long the cord will stretch. Since the origin of the graph is from zero, the value plotted on the vertical axis would be just the length caused by each weights. Thus, <u>the original length would have to be subtracted from the measured length to determine the actual length caused by the weight added to the cord</u>.
