1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.
<u>Explanation</u>:
We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the 
and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:
Better understood from numerical example as given:
If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?
This can be solved as follows:
It shows that man A will have more K.E.
Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.
Answer:
Batteries are systems that store chemical energy and then release it as electrical energy when they are connected to a circuit. Batteries can be made from many materials, but they all share three main components: a metal anode, a metal cathode and an electrolyte between them. The electrolyte is an ionic solution that allows charge to flow through the system. When a load, such as a light bulb, is connected, an oxidation-reduction reaction occurs that releases electrons from the anode while the cathode gains electrons
Explanation:
Answer:
(a) - 42700 m/s
(b) - 6.8 x 10^-4 m/s^2
Explanation:
initial velocity of star, u = 20.7 km/s
Final velocity of star, v = - 22 km/s
time, t = 1.99 years
Convert velocities into m/s and time into second
So, u = 20700 m / s
v = - 22000 m/s
t = 1.99 x 365.25 x 24 x 3600 = 62799624 second
(a) Change in planet's velocity = final velocity - initial velocity
= - 22000 - 20700 = - 42700 m/s
(b) Accelerate is defined as the rate of change of velocity.
Acceleration = change in velocity / time
= ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2
Answer:
(a) V = 0.75 m/s
(b) V = 0.125 m/s
Explanation:
The speed of the flow of the river can be given by following formula:
V = Q/A
V = Q/w d
where,
V = Speed of Flow of River
Q = Volume Flow Rate of River
w = width of river
d = depth of river
A = Area of Cross-Section of River = w d
(a)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 20 m
d = 20 m
Therefore,
V = (300 m³/s)/(20 m)(20 m)
<u>V = 0.75 m/s</u>
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(b)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 60 m
d = 40 m
Therefore,
V = (300 m³/s)/(60 m)(40 m)
<u>V = 0.125 m/s</u>
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
