Answer:
Explanation:
for baseball
(a) Let the mass of the baseball is m.
radius of baseball is r.
Total kinetic energy of the baseball, T = rotational kinetic energy + translational kinetic energy
T = 0.5 Iω² + 0.5 mv²
Where, I be the moment of inertia and ω be the angular speed.
ω = v/r
T = 0.5 x 2/3 mr² x v²/r² + 0.5 mv²
T = 0.83 mv²
According to the conservation of energy, the total kinetic energy at the bottom is equal to the total potential energy at the top.
m g h = 0.83 mv²
where, h be the height of the top of the hill.
9.8 x h = 0.83 x 6.8 x 6.8
h = 3.93 m
(b) Let the velocity of juice can is v'.
moment of inertia of the juice can = 1/2mr²
So, total kinetic energy
T = 0.5 x I x ω² + 0.5 mv²
T = 0.5 x 0.5 x m x r² x v²/r² + 0.5 mv²
m g h = 0.75 mv²
9.8 x 3.93 = 0.75 v²
v = 7.2 m/s
The metals will start to rust lol. i think. because this messes up how the metals conduct the flow of the electricity.
Answer:
Oi, mate its false
Explanation:
because if an leaf floats down from a tree it is not considered an object for a free-fall
The best explanation for the difference in time is: A. The difference in weight doesn't affect the time, but they are affected differently by air resistance.
<h3>What is weight?</h3>
Weight can be defined as the force acting on an object or a physical body due to the effect of gravity. Also, the weight of an object (body) is typically measured in Newton.
<h3>The factors that affect weight.</h3>
Some of the factors that affect the weight that is possessed by an object or a physical body include the following:
In conclusion, the weight possessed by the shoe and shirt has no effect on time but would be affected differently by air resistance.
Read more on weight here: brainly.com/question/13833323
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
