Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)
The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)
Moment of Inertia refers to:
- the quantity expressed by the body resisting angular acceleration.
- It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.
The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)
here We note that the,
In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.
The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:
I(edge) = I (center of mass) + md^2
d be the distance from an axis through the object’s center of mass to a new axis.
I2(edge) = 1/3 (m*L^2)
learn more about moment of Inertia here:
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<h2>The frequency of driver is 700 Hz</h2>
Explanation:
The frequency of wave in a string is given by the relation
n = 
here n is the frequency
p is the number of antinodes and l is the length of string .
T is the tension in string and m is the mass per unit length
Thus 420 =
I
Now if there is 5 antinodes , the value of p = 5
Thus n =
II
Dividing II by I , we have
n/420 = 5/3
or n = 5/3 x 420 = 700 Hz
Using Newton's Second Law, F = ma, where F is the net force
So the net force is:
F = (6kg)(4m/s^2) = 24N
Since you are applying a horizontal force of 30N, we can find the force of friction by the difference of the net force and the applied force.
30N-24N = 6N
Answer:
the change in thermal energy of the projectile is 43.8 kJ
Explanation:
Given;
mass of the object, m = 5kg
initial velocity of the projectile, v₁ = 200 m/s
final velocity of the projectile, v₂ = 150 m/s
To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.
Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²
KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)
KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ
Therefore, the change in thermal energy of the projectile is 43.8 kJ
Answer:
Get turned Into Water.
Explanation:
Combustion of Hydrogen involves combining oxygen and hydrogen essentially so when oxygen and hydrogen combine water is produced, following chemical equation describes this process.
.
Resulting product is two molecules of water.