Answer:
Explanation:
The problem is related to rotational motion . So we shall find out rotational kinetic energy .
K E = 1/2 x I ω²
ω is the final angular velocity
Moment of inertial of the disk
I ₁ = 1/2 m r²
= .5 x 165 x 2.93²
= 708.25 kgm²
Moment of inertial of the person
I₂ = mr²
= 62.5 x 2.93²
= 536.55 kgm²
ω₂ = v / R
= 3.11 / 2.93 rad /s
At the time of jumping , law of conservation of angular momentum will apply
I₁ ω₁ + I₂ω₂ = (I₁ + I₂)ω
708.25 x0.691 + 536.55 x ( 3.11 / 2.93 ) = ( 708.25 + 536.55 ) ω
ω = 0 .85 rad/ s
K E = 1/2 x I ω²
= .5 x ( 708.25 + 536.55 ) ( .85 )²
449.68 J
To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as
Here,
k = Coulomb's constant
q = Charge of proton and electron
r = Distance
Replacing we have that,
The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.
The acceleration of the electron is given as
The acceleration of the proton is given as,
Answer:
7kgm/s
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Let P1A and P1B be the initial momentum of the bodies A and B respectively
Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.
Based on the law:
P1A+P2A = P1B + P2B
Given P1A = 5kgm/s
P2A = 0kgm/s(ball B at rest before collision)
P2A = -2.0kgm/s (negative because it moves in the negative x direction)
P2B = ?
Substituting the values in the equation gives;
5+0 = -2+P2B
5+2 = P2B
P2B = 7kgm/s
Answer:A i think or D but its not c or b
Explanation:
