Answer:
280 N
Explanation:
Applying Newton's third second law of motion,
F = m(v-u)/t................... Equation 1
Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.
Note: Let the direction of the initial velocity of the ball be positive
Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s
Substitute into equation 1
F = 4(-4-3)/0.1
F = 4(-7)/0.1
F = -28/0.1
F = -280 N.
Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball
Answer:
When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.
(a) The ball's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :
0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )
<em>t</em> = 0 or (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 <em>g t</em>
<em>t</em> = (40 m/s) sin(40º) / <em>g</em>
<em>t</em> ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So
0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>
where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :
<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)
<em>y</em> ≈ 8.4 m
Answer:
Part a)
Part b)
Part c)
Explanation:
Part a)
As we know that charge density is the ratio of total charge and total volume
So here the volume of the charge ball is given as
now the charge density of the ball is given as
Part b)
Now the charge enclosed by the surface is given as
at radius of 5 cm
at radius of 10 cm
at radius of 20 cm
Part c)
As we know that electric field is given as
so we have electric field at r = 5 cm
electric field at r = 10 cm
electric field at r = 20 cm
Because they are different they all show different traits.
