Answer:
a) about 20.4 meters high
b) about 4.08 seconds
Explanation:
Part a)
To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.
In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:
Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:
Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)
To solve for "t" in this quadratic equation, we can factor it out as shown:
Therefore there are two possible solutions when each of the two factors equals zero:
1) t= 0 (which is not representative of our case) , and
2) the expression in parenthesis is zero:
Answer:
The electric flux remains unchanged
Explanation:
From Gauss law the Electric flux is directly proportional to the number of electric field lines passing through a surface. The number of field lines passing through a surface become if the radius is doubled becomes 1/4th that is when radius of the Gaussian surface is doubled, but at the same time, the surface area has increased 4 times , so the electric flux remains unchanged
Lets use the expression
F = k*x
Where k is the spring constant [N/m]
And x the distance from the equilibrium position.
This force is equal to the force due to the acceleration of gravity
F = k*x = m*g
F = 525[N/m]* 0.30 [m] = 157.5 [N]
157.5 [N] = m *10[m/s**2] ...................> m = 15.75 kg
convergent and counterclockwise
hope it helps :)