They were both odd and obsessive
Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.
Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,
Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules
Just ignore the negative value for the final result because work is a scalar quantity.
The answer to this question is 2400Hz
Answer:
The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.
Explanation:
Given that,
Velocity of ship = 2.00 m/s due south
Velocity of boat = 5.60 m/s due north
Angle = 19.0°
We need to calculate the component
The velocity of the ship in term x and y coordinate
The velocity of the boat in term x and y coordinate
For x component,
Put the value into the formula
For y component,
Put the value into the formula
We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat
For x component,
Put the value into the formula
For y component,
Put the value into the formula
Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.
Answer:
a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2
b) See the picture
Explanation:
We can use Gauss's law to find the electric field in all the regions:
EA = qen/e0 where qen is the enclosed charge
Remember that the electric field everywhere outside a sphere is:
E(r) = q/(4*pi*eo*r^2) = Kq/r^2
a)
- For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
- For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
- For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2
b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance