Answer:
Explanation:
It is given that,
The radius of the solid sphere, R = 50 cm = 0.5 m
Charge on the sphere, ![q=40\ \mu C=40\times 10^{-6}\ C](https://tex.z-dn.net/?f=q%3D40%5C%20%5Cmu%20C%3D40%5Ctimes%2010%5E%7B-6%7D%5C%20C)
We need to find the magnitude of the electric field r = 10.0 cm away from the center of the sphere. The electric field at point r away form the center of the sphere is given by :
![E=\dfrac{kqr}{R^3}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkqr%7D%7BR%5E3%7D)
![E=\dfrac{9\times 10^9\times 40\times 10^{-6}\times 0.1}{0.5^3}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%2040%5Ctimes%2010%5E%7B-6%7D%5Ctimes%200.1%7D%7B0.5%5E3%7D)
![E=2.88\times 10^5\ N/C](https://tex.z-dn.net/?f=E%3D2.88%5Ctimes%2010%5E5%5C%20N%2FC)
So, the electric field 10.0 cm away from the center of the sphere is
. Hence, this is the required solution.
I believe the answer is C
Answer:
1. 10.0 J
2. 0.742 m/s
3. 0.494 m/s
4. 0.0249 m
Explanation:
(1/4) The elastic energy in a spring is:
EE = ½ k x²
EE = ½ (502 N/m) (0.2 m)²
EE ≈ 10.0 J
(2/4) The energy in the spring is converted to kinetic energy in the block and work by friction.
EE = KE + W
EE = ½ m v² + Fd
10.0 J = ½ (8 kg) v² + (8 kg × 9.8 m/s² × 0.5) (0.2 m)
v ≈ 0.742 m/s
(3/4) Momentum is conserved.
Momentum before = momentum after
(8 kg) (0.742 m/s) = (8 kg + 4 kg) v
v ≈ 0.494 m/s
(4/4) The kinetic energy of the blocks is converted to work by friction.
KE = W
½ m v² = Fd
½ (8 kg + 4 kg) (0.494 m/s)² = ((8 kg + 4 kg) × 9.8 m/s² × 0.5) d
d ≈ 0.0249 m
Answer:
Explanation:
A positive v time graph is a velocity graph. If the position/time is negative, then the slope of the line is negative. If the line is indicative of the velocity, which it is, that means that the velocity is negative (decreasing). The first choice is the one you want.