Answer:
313.92w
Explanation:
Formula for power:
P=W/∆t = Fv
Givens:
m=20kg
∆y=4.0m
∆t=2.5s
a=9.81m/s²
In order to find power, we first need to solve for work.
W=Fd (force*displacement), f=mg
W=mg∆y
W=(20kg)(9.81m/s²)(4.0m)
W=784.8J
P=W/∆t
P=784.8J/2.5s
P=313.92 watts
Suppose car A is moving with a velocity Va, and car b with a velocity Vb,
According the principle of conservation of momentum:
Va x Ma + Vb x Mb = (Ma + Mb) V
V = (Va x Ma + Vb x Mb)/(Ma +Mb)
V = speed of cars after coupling
V = (Va x 20 mg + Vb x 15 mg)/(20 mg + 15 mg)
Put in the values of Va and Vb, and get the V
Answer:
Both the astronauts and photographer have the same displacement
Explanation:
Displacement is the minimum distance between two point. The initial point of both the astronauts and the photographer was Florida and the final point was California. So, the minimum distance for both of the astronauts and the photographer would be the distance between Florida and California would be the same.
Hence, both the astronauts and photographer will have the same displacement.
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:

- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:

- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.