Answer:
A) Earth and the other inner planets have higher average surface temperatures than the outer planets.
Explanation:
the earth and the other inner planets have higher average surface temperatures than the outer planets.
The reason for this response is due to the distance between the sun and the respective planet, the source of energy generation is the sun and the only way in which the temperature increase of each planet is guaranteed is by radiation, the further away a planet is from its star, its temperature will decrease. Although it is also important to highlight the atmospheric composition of the planet if this planet in its stratosphere has high density clouds that do not allow the entry of solar radiation, the temperature of the planet's surface will not increase, independent of the distance from the sun, but these are more complex cases where specialists in that area enter to perform a study in detail.
It depends on the direction in which the shell is launched. The time can be anything from 3.6 seconds to never.
Answer:
Magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Given
Contact Time = t = 0.05 seconds
Mass (of ball) = 0.80kg
Initial Velocity = u = 25m/s
Final Velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is given by;
F = ma
Where m = 0.8kg
a = Average Acceleration
a = (u + v)/t
a = (25 + 25)/0.05
a = 50/0.05
a = 1000m/s²
Average Force = Mass * Average Acceleration
Average Force = 0.8kg * 1000m/s²
Average Force = 800kgm/s²
Average Force = 800N
Hence, the magnitude of the average force exerted on the wall by the ball is 800N
The radius, r, of the child from the center of the wheel is
r = 1.3 m
The wheel makes one revolution in 4.2 s. Its angular velocity is
ω = (2π rad)/(4.2 s) = 1.496 rad/s
The linear speed of the child is the tangential velocity, given by
v = rω
= (1.3 m)*(1.496 rad/s)
= 1.945 m/s
Answer: 1.95 m/s (nearest hundredth)
Answer:
1) the capacitance of the capacitor increases. This is due to the induction of opposite charges on the two surfaces of the dielectric by the plate, this increased the charge in the field, from C =Q/v, it is seen that capacitance C will increase with increase in Q since v is constant.
2) the electric field intensity will also increase with increase in electric charges provided plate separation d remains constant.
