B. At the equator
Explanation:
The energy coming from the Sun hits the Earth's surface at different angles, depending on the latitude of the place. The more perpendicular the ray of lights hit the surface, the more the energy transmitted to the Earth's surface, the warmer the location.
The angle at which the ray of lights hit the Earth is related to the latitude: in particular, the ray of lights arrive perpendicular at the equator (
), they arrive at larger angle in the United States (which is located at intermediate latitudes) and they arrive at the largest angles at the poles. For this reason, the sun's most energy is concentrated at the equator.
To solve this problem it is necessary to apply the trigonometric ratios of the given velocity components.
If we make a graph of the velocity vectors in their respective velocities according to the given description we will have something similar to the attached graph.
The angle could be obtained from the components of the opposite leg and the adjacent leg so that
The opposite leg value (y) is 40cm / s and the adjacent leg (x) is 30cm / s
Therefore the final direction that does the first ball is 36.87°
On the dog's return trip (between <em>t</em> = 10 and <em>t</em> = 12.5 seconds), the slope of the position function is steeper than during the first 5 seconds, which means the dog ran home faster. The only option that captures this is D.
You can check to make sure that the dog indeed runs twice as fast on the return trip. The slope of the position function during the first 5 seconds is
(change in position) / (change in time) = (5 - 0) / (5 - 0) = 5/5 = 1
while during the return trip, it is
(0 - 5) / (12.5 - 10) = -5/2.5 = -2
Ignoring the sign (which only indicates the direction in which the dog was running), we see that the dog's speed on the return trip was indeed twice as high as during the first 5 s.
Answer:
b . Move the conductor near a magnet
Answer:
Δt=0.85 seconds
Explanation:
In this chase the speed does not change as the mass change.So we can use the follow equation to find the required time
Δt=Δv/gμ
To stop the final speed will be zero therefore the change in speed will be
Δv= vf-vi
Δv=0-5 m/s
Δv= -5 m/s
Now we plug our values for Δv,g and μ to find time
Δt=Δv/gμ
Δt=(-5m/s) ÷(9.8m/s² × 0.6)
Δt=0.85 seconds
