Answer:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
Explanation:
The graph shown in the figure is a velocity-time graph, which means that:
- On the x-axis, the time is plotted
- On the y-axis, the velocity is plotted
Therefore, this means that the object is not moving when the line is horizontal (because at that moment, the velocity is constant, so the object is not moving). This occurs in the following intervals:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
From the graph, it would be possible to infer additional information. In particular:
- The area under the graph represents the total distance covered by the object
- The slope of the graph represents the acceleration of the object
Downstream, the current is helping the swimmer go faster relative to the land
Using Ampere's Law, the magnetic field produced inside this solenoid is given by
B = uo N I / h
where uo is the vacuum permeability, N is the number of turns in the solenoid and h is the length of the solenoid. Earth's magnetic field is around 50 microteslas in North America thus the current needed in the solenoid is
I = B h / (uo N) = (50 E-6 ) (4) / ((4 pi E-7)(6000) ) = 0.026 A
I = 26 mA
So you need a current of around 26 mA.
Answer:
20.1 m
Explanation:
First, find the time it takes for the cannonball to travel the horizontal distance of 50.0 m.
Given (in the x direction):
Δx = 50.0 m
v₀ = 68 cos 25° m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
(50.0 m) = (68 cos 25° m/s) t + ½ (0 m/s²) t²
t = 0.811 s
Now find the vertical displacement after that time.
Given (in the y direction):
v₀ = 68 sin 25° m/s
a = -9.8 m/s²
t = 0.811 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (68 sin 25°) (0.811 s) + ½ (-9.8 m/s²) (0.811 s)²
Δy = 20.1 m