Answer:
The induced current is 26.7 mA
Explanation:
Given;
area of the loop, A = 0.078 m²
initial magnetic field, B₁ = 3.8 T
change in the magnetic field strength, dB/dt = 0.24 T/s
The induced emf is calculated as;
The resistance of the loop = 0.7 Ω
The induced current is calculated as;
Answer:
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Answer:
t = 4.08 s
R = 40.8 m
Explanation:
The question is asking us to solve for the time of flight and the range of the rock.
Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:
We have these known variables:
- (v_0)_y = 0 m/s
- a_y = -9.8 m/s²
- Δx_y = -20 m
And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.
- -20 = 0t + 1/2(-9.8)t²
- -20 = 1/2(-9.8)t²
- -20 = -4.9t²
- t = 4.08 sec
The time it takes for the rock to reach the ground is 4.08 seconds.
Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.
List out known variables:
- v_0 = 10 m/s
- t = 4.08 s
- a_x = 0 m/s
We are trying to solve for:
By using the same equation, we can plug these known values into it and solve for Δx.
- Δx = 10 * 4.08 + 1/2(0)(4.08)²
- Δx = 10 * 4.08
- Δx = 40.8 m
The rock lands 40.8 m from the base of the cliff.
To look for displacement, just draw a vector from your beginning stage to your last position and settle for the length of this line. So we begin by drawing a line to the north which is 30 ft, since it is north, the line is going up, then it move 5 ft to the south, so put a line going down, so we are in 25 ft, North so that would be the answer.
