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QveST [7]
3 years ago
13

Magnetic lines of force are unaffected by paper, glass, or plastic. True False

Physics
1 answer:
myrzilka [38]3 years ago
3 0

I believe the answer is true.

I hope this helps :)

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A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o
Margarita [4]

Answer:

V_{average} = \frac{1}{2}  V_o  ,     V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

            V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

             V_{average} = \frac{1}{2}  V_o

we evaluate  V₀ = 4 V

             V_{average} = 4 / 2)

             V_{average} = 2 V

6 0
3 years ago
Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required:  vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE     (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0
3 years ago
Does mars has a bulge near its equator ?
Keith_Richards [23]
Yessir it sure does
7 0
3 years ago
IMPORTANTT!!! SOLVEE!!! and EXPLAIN!!!
Pepsi [2]
My answer -

I believe that the answer is (A).

P.S

Happy to help you have an AWESOME!! day
5 0
3 years ago
Calculate the speed of the car at each checkpoint by
lana66690 [7]

Answer: The speed at the first quarter checkpoint is 0.74 m/s. The speed at the second quarter checkpoint is 1.40 m/s. The speed at the third quarter checkpoint is 1.61 m/s. The speed at the finish line is 1.89 m/s.

Explanation: I did the assignment and got it correct :)

3 0
3 years ago
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