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4 years ago

With the switch closed, how does the voltage across the 20-ω resistor compare to the voltage across the 10-ω resistor?

1 answer:

5 0

In electricity, the most famous and basic equation is the Ohm's Law which relates the parameters voltage, current and resistance. One form of this law as written in equation is V = IR, where V is the voltage in volts, I is the current in amperes and R is the resistance in ohms. These parameters depends in the arrangements, whether it's series or parallel.

In a series connection, the voltage is greater across a high-resistance resistor. Therefore, the voltage is much greater for the 20-ohm resistor. However,if it is a parallel circuit, the voltage is just the same for both resistors.

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Electromagnetic waves are sometimes demonstrated by rippling waves in water. However, electromagnetic waves are different from w

algol [13]

The answer is D: all electromagnetic waves can travel in a vacuum

3 0

3 years ago

Read 2 more answers

A ball is thrown horizontally from the top of a building 54 m high. The ball strikes the ground at a point 35 m horizontally awa

Ivanshal [37]

Answer:

V=34.2 m/s

Explanation:

Given that

Height , h= 54 m

Horizontal distance , x = 35 m

Given that , the ball is thrown horizontally , therefore the initial vertical velocity will be zero.

In vertical direction :

We know that

$V^2_y=U^2_y+2 g h$

Now by putting the values in the above equation we got

$V^2_y=U^2_y+2 g h$

$V^2_y=0^2+2\times 9.81 \times 54$

Assume $g= 9.81 m/s^2$

Thus

$V^2_y=1059.48$

$V_y=\sqrt{1059.48}\ m/s$

$V_y=32.54 m/s$

We also know that

$V_y=U_y+ g\times t$

$32.54=9.81\times t$

$t=\dfrac{32.54}{9.81}=3.31\ s$

In horizontal direction :

$x=U_x\times t$

$U_x=\dfrac{35}{3.31}=10.54\ m/s$

Thus the resultant velocity

$V=\sqrt{V^2_y+U^2_x}$

$V=\sqrt{32.54^2+10.54^2}=34.2\ m/s$

V=34.2 m/s

Therefore the velocity will be 34.2 m/s.

5 0

4 years ago

A car drives over a hilltop that has a radius of curvature 120 m at the top of the hill. at what speed would the car be travelin

FinnZ [79.3K]

The speed of a car travelling over a hill that has a radius of curvature should not exceed a certain speed other it will topple. This speed is related to the radius of curvature and the gravitational acceleration as shown below:

V^2 = Rg, where V = maximum speed, R = Radius of curvature, g = gravitational acceleration.

Substituting;
V = Sqrt (Rg) = Sqrt (120*9.81) = 34.31 m/s

6 0

3 years ago

Physics question 28 plz help me

Alexus [3.1K]

Answer:

a. I = 30 A

b. E = 1080000 J = 1080 KJ

c. ΔT = 12.86°C

d. Cost = $ 4.32

Explanation:

The current in the coil is given by Ohm's Law:

$V = IR\\I = \frac{V}{R}$