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If the total dissipated power is to be reduced by 10%, how much should the voltage be reduced to maintain the same leakage curre

harkovskaia [24]

<span>Since P = V x I, a 10% reduction of power would lead to a 10% reduction in the product of voltage and current. What is left is 90% of the original power: .9P = .9(V x I). If we assume that current must be the same, then we can regroup the terms on the right-hand side as follows: .9P = (.9V) x I In this case, voltage is also reduced by 10% (100% - 90% = 10%).</span>

6 0

3 years ago

A bus is moving with an acceleration of 4 m/s^2. If it was initially at rest,

shusha [124]

Probably b) 6 m/s
Not sure

6 0

2 years ago

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. A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00m above the ground. At t

dedylja [7]

Answer:

Explanation:

Height of building

H = 6m

Horizontal speed of first balloon

U1x = 2m/s

Second ballot is thrown straight downward at a speed of

U2y = 2m/s

Time each gallon hits the ground

Balloon 1.

Using equation of free fall

H = Uoy•t + ½gt²

Uox = 0 since the body does not have vertical component of velocity

6 = ½ × 9.8t²

6 = 4.9t²

t² = 6 / 4.9

t² = 1.224

t = √1.224

t = 1.11 seconds

For second balloon

H = Uoy•t + ½gt²

6 = 2t + ½ × 9.8t²

6 = 2t + 4.9t²

4.9t² + 2t —6 = 0

Using formula method to solve the quadratic equation

Check attachment

From the solution we see that,

t = 0.9211 and t = -1.329

We will discard the negative value of time since time can't be negative here

So the second balloon get to the ground after t ≈ 0.92 seconds

Conclusion

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

4 0

2 years ago

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Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i

spayn [35]

Answer:

$v = 15.56 m/s$

$v = 56 km/h$

Explanation:

When coefficient of friction is approximately zero then we have

$F_ncos\theta = mg$

$F_n sin\theta = \frac{mv^2}{R}$

$tan\theta = \frac{v^2}{Rg}$

here we know that

$v = 40 km/h = 11.11 m/s$

R = 30 m

$tan\theta = \frac{11.11^2}{30\times 9.81}$

$\theta = 22.75 degree$

now when friction coefficient is 0.30 then we have

$F_n cos\theta = mg + F_f sin\theta$

$F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}$

now we have

$v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}$

$v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}$

$v = 15.56 m/s$

$v = 56 km/h$

3 0

2 years ago

Which of the following provides evidence that earth is rotating? The sun rises and sets, The plane of a foucault pendulum appear

bagirrra123 [75]

The answer would be '<span>The plane of a Foucault pendulum appears to shift its orientation' because it is an experiment to demonstrate the rotation of the earth.

Hope this helps.</span>

4 0

2 years ago