Ask question

Ask question

All categories

4 years ago

10

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

nt? note: power is defi ned as the product of voltage and current

1 answer:

Mandarinka [93]4 years ago

5 0

Let
I₁ = initial current, A
V₁ = initial voltage, V

The power is
P₁ = I₁*V₁ W

If P is reduced by 10% and the current is kept the same, then
P₂ = 0.9P₁
I₂ = I₁

The new voltage, V₂, is given by
P₂ = I₂V₂
0.9P₁ = I₁V₂
0.9I₁V₁ = I₁V₂
V₂ = 0.9V₁

Answer: The voltage should be reduced by 10%

You might be interested in

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

6 0

4 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

3 years ago

Carlos uses a rope to pull his car 30 m to a parking lot because it ran out of gas. If Carlos exerts 2,000 N of force to pull th

Fiesta28 [93]

<span>5.8 × 104 J
i already checked it on edge

</span>

6 0

3 years ago

Read 2 more answers

Because so little can be found of the first rocks to form on

fgiga [73]

Answer:

B is the answer

Explanation:

0-/-<

7 0

3 years ago

The blackbody radation emmitted from a furnace peaks at a wavelength of 1.9 x 10^-6 m (0.0000019 m). what is the temperature ins

krek1111 [17]

Answer:

Temperature, T = 1542.10 K

Explanation:

It is given that,

The black body radiation emitted from a furnace peaks at a wavelength of, $\lambda=1.9\times 10^{-6}\ m$

We need to find the temperature inside the furnace. The relationship between the temperature and the wavelength is given by Wein's law i.e.

$\lambda\propto \dfrac{1}{T}$

or

$\lambda=\dfrac{b}{T}$

b = Wein's displacement constant

$\lambda=\dfrac{2.93\times 10^{-3}}{T}$

$T=\dfrac{2.93\times 10^{-3}}{\lambda}$

$T=\dfrac{2.93\times 10^{-3}}{1.9\times 10^{-6}\ m}$

T = 1542.10 K

So, the temperature inside the furnace is 1542.10 K. Hence, this is the required solution.

3 0

3 years ago