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Svetllana [295]
3 years ago
14

During energy conversions, some energy is always lost as _____. light , heat , electricity or chemical energy?

Physics
2 answers:
masya89 [10]3 years ago
3 0
Heat because in a machine heat must come out to follow the law of conservation of energy
Keith_Richards [23]3 years ago
3 0

Answer: It is heat because the two objects are rubbing together

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A certain wave has a compressions and rarefactions.How should this wave be classified?
AlexFokin [52]
A) as a longitudinal wave
7 0
4 years ago
An inductor is connected to an AC power supply having a maximum output voltage of 3.00 V at a frequency of 280 Hz. What inductan
Sergio [31]

Answer:

L = 0.48 H

Explanation:

let L be the inductance, Irms be the rms current, Vrms be the rms voltage and Vmax  be the maximum voltage and XL be the be the reactance of the inductor.

Vrms = Vmax/(√2)

         = (3.00)/(√2)

         = 2.121 V

then:

XL = Vrms/I  

     = (2.121)/(2.50×10^-3)

     = 848.528 V/A

that is L = XL/(2×π×f)

              = (848.528)/(2×π×(280))

              = 0.482 H

Therefore, the inductance needed to kepp the rms current less than 2.50mA is 0.482 H.

6 0
3 years ago
if a person can jump 2m in earth surface how high can he jump in the moon (g of moon = 1.66m/s, g of earth = 9.8 m/s) [hint: use
julia-pushkina [17]

Answer:

h_{moon} = 11.8\ m

Explanation:

Since the work done is same everywhere in the universe. Hence, the work done in jumping will be same for the person on moon and earth:

W_{moon} = W_{earth}\\\\P.E_{moon} = P.E_{earth}\\\\mg_{moon}h_{moon} = mg_{earth}h_{earth}\\\\g_{moon}h_{moon} = g_{earth}h_{earth}\\\\(1.66\ m/s^2)h_{moon} = (9.8\ m/s^2)(2\ m)\\\\h_{moon} = \frac{(9.8\ m/s^2)(2\ m)}{1.66\ m/s^2}\\\\h_{moon} =  11.8\ m

5 0
3 years ago
A 2400-kg satellite is in a circular orbitaround a planet. The
Fofino [41]

Answer

given,

mass of satellite = 2400 Kg

speed of the satellite =  6.67 x 10³ m/s

acceleration of satellite = ?

gravitational force of the satellite will be equal to the centripetal force

F = \dfrac{mv^2}{r}

F = \dfrac{2400\times (6.67\times 10^3)^2}{r}

Assuming the radius of circular orbit = 8.92 x 10⁶ m

now,

F = \dfrac{2400\times (6.67\times 10^3)^2}{8.92\times 10^6}

F = 11970.11 N

acceleration,

a = \dfrac{F}{m}

a = \dfrac{11970.11}{2400}

  a = 4.98 m/s²

4 0
3 years ago
A commuter backs her car out of her garage with an acceleration of . (a) How long does it take her to reach a speed of 2.00 m/s
aliya0001 [1]

Question: A commuter backs her car out of her garage with an acceleration of 1.4 m/s² (a) How long does it take her to reach a speed of 2.00 m/s

Answer:

1.43 s

Explanation:

Applying,

a = (v-u)/t........... Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t = time

make t the subject of the equation

t = (v-u)/a........... Equation 2

From the question,

Given: v = 2 m/s, u = 0m/s (from rest), a = 1.4 m/s²

Substitute into equation 2

t = (2-0)/1.4

t = 1.43 s

3 0
3 years ago
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