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Orlov [11]
2 years ago
10

Calculate the energy used by a radio of power 30W in 1 minute.

Physics
2 answers:
Paul [167]2 years ago
7 0

Answer:

1800 joule (1.8kJ)

Explanation:

Formula:

Power = Work / Time

30 = Work / 60

Work = 1800 joule

Work = 1.8kJ

Energy = Work done = 1.8kJ

Hence, Energy used is 1.8kJ.

Alekssandra [29.7K]2 years ago
5 0

Answer:

<u>1.8kJ</u>

Explanation:

Formula :

<u>Energy used = Power x time</u>

<u />

===============================================================

Given :

⇒ Power = 30 W

⇒ Time = 1 minute = 60 seconds

=============================================================

Solving :

⇒ Energy used = 30 W × 60 s

⇒ Energy used = 1,800 J

⇒ Energy used = <u>1.8kJ</u>

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Answer:

Comparison Microscope

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It is primarily used in criminology for ballistics which makes it ideal to find out if bullets, shells, or cartridge cases were fired from a specific weapon.

7 0
3 years ago
Use the H-R diagram to answer the following questions.
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Explanation:

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A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of 2 Newton’s. What is the acce
Zina [86]

PHYSICS

Mass = 10 kg

Force = 2 N

Acceleration = ____?

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f \:  = m \times a

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7 0
3 years ago
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

4 0
3 years ago
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