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3 years ago

How is energy and work related.

Physics

2 answers:

mel-nik [20]3 years ago

7 0

<h3>The capacity of doing work.</h3>

every form of energy is work. Work done on the body is stored in the form of energy. To do more work, more energy is required.

Explanation:

Anna007 [38]3 years ago

5 0

Answer:

Varies

Explanation:

They both relate to the process of doing something.

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Shanai drove 40 miles to the west, then turned around and drove 70 miles to the east. Draw vectors that show each segment of her

Dmitry [639]

Well she drove 30 more miles to the east than the west but I don’t understand what u are asking

4 0

3 years ago

The half-wave rectifier circuit of vs(t) = 170 sin(377t) V and a load resistance R = 15Ω. Determine: a. The average load current

irina1246 [14]

Answer:

a average load current = 11.33 A

b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

Explanation:

Vs (t) = 170 sin(377t) v

Vm =170v

Vrms = 170/√2 =120.23 v

Im = Vm/R = 170/15 = 11.33 A

Irms = Im/ √2 = 11.33/√2 =8.02A

Resistors are electronic components that consume energy

the power in a resistor is given by P =IVcosθ ; in a resistor cosθ =1

P =IV

The electrical power consumed by a resistance, (R) is called the true or real power

and is obtained by multiplying the rms voltage with the rms current.

P= Vrms × Irms

120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

For the purely resistive circuit, the power factor is 1 , because the reactive power is equal to zero (0).

6 0

4 years ago

Read 2 more answers

On a cloudless day, the sunlight that reaches the surface of the earth has an intensity of about W/m². What is the electromagnet

Otrada [13]

Answer:

1.907 x 10⁻⁵ J.

Explanation:

Given,

Volume of space, V = 5.20 m³

Assuming the intensity of sunlight(S) be equal to 1.1 x 10³ W/m².

Electromagnetic energy = ?

$E = \mu V$

$E = (\dfrac{S}{c})\times V$

where c is the speed of light.

$E = (\dfrac{1.1\times 10^3}{3\times 10^8})\times 5.20$

$E = 1.907\times 10^{-5}\ J$

Hence, Electromagnetic energy is equal to 1.907 x 10⁻⁵ J.

4 0

4 years ago

It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you

Debora [2.8K]

You'd get an extra 40/60 of the energy, or 2/3. Multiply 5/3 by the required energy to get the actual consumption.

7 0

3 years ago

A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

Gnom [1K]

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

8 0

3 years ago

How is energy and work related.​

How is energy and work related.