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Arisa [49]
3 years ago
8

If you have a 100-watt light bulb, how much energy does it use each minute?

Physics
1 answer:
Mamont248 [21]3 years ago
3 0
A 100-watt light bulb uses 6,000 joules per minute.
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Question 8
FinnZ [79.3K]

Answer:

A hope this helps

Explanation:

8 0
3 years ago
Hurricanes are especially destructive because of the abrupt decrease in air pressure that they bring with the winds. Consider a
ivann1987 [24]

Answer:

2.96 × 10^4 N

Explanation:

1 atm = 101325 N/m², pressure inside the airtight room = 1.02 atm, pressure outside due to hurricane = 0.91 atm

net pressure directed outward = P inside - P  outside

net pressure = 1.02 - 0.91 = 0.11 atm

where 1 atm = 101325N/m²

0.11 atm = 0.11 × 101325 N/m² = 11145.75 N/m²

area of the square wall = l × l where l is the length of the wall in meters = 1.63 × 1.63 = 2.6569

net pressure = net force / area

make net force subject of the formula

net force = net pressure × area = 11145.75 × 2.6569 = 2.96 × 10 ^4 N

8 0
3 years ago
The electroscope is an apparatus used to detect electric charge. The electroscope consists of a plate, a support and a free to r
Marat540 [252]

Answer:

Positive

Explanation:

The leaves will diverge further: The positive charge on the leaves has increased further. This occurs when positive charge is produced on the leaves by the charged object. This is quite possible only when the object is positively charged.

4 0
3 years ago
A coil of area 0.320 m2 is rotating at 100 rev/s with the axis of rotation perpendicular to a 0.430 T magnetic field. If the coi
weqwewe [10]

Answer:

The maximum  emf generated in the coil is 60527.49 V

Explanation:

Given;

area of coil, A = 0.320 m²

angular frequency, f = 100 rev/s

magnetic field, B = 0.43 T

number of turns, N = 700 turns

The maximum emf generated in the coil is calculated as,

E = NBAω

where;

ω is the angular speed = 2πf

E = NBA(2πf)

Substitute in the given values and solve for E

E = 700 x 0.43 x 0.32 x 2π x 100

E = 60527.49 V

Therefore, the maximum  emf generated in the coil is 60527.49 V

8 0
3 years ago
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
Sati [7]

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

5 0
3 years ago
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