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never [62]
3 years ago
8

A light, a heater, and a hair dryer were all running in the bathroom at the same time. Suddenly, they all turn off. After all, t

he light usually stays on just fine whenever you use the bathroom. Why would it go off now?
Please answer this question assuming the circuit breaker or fuse for the bathroom has a 40-ampere capacity. Use the terms “circuit,” and “overload.” Then think of at least two solutions to prevent this from happening again.
Physics
1 answer:
Ratling [72]3 years ago
3 0
Well depending on what current the heater pulls im going to assume about 13, and 13A for the hair dryer, thats 26A on the 40A circuit.
I dont see how a lightbulb could overload the circuit. 

Anyway, assuming the circuit is overloaded by some really big heater- the circuit would trip, the fuse would go and remain off. Most houses are fitted with seperate circuits for lights and sockets, so the light may remain on depending on the breaker board. - the reason for them all being able to run with the sudden overload may be due to a surge.

One solution to this is not to put such a large heater on the circuit with other appliances. 

Another may be to dry your hair in the dark
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An ordinary ruler is used to measure the area and its error of a rectangle. It is found that their sides are 5.0 cm long and 2.0
Elina [12.6K]

Answer:

You need to know the accuracy to which you can read the ruler:

Suppose that you can read the read the ruler to the nearest milimeter

A = L * W     your calculated area of the rectangle

A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA

Or ΔA =  L ΔW + W ΔL

Where we have subtracted A = L * W and the term ΔL * ΔA is very small

So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2

Then you report A = 10 cm^2 +- .7 cm^2    including the - sign for completeness

3 0
3 years ago
A red cart has a mass of 4 kg and a velocity of 5 m/s. There is a 2-kg blue cart that is parked and not moving, thus its velocit
IRISSAK [1]

Answer:dam hold up

Explanation:

5 0
3 years ago
What is the difference in KE between a 52.5 kg person running 3.50 m/s and a 0.0200 kg bullet flying 450 m/s?
rusak2 [61]

Answer:

Ek = 1705.28 [J]

Explanation:

In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.

E_{k}=\frac{1}{2} *m*v^{2}

where:

m = mass [kg]

v = velocity [m/s]

Ek = kinetic energy [J] (Units of Joules)

<u>For the person running</u>

<u />E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]<u />

<u />

<u>For the bullet</u>

<u />E_{k} =\frac{1}{2} *m*v^{2}<u />

<u />E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]<u />

<u />

The difference in Kinetic energy is equal to:

Ek = 2025 - 319.72

Ek = 1705.28 [J]

8 0
3 years ago
/
Mumz [18]
No se ha da han dicho nada más de lo dicho y han ido de vuelta y han dicho nada más de que se pueda hacer el favor del niño
4 0
3 years ago
Read 2 more answers
14. A rocket is shot up into the air and then comes back down and hits the ground 9.2 second later.
sineoko [7]

Answer:

105.8 m

46 m/s

Explanation:

From the time the rocket is launched to the time it reaches its maximum height:

v = 0 m/s

a = -10 m/s²

t = 9.2 s / 2 = 4.6 s

Find: Δy and v₀

Δy = vt − ½ at²

Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

v = at + v₀

0 m/s = (-10 m/s²) (4.6 s) + v₀

v₀ = 46 m/s

3 0
3 years ago
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