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never [62]
2 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]2 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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8 0
2 years ago
Describe a situation that includes no less than four charges of any magnitude, but they combine so that another location, p, has
Degger [83]

Answer:

Four charges of equal magnitude sitting at the vertices of a square

Explanation:

We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.

Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.

We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.

6 0
2 years ago
It is recommended that adults should get at least 150 minutes of moderate-intensity physical activity each week. they should als
shepuryov [24]
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4 0
3 years ago
Derive a relation between Universal Gravitational Constant (G) and Acceleration due to gravity(g)?
snow_tiger [21]
Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2                             ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2




4 0
3 years ago
Read 2 more answers
Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of 2.0A. The dis
Veronika [31]

Answer:

Explanation:

Two straight wires

Have current in opposite direction

i1=i2=i=2Amps

Distance between two wires

r=5mm=0.005m

Length of one wire is ∞

Length of second wire is 0.3m

Force between the wire,

The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by

F/l = μoi1i2/2πr

F/l=μoi²/2πr

μo=4π×10^-7 H/m

The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

F/l = μoi1i2/2πr

F/0.3=4π×10^-7×2²/2π•0.005

F/0.3=1.6×10^-4

Cross multiply

F=1.6×10^-4×0.3

F=4.8×10^-5N

3 0
3 years ago
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