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3 years ago

A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is:

Physics

1 answer:

kondaur [170]3 years ago

5 0

Answer:

Total charge = 1800C

Explanation:

Q= IT

I = current
t = time
Q = charge

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A loop of wire is placed in a magnetic field such that it has a flux, LaTeX: \phi ϕ, through it. The loop is compressed so that

Juliette [100K]

Answer:

B'= 3.333 B

Explanation:

Lets take

Initial area = A

Magnetic field = B

The area after compression

A'=0.3 A

Magnetic field = B'

We know that flux ,Ф

Ф = B A

Given that flux is constant so

B A = B' A'

B A=B' x 0.3 A

B'= 3.333 B

It means that magnetic field will increase.

6 0

2 years ago

Read 2 more answers

The vertical force f acts downward at A on the two membered frames. Determine the magnitude of the two components of F directed

galben [10]

Answer:

The magnitude will be "353.5 N". A further solution is given below.

Explanation:

The given values is:

F = 500 N

According to the question,

In ΔABC,

⇒ $\angle BCA = (90-30)$

⇒ $=60^{\circ}$

then,

⇒ $\angle BAC=(180-45-60)$

⇒ $=75^{\circ}$

Now,

The corresponding angle will be:

⇒ $\angle FAC=60^{\circ}$

⇒ $\angle FAB=70+60$

⇒ $=135^{\circ}$

Aspect of F across the AC arm will be:

= $F\times cos(60)$

On putting the values of F, we get

= $500\times (.5)$

= $200 \ Newton$

Component F along the AC (in magnitude) will be:

= $F\times cos(135)$

= $500\times (-.707)$

= $-353.5 \ N \$

4 0

2 years ago

A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1 480 kg. It has strayed too

maks197457 [2]

Answer:

2352645198509.9604 m/s²

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of black hole = $90\times 1.989\times 10^{30}\ kg$

$R_{f}$ = 10000+100 m

$R_{sb}$ = Distance between the nose and the center of the black hole = 10000 m

The difference in the gravitational field in this system is given by

$\Delta F=\dfrac{GMm}{R_{f}^2}-\dfrac{GMm}{R_{sb}^2}\\\Rightarrow \Delta g=GM\left(\frac{1}{R_f^2}-\frac{1}{R_{sb}^2}\right)\\\Rightarrow g=6.67\times 10^{-11}\times 90\times 1.989\times 10^{30}\left(\frac{1}{(10000+100)^2}-\frac{1}{10000^2}\right)\\\Rightarrow \Delta g=-2352645198509.9604\ m/s^2$