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hichkok12 [17]
3 years ago
13

If 1.8 1016 electrons enter a light bulb in 3 milliseconds, what is the magnitude of the electron current at that point in the c

ircuit?
Physics
1 answer:
Mice21 [21]3 years ago
7 0

Answer:

I = 0.96 A

Explanation:

No of electrons, n=1.8\times 10^{16}

Time, t = 3 ms = 3\times 10^{-3}\ s

We need to find the electric current. We know that electric charge per unit time is equal to the electric current.

I=\dfrac{q}{t}

q = ne (Quantization of electric charge)

I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A

So, the electric current is 0.96 A.

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A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it
asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0
3 years ago
Define fractional distillation
Serga [27]

Answer:

separation of a liquid mixture into fractions differing in boiling point (and hence chemical composition) by means of distillation, typically using a fractionating column.

7 0
2 years ago
Read 2 more answers
A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

8 0
3 years ago
Can you help me doing an essay about actual self​
Rudik [331]

Answer:

yes absolutely

Explanation:

why not

4 0
2 years ago
during a test crash an airbag inflates to stop a dummies forward motion. the dummies mass is 75 kg. if the net force on the dumm
den301095 [7]
Use Force=Mass x Acceleration (newtons second law states force is directly proportional to the acceleration) so you can say that the force is negative and solve for Acceleration.
5 0
3 years ago
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