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3 years ago

What is the S.I. unit of drift velocity and electron mobility?

1 answer:

7 0

Drift velocity is equal to displacement of the moving object per unit time. The SI unit for displacement is meters while that of time is second. Hence the derived SI unit of velocity is meter per second. This also applies to electron mobility which relates to the displacement per unit time of a moving electron

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A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective val

Lostsunrise [7]

Answer:

$1.25\ \mu\text{g/m}^3$

Explanation:

v = Velocity of the breeze = 4 m/s

w = Width of the valley = 5000 m

h = Height of the valley = 1000 m

Volumetric flow rate is given by

$\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}$

$\dot{m}$ = Mass flow rate of pollutant = 25 g/s = $25\times 10^6\ \mu\text{g/s}$

Concentration is given by

$C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3$

The steady state concentration of pollutants in the valley, is $1.25\ \mu\text{g/m}^3$ .

6 0

2 years ago

The moon has a weaker gravitational force than earth. sofia weighs 50 lbs. on earth. how much will she weigh on the moon?

kupik [55]

If Sofia weighs 50 lbs on Earth, then she would weigh 8.3 lbs, because you are 6 times lighter on the moon than you are on Earth!

3 0

3 years ago

List two industrial uses for nonmetallic mineral resources

Anni [7]

Answer:

To make useful chemicals and abrasives

Explanation:

4 0

2 years ago

Please help ASAP! Thank you :)

puteri [66]

Answer:

magnitude of gravitational force between the Earth and the Sun at B is greater than that at A

Explanation:

Formula of gravitational force:

F = GMm/r^2

(r is the distance between 2 objects)

We see that r(B) < r(A) since at B, the Earth is closer to the Sun than at A

According to the Formula, the smaller r is, the greater F is

So, F(B) > F(A)

8 0

3 years ago

1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it

spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

PE = (2.5)(98)(14) and

PE = 340 J

b. $KE=\frac{1}{2}mv^2$ so

$KE=\frac{1}{2}(2.5)(14)^2$ and

KE = 250 J

c. TE = KE + PE so

TE = 340 + 250 and

TE = 590 J

d. PE at 8.7 m:

PE = (2.5)(9.8)(8.7) and

PE = 210 J

e. The KE at the same height:

TE = KE + PE and

590 = KE + 210 so

KE = 380 J

f. The velocity at that height:

$380=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(380)}{2.5} }$ so

v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

590 = KE + PE and

PE = (2.5)(9.8)(11.6) so

PE = 280 then

590 = KE + 280 so

KE = 310 then

$310=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(310)}{2.5} }$ so

v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

$v=v_0+at$ and

26 = 0 + 9.8t and

26 = 9.8t so the time at 26 m/s is

t = 2.7 seconds. Now we use that in the equation for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in the time the object was at 26 m/s:

Δx = 0t + $\frac{1}{2}(-9.8)2.7)^2$ so

Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0

2 years ago