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3 years ago

What is the independent variable of the graph above?

Physics

1 answer:

Makovka662 [10]3 years ago

6 0

Answer:

Height of 15 kg Object is the correct answer.

Explanation:

Because the IV (Independent Variable) is always on the x-axis.

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A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of

Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

$E = \frac{mg}{q}$

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

$q = \frac{mg}{E}$

Replacing,

$q = \frac{(3.37*10^{-9})(9.8)}{11000}$

$q = 3.002*10^{-12}C$

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

$q = ne$

Here,

n = Number of electrons

e = Charge of each electron

$n = \frac{q}{e}$

Replacing,

$n = \frac{3.002*10^{-12}}{1.6*10^{-19}}$

$n = 2.44*10^7$

Therefore the number of electrons that reside on the drop is $2.44*10^7$

5 0

3 years ago

An aquifer pump test was conducted in a confined aquifer where the initial piezometric surface was at elevation 12.45 m, and wel

murzikaleks [220]

Answer:

T = 0.0088 m²/s

Explanation:

given,

initial piezometric elevation = 12.5 m

thickness of aquifer = 14 m

discharge = 28.24 L/s = 0.02824 m³/s

we know

$k = \dfrac{qln(\dfrac{R_2}{R_1})}{2\pi D(H_2-H_1)}$

$k = \dfrac{0.02824 \times ln(\dfrac{75}{25})}{2\pi \times 14 (11.45-10.89)}$

k = 0.629 mm/sec

Transmissibilty

T = k × H

T = 0.629 × 14 × 10⁻³

T = 0.0088 m²/s

4 0

3 years ago

A little girl riding a train rolls a ball toward the back of the train at 1.25 m/s NE. The train is traveling at a velocity of 1

Lelechka [254]

Answer:

Answer: 2.70m/s NE

Explanation:

Just did it.

4 0

3 years ago

How could two waves on a rope interfere so the rope does not move at all?

coldgirl [10]

Answer:

If a crest formed by one wave interferes with a trough formed by the other wave then the rope will not move at all.

Explanation:

Assume a straight rope tied to both ends is at rest. When a wave is created at one end of the rope, it travels to the other end of the rope through formation of alternative crest and trough. Due to these crest and trough the rope shifts up and down.

But when there are two waves travelling through the rope and both have opposite direction (directed towards one another) in such a way that crest formed by one wave is interfering with the trough formed by the other wave then due to this interference the waves will cancel the effects of each other on the rope and rope will be stable.

5 0

3 years ago

A boy standing on a bridge above a river throws stone A vertically upward with an initial velocity =15m/s.2 seconds later he dro

Anastaziya [24]

I'm not sure if this is correct but it's what I'll do

This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.

Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )

Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12

Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2

4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t

Time for Stone B is 4s
Time for Stone A is 6s

7 0

3 years ago