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Butoxors [25]
2 years ago
11

Burning coal, which is how many power plants generate electricity, releases a number of harmful byproducts. Particulate pollutio

n (i.e., soot or smoke) is the most visually obvious. Modern coal-burning power plants make use of electrostatic precipitators (ESPs) to remove most of the particulate pollution. As the hot, impure air heads out of the plant, it passes through an area where soot particles can pick up some electric charge. The air then passes through an area with a strong electric field to pull the pollutant particles away to an area where they can be safely removed.
Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105 newtons per coulomb while acted upon by a drag force of 7.25×10−11 newtons. What is the charge q1 on the particle? Ignore the effects of gravity.

Express your answer in coulombs to three significant figures.
Physics
1 answer:
kiruha [24]2 years ago
8 0

Answer:

7.25\times 10^{-16}\ C

Explanation:

F = Force = 7.25\times 10^{-11}\ N

E = Electric field = 1\times 10^5\ N/C

Force is given by

F=qE

\Rightarrow q=\dfrac{F}{E}

\Rightarrow q=\dfrac{7.25\times 10^{-11}}{1\times 10^5}

\Rightarrow q=7.25\times 10^{-16}\ C

The charge on the particle is 7.25\times 10^{-16}\ C

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A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
tekilochka [14]
This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
We know that the normal component of Fg is just Fg, which is equal to as 1110N. 
From the geometry, the normal component of Fp can be calculated: 
Fpn = Fp * cos(θp) 
= 1016.31 N * cos(53) 
= 611.63 N 

The total normal force Fn then is: 
Fn = Fg + Fpn 
= 1110 + 611.63
= 1721.63 N</span>

 

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely,  so we can say that the horizontal component is zero: 

Fh = Fph + Fgh 
= (Fp * sin(θp)) + 0 
= 1016.31 N * sin(53) 
= 811.66 N</span>

 

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

 

μs = Fh / Fn

= 811.66 N / 1721.63 N

<span>= 0.47</span>

8 0
2 years ago
An astronaut holds a rock 100m above the surface of Planet X . The rock is then thrown upward with a speed of 15m/s , as shown i
Butoxors [25]

Answer:5 m/s^{2}

Explanation:

The described situation is is related to vertical motion (and free fall). So, we can use the following equation that models what happens with this rock:

y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2} (1)

Where:

y=0m is the rock's final height

y_{o}=100 m is the rock's initial height

V_{o}=15 m/s is the rock's initial velocity

\theta=90\° is the angle at which the rock was thrown (directly upwards)

t=10 s is the time

g is the acceleration due gravity in Planet X

Then, isolating g and taking into account sin(90\°)=1:

g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t) (2)

g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s)) (3)

Finally:

g=5 m/s^{2} (4) This is the acceleration due gravity in Planet X

7 0
2 years ago
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
Absorbance measurements in the range of a = 0.3-2 are considered the most accurate. why would absorbance measurements of 0.05 an
zhenek [66]
Please provide the choices to select the possible choices.
6 0
1 year ago
A stone is held at a height h above the ground. A second stone with four times the mass of the first one is held at the same hei
QveST [7]

gravitational potential energy is given by formula

U = mgh

here we need to compare the gravitational potential energy of stone 2 with respect to stone 1

so we will say

\frac{U_2}{U_1} = \frac{m_2gh}{m_1gh}

\frac{U_2}{U_1} =\frac{m_2}{m_1}

given that

m_2 = 4 m_1

now we have

\frac{U_2}{U_1} = 4

5 0
2 years ago
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