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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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Determine the maximum weight of the bucket that the wire system can support so that no single wire develops a tension exceeding
stellarik [79]
Let there be N number of wires.

Maximum tension a wire can withstand = 100 lb

so, Total tension N wires can withstand =  100 N

now, total tension in N wires = Maximum weight of bucket

100 N  = W

so, W = 100N

W is the weight of bucket and 100N is its maximum value.
8 0
2 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
2 years ago
If the wire is replaced by an infinite current sheet with density Js = 0.40 A/m, what would be the magnetic flux (in T · m2) thr
oksian1 [2.3K]

Answer:

\phi _{B} =0.855 T-m^{-2}

Explanation:

given data

density of current sheet = 0.40 A/m

length a = 0.27 m

width b = 0.63 m

For infinite sheet, magnetic field is given as

B = \mu _{O}J

magnetic flux is given as

\phi _{B} = BA

                   = \mu _{O}Jab

                   = 4\pi *0.40*0.27*0.63

\phi _{B} =0.855 T-m^{-2}

6 0
2 years ago
A reel of flexible power cable is mounted on the dolly, which is fixed in position. There are 190 ft of cable weighing 0.402 lb
White raven [17]

Answer:

Explanation:

Total mass of cable m = 190 x .402 = 76.38 lb

moment of inertial due to this cable = m r²

= 76.38 x (14/12)²

= 103.96 lb ft²

moment of inertia of empty spoon

= mR² where R is radius of gyration

= 65 x (11 / 12 )²

= 54.61 lb ft²

Total moment of inertia I = 158.57 lb ft²

Net force applied = force applied - frictional force

= 33 - 15 = 18 lb

= 18 x 32 poundal

= 576 poundal

Torque applied = force x radius

= 576 x 14/12

= 672 unit

Angular acceleration = torque / total moment of inertia

= 672 / 158.57

= 4.238 radian / s²

5 0
2 years ago
A firecracker is thrown downward from a height of 2.75m above the ground, with a speed of 3.15m/s. Ignore air resistance, determ
3241004551 [841]

Here in this question as we can see there is no air friction so we can use the principle of energy conservation

PE_i + KE_i = PE_f + KE_f

mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2

now here we know that

h_1 = 2.75 m

v_i = 0

v_f = 5.23 m/s

now plug in all values in above equation

mg*2.75 + 0 = mgh + \frac{1}{2}m(5.23)^2

divide whole equation by mass "m"

9.8*2.75 = 9.8*h + \frac{1}{2}*27.35

9.8*h = 13.27

h = 1.35 m

so height of the ball from ground will be 1.35 m

4 0
2 years ago
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