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aivan3 [116]
3 years ago
9

Pls help lsplsplsplsplsplsplsplps

Physics
2 answers:
Elina [12.6K]3 years ago
6 0

hey there! the correct answer would be ozone! hope this helps!!

Leno4ka [110]3 years ago
6 0

b. Ozone

I hope it's help you...

Thanks♥♥

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When a 4-μf capacitor has a potential drop of 20 v across its plates, how much electric potential energy is stored in this capac
Genrish500 [490]

Thew energy stored in a capacitor of capacitance C and voltage between the plates V is

E=\frac{1}{2} CV^2.

Substituting numerical value

E=\frac{1}{2}(4*10^{-6}) (20)^2\\ E=800\; \mu J

7 0
4 years ago
at a certain instant an object is moving to the right with speed 1.0 m/s and has a constant acceleration to the left of 1.0 m/s^
chubhunter [2.5K]
For this problem, you would use the equation v=u+at. In this case, u=1 v=0 (when the object is at rest) a=-1
v=u+at
0=1+(-1)t
t=1 second
3 0
4 years ago
A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the t
aliina [53]

The question seems a bit incomplete. The question should be as follow:

A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well. A crank with a turning radius of 0.25 m is attached to the end of the cylinder. What minimum force directed perpendicular to the crank handle is required to just raise the bucket? (Assume the rope's mass is negligible, that cylinder turns on frictionless bearings, and that g= 9.8 m/s2.)

Answer:

45.08N

Explanation:

The question involves moment topic. Note that the equation of moment, M is

Moment, M = Force, F x Perpendicular distance from the turning point, d

M = F x d

Moment involves turning movement along the pivot. in this case, we have 2 pivots. The first one the attached near the rope, while the other is the crank.

To raise the bucket, minimum force required must be able to provide at least the same moment as the moment due to the bucket of water. i.e.

Moment due to bucket = moment due to force on the crank

First find the moment due to bucket,

Mb = F (weight) x d (radius of cylinder on top of well)

   =  (23 x 9.8) x 0.05

   = 11.27 Nm

Next, Find the moment due to force on the crank. Note that the force is the minimum required force for this equation.

Mc = F (min Force) x d (radius of crank)

     = F x  0.25  = 0.25F

Now we get a simple equation of

11.27 = 0.25F

Using Algebra, we'll get the minimum Force, F:

F = 11.27 / 0.25

  = 45.08 N

4 0
4 years ago
Read 2 more answers
Find the volume of the tire with dimensions
Vinvika [58]

The volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

<h3>What is volume?</h3>

Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.

<h3>Volume of the tire</h3>

The volume of the tire is the measure of the product of area and thickness of the tire.

The volume of the tire is calculated as follows;

Radius of the tire = 0.5 x 26" = 13"

Volume of the tire = Area x thickness

Volume of the tire = πr² x h

where;

  • r is the radius of the tire
  • h is the thickness of the tube

Volume of the tire = π(13)² x (2.125)

Volume of the tire =  1,128.2 cubic inch

Thus, the volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

Learn more about Volume of tire here: brainly.com/question/1972490

#SPJ1

7 0
2 years ago
A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9
Andreyy89

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

U + K = constant

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

6 0
3 years ago
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