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prohojiy [21]
2 years ago
10

Several years from now you have graduated with an engineering/physics degree from OSU and have been hired by a nanoengineering f

irm as an intern. You have been assigned to work under a top engineer from the company. Their current project is to design a microscopic oscillator as a time keeping device. The engineering design involves placing a negative charge at the center of a very small positively charged metal ring. Your boss claims that the negative charge will undergo simple harmonic motion of displaced away from the center of the ring. Furthermore, they claim they can change the period (timing) of oscillation by adjusting the amount of charge on the ring. The first task they give you is to check the validity of their design.Consider a charge −???? located a small distance z above the center of a positively charged ring with total charge +Q and radius R. Write an expression for the net force exerted on the charge −???? due to the ring of charge. What is the magnitude of the force on the charge −???? if it is at the location z = 0?

Physics
1 answer:
spayn [35]2 years ago
5 0

Complete Question in order

See the first image attached

Answer and Explanation:

The three images attached discussed the solution to this question

second Image                                                                                      

Third Image                                                                                          

and Fourth Image

If the images are not clear enough right-click on it and open in a new tab

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Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete
Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m      

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(15)^2}{22.5}

a=10\ m/s^2

We know that, g=9.8\ m/s^2

a=\dfrac{10\times g}{9.8}

a=1.02\ g

Hence, this is the required solution.                                              

5 0
2 years ago
A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener
Bingel [31]

The ball was dropped from a height 20 meters

Explanation:

The given is

1. A ball is dropped from the top of a cliff

2. By the time it reaches the ground, all the energy in its gravitational

   potential energy store has been transferred into its kinetic energy

   store, that mean K.E = P.E

3. The ball is travelling at 20 m/s when it hits the ground

4. The gravitational field strength is 10 N/kg

We need to find the height that the ball dropped from it

The ball dropped from the top of a cliff means the initial speed is 0

→ K.E = \frac{1}{2}m(v^{2}-v_{0}^{2})

where v is the final speed, v_{0} in the initial speed and m

is the mass

→ v = 20 m/s and v_{0} = 0 m/s

→ K.E = \frac{1}{2}m(20^{2}-0^{2})

→ K.E = \frac{1}{2}m(400)

→ K.E = 200 m joules ⇒ when the ball hits the ground

→ P.E = m g h

where g is the gravitational field strength, m is the mass and h is

the height

→ g = 10 N/kg

→ P.E = m(10)(h)

→ P.E = 10 m h joules

→ P.E = K.E

→ 10 m h = 200 m

Divide both sides by 10 m

→ h = 20 meters

The ball was dropped from a height 20 meters

Learn more

You can learn more about gravitational potential energy in brainly.com/question/1198647

#LearnwithBrainly

8 0
2 years ago
A wagon full of manure accidentally rolls down a driveway for 5.0m while a person pushes against the wagon with a force of 420 N
Cerrena [4.2K]

Answer:

2100 J

Explanation:

Parameters given:

Force acting on the object, F = 420 N

Distance moved by object, d = 5m

The change in kinetic energy of an object is equal to the work done by a force acting on the object:

W = F * d

∆KE = F * d

∆KE = 420 * 5

∆KE = 2100 J

8 0
2 years ago
Read 2 more answers
A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
ankoles [38]
Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
4 0
2 years ago
A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds 60.0 N. T
AleksandrR [38]

Answer:

v=8.2158m/s

Explanation:

(a) Free-body diagram attached.

(b) The stone attached with the string experiences both centripetal (towards the center) and centrifugal (away from the center) forces. The tension of the string counters the centrifugal force until it breaks.

We know that,

Centrifugal force = \frac{mv^2}{r}

where,

m = mass of the stone

v = velocity of the stone

r = length of the string

To find the maximum speed attained by the stone without the string breaking, we must equate:

\frac{mv^2}{r} =60

or, v=\sqrt \frac{{r\times 60}}{m} } =\sqrt{\frac{0.90\times60}{0.80} } =8.2158m/s

5 0
2 years ago
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