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Marianna [84]
3 years ago
7

Question must be solved using only symbolic algebra.During the experiment if you could triple the breakaway magnetic force with

all other quantities left unchanged, what is the new value for the critical velocity if it was v0 (initial velocity), initially? (b) Now if you halved the radius with all other quantities left unchanged, what is the new critical velocity if it was v0 (initial velocity), initially? (c) If during the experiment, critical velocity quadrupled with all other quantities left unchanged, what is the new breakaway force if its magnitude was initially F0,?

Physics
1 answer:
Papessa [141]3 years ago
3 0

Answer:

An attachment has been added

Explanation:

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The answer is d. experiment
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an always be used to calculate the electric field. relates the electric field at points on a closed surface to the net charge en
Temka [501]

Complete Question:

Gauss's law:

Group of answer choices

A. can always be used to calculate the electric field.

B. relates the electric field throughout space to the charges distributed through that space.

C. only applies to point charges.

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

E. relates the surface charge density to the electric field.

Answer:

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.

This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.

This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.

Mathematically, Gauss's law is given by this formula;

ϕ = (Q/ϵ0)

Where;

ϕ is the electric flux.

Q represents the total charge in an enclosed surface.

ε0 is the electric constant.

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2 years ago
This is physics 11th grade and a homework question I don’t understand how to do this or what the question is asking me
Alexxx [7]

a) Frequency is the number of complete oscillations per second. Looking at the graph, there are 9 complete oscillations in 5 seconds. Thus,

Frequency = 9/5 = 1.8 oscillations per second

Frequency = 1.8 Hz

Period = 1/frequency = 1/1.8

Period = 0.056 s

b) When we differenctiate displacement with respect to time, the result is velocity.

Recall, period = 1/f = 5/9 cycles

1/4 cycle behind = 1/4 x 5/9 = 5/36

It is delayed with 5/36 sec with respect to displacement.

5/36 sec = 0.139 sec

Acceleration = first derivative of velocity = second derivative of displacement = 1/4 cycle behind velocity = 1/2 cycle behind displacement =

5/36 = 0.139 sec delayed with respect to velocity

= 5/18 = 0.2777 secs delayed with respect to displacement

Thus, the number of seconds out of phase with the displacements is 0.278 seconds

c) The formula for calculating the period of an ideal pendulum anywhere is

T = 2π√length/local gravity). We would calculate the local gravity.

From the information given,

length = 0.2

T = P = 5/9

Thus,

5/9 = 2π√0.2/local gravity)

(5/9)/2π = √0.2/local gravity

Square both sides. It becomes

[(5/9)/2π]^2 = 0.2/local gravity

local gravity = 0.2/[(5/9)/2π]^2

local gravity = 25.56 m/s^2

Thus,

acceleration due to gravity = 25.56 m/s^2

Recall, earth's gravity = 9.8 m/s^2

number of g forces = 25.56/9.8

number of g forces = 2.61

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1 year ago
True or false. there would be no like in earth without the sun
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The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit
Stels [109]

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, P=2.2\times 10^5\ Pa

Area of each tire, A=0.023\ m^2

Area of 4 tires,  A=0.092\ m^2

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

P=\dfrac{F}{A}

F=P\times A

F=2.2\times 10^5\times 0.092

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

7 0
3 years ago
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