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sineoko [7]
3 years ago
11

Type A.

Physics
1 answer:
pantera1 [17]3 years ago
5 0

Answer:

a) 750W

b) 800W

  1018.9W

c) the circular frame

Explanation:

a) To find the electric flux you use the following formula:

\Phi_E=EAcos\theta=EA

where you have assumed that the plane of the loop and B are perpendicular between them.

By replacing you obtain:

\Phi_E=(2*10^4N/C)(0.25m)(0.15m)=750W

b) In order to calculate the electric flux in a square frame you first calculate the perimeter of the rectangular frame of wire:

P = 0.25m+0.25m+0.15m+0.15m=0.8m

A square with this length will have sides of 0.2m

Hence. you have that the electric flux is:

\Phi_E=(2*10^4N/C)(0.2m)^2=800W

for a circular frame you have that the radius is:

s=2\pi r\\\\r=\frac{s}{2\pi}=\frac{0.8m}{2\pi}=0.127m

Then, the electric flux will be:

\Phi_E=(2*10^4N/C)(\pi)(0.127m)^2=1018.9W

c) The electric flux is maximum in the circular frame.

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_____ electrons are the most important electrons to an atom .
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Answer:

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Explanation:

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8 0
3 years ago
A crate of oranges is shoved across a grocery store floor with a force of 100 N for a distance of 1 meter. The crate travels an
navik [9.2K]

The statements that are true are;

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  • the magnitude of work done by the applied force on the crate is 100 J
<h3>What is work done?</h3>

Work is said to be done when te force applied travels a distance in the direction of the force. Now we can see that when the force is applied by shoving the crate, work is done, an additional work is done by friction to bring the crate to a stop.

Hence, the following are true;

  • the magnitude of work done by frictional forces on the crate is 100 J
  • the magnitude of work done by the applied force on the crate is 100 J

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4 0
2 years ago
Air is being pumped into a spherical balloon so that its volume increases at a rate of 150 cm3/s. How fast is the radius of the
Slav-nsk [51]

0.119cm/s is the radius of the balloon increasing when the diameter is 20 cm.

<h3>How big is a circle's radius?</h3>

The radius of a circle is the distance a circle's center from any point along its circumference. Usually, "R" or "r" is used to indicate it.

A circle's diameter cuts through the center and extends from edge to edge, in contrast to a circle's radius, which extends from center to edge. Essentially, a circle is divided in half by its diameter.

dv/dt = 150cm³/s

d = 2r = 20cm

r = 10cm

find dr/dt

Given that the volume of a sphere is calculated using

v = 4/3πr³

Consider both sides of a derivative

d/dt(v) = d/dt( 4/3πr³)

dv/dt =  4/3π(3r²)dr/dt = 4πr²dr/dt

Hence,

dr/dt = 1/4πr².dv/dt

dr/dt = 1/4π×(10)²×150

dr/dt = 1/4π×100×150

dr/dt = 0.119cm/s.

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3 0
1 year ago
A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the
Elina [12.6K]

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

          T = 2\pi  \sqrt{ \frac{L}{g} }

          T² =  4\pi ^2 \frac{L}{g}4pi2 L / g

          g = 4\pi ^2   \frac{L}{T^2}

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

           g = 4\pi ^2  \frac{0.823}{1.823^2}4 pi2 0.823 / 1.823 2

            g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

             T = t / n

             ΔT = \frac{dT}{dt} Δt + \frac{dT}{dn} ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

              ΔT = Δt / n

              ΔT = Δt

now let's look for the uncertainty of g

             Δg = \frac{dg}{dL} ΔL + \frac{dg}{dT}  ΔT

             Δg = 4\pi ^2 \frac{1}{T2}   ΔL + 4π²L  (-2  T⁻³) ΔT

           

a more manageable way is with the relative error

             \frac{\Delta g}{g}   = \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}

we substitute

              Δg = g ( \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}DL / L + ½ Dt / T)

the error in time give us the stanndard deviation  

let's calculate

               Δg = 9.7766 (\frac{0.001}{0.823} + \frac{1}{2}  \ \frac{0.671}{1.823})

               Δg = 9.7766 (0.001215 + 0.0184)

               Δg = 0.19 m / s²

the absolute uncertainty must be true to a significant figure

                Δg = 0.2 m / s2

therefore the correct result is

               g ±Δg = (9.8 ± 0.2) m / s²

5 0
3 years ago
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