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Anuta_ua [19.1K]

2 years ago

7

A solid conducting sphere with radius 0.75 m carries a net charge of 0.13 nC. What is the magnitude of the electric field at a p

oint located 0.50 from the sphere's center 0.25 beneath the sphere's surface)?

2 answers:

ludmilkaskok [199]2 years ago

5 0

Answer:

Magnitude of Electric field E at at point 0.50m which is within the sphere is Zero( i.e E = 0)

Explanation:

It is understand from Guass' law, the electric field in a region enclosed by a conducting sphere is Zero.

It is given that the radius of the sphere is 0.75m, therefore, a point located at 0.50m from the sphere centre 0.25m before the sphere surface still falls inside the sphere, therefore making the electric field at that point zero in magnitude.

Delicious77 [7]2 years ago

3 0

Answer:

Explanation:

given that

Radius =0.75m

Cnet=0.13nC

a. Electric field inside the sphere located 0.5m from the center of the sphere.

The electric field located inside the sphere is zero.

b. The electric field located 0.25m beneath the sphere.

Since the radius is 0.75m

Then, the total distance of the electric field from the centre of the circle is 0.75+0.25=1m

Then

E=kq/r2

K=9e9Nm2/C2

q=0.13e-9C

r=1m

Then,

E= 9e9×0.13e-9/1^2

E=1.17N/C. Q.E.D

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27. (a) What magnification is produced by a 0.150 cm focal length microscope objective that is 0.155 cm from the object being vi

denpristay [2]

Answer:

magnification = - 30

overall magnification = -240

Explanation:

given data

Focal length of microscope objective f = 0.150 cm

Object distance from microscope objective do = 0.155 cm

magnification by eyepiece = 8 ×

to find out

What magnification is produced and overall magnification

solution

we consider here Image distance from microscope objective is = di

so that

Magnification produced by objective will be = - $\frac{di}{do}$

so we find here di by given equation that is

$\frac{1}{do} +\frac{1}{di} = \frac{1}{f}$ ..................1

$\frac{1}{di} = \frac{1}{0.150} - \frac{1}{0.155}$

di = 4.65 cm

so that magnification by object will be

magnification = - $\frac{di}{do}$

magnification = - $\frac{4.65}{0.155}$

magnification = - 30

and

overall magnification will be

overall magnification = magnification by objective × magnification by eyepiece ........................2

overall magnification = -30 × 8

overall magnification = -240

7 0

2 years ago

If the magnitude of the initial velocity of the ball v0 = 7.94 ± 0.03, and the "gun" is tilted 31 ± 0.4º upwards, what is the un

ycow [4]

Answer:

0.05

Explanation:

Given a variable C which is the product of two variables A, B:

$C=A\cdot B$

Then the absolute error on C is given by:

$\frac{\sigma_C}{C}=\frac{\sigma_A}{A}+\frac{\sigma_B}{B}$

where $\sigma_A, \sigma_B, \sigma_C$ are the uncertanties on the measure of A, B and C, respectively.

In this problem, the horizontal component of the velocity $v_x$ is given by

$v_x = v_0 cos \theta$

Therefore, the uncertainty on vx is given by:

$\frac{\sigma_{v_x}}{v_x}=\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta}$ (1)

where we have:

$v_0 = 7.94$

$\sigma_{v_0}=0.03$

$\theta=31^{\circ}$ , so

$cos \theta=cos 31^{\circ}=0.857$

The uncertainty on $cos \theta$ is given by:

$\sigma_{cos \theta}=|sin \theta|\sigma_\theta$

where:

$|sin \theta| = |sin 31^{\circ}|=0.515$

and

$\sigma_\theta=0.4^{\circ}=0.007 rad$

So

$\sigma_{cos \theta}=|0.515|\cdot 0.007 = 0.0036$

Also,

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So, combininb everything into (1), we find:

$\sigma_{v_x}=(\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta})v_x=(\frac{0.03}{7.94}+\frac{0.0036}{0.857})(6.80)=0.05$

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