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Effectus [21]
3 years ago
12

20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot

ential?
a. Varies in magnitude and direction, parallel to the equipotential.
b. Constant magnitude, but varying direction, perpendicular to the equipotential.
c. Varies in magnitude and direction, perpendicular to the equipotential.
d. Constant direction, but varying magnitude, perpendicular to the equipotential.
e. Constant direction, but varying magnitude, parallel to the equipotential.
f. Constant magnitude, but varying direction, parallel to the equipotential.
Physics
1 answer:
Alex3 years ago
6 0

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

\Delta E = -\frac{dV}{dr}

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

\Delta E = 0

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

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A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning
Lesechka [4]

Answer:

speed of current is 5 mile/hr

Explanation:

GIVEN DATA:

speed of motorboat = 15 miles/hr relative with water

let c is speed of current

15-c is speed of boat at  upstream

15+c is speed of boat at downstream

we know that

travel time=distance/speed

\frac{10}{15-c} +\frac{10}{15+c} = 1.5

150+10c+150-10c=1.5(15-c)(15+c)

300=1.5(225-c^2)

300=337.5-1.5c^2

200=225-c^2

c^2=25

c = 5

so speed of current is 5 mile/hr

6 0
3 years ago
A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick
DerKrebs [107]
Dispersion angle = 0.3875 degrees. 
Width at bottom of block = 0.09297 cm 
Thickness of rainbow = 0.07038 cm 
 Snell's law provides the formula that describes the refraction of light. It is:
 n1*sin(θ1) = n2*sin(θ2)
 where
 n1, n2 = indexes of refraction for the different mediums
 Î¸1, θ2 = angle of incident rays as measured from the normal to the surface. 
 Solving for θ2, we get
 n1*sin(θ1) = n2*sin(θ2)
 n1*sin(θ1)/n2 = sin(θ2)
 asin(n1*sin(θ1)/n2) = θ2 
 The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
 Red:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.641) = θ2
 asin(1.00029*0.653420604/1.641) = θ2
 asin(0.398299876) = θ2
 23.47193844 = θ2 
 Violet:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.667) = θ2
 asin(1.00029*0.653420604/1.667) = θ2
 asin(0.39208764) = θ2
 23.08446098 = θ2 
 So the dispersion angle is:
 23.47193844 - 23.08446098 = 0.38747746 degrees. 
 Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
 tan(θ) = X/11.6
 11.6*tan(θ) = X
 So for Red:
 11.6*tan(θ) = X
 11.6*tan(23.47193844) = X
 11.6*0.434230136 = X
 5.037069579 = X 
 And violet:
 11.6*tan(θ) = X
 11.6*tan(23.08446098) = X
 11.6*0.426215635 = X
 4.944101361 = X 
 So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm 
 The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
 cos(θ) = X/0.092968218
 0.092968218*cos(θ) = X 
 0.092968218*cos(40.80) = X
 0.092968218*0.756995056 = X
 0.070376481 = X 
 So the distance between the red and violet rays is 0.07038 cm.
7 0
2 years ago
1. To calulate the molarity of a solution, you need to know the moles of the solute and the?
Fofino [41]
1. Volume of the solution (B)

2. Celery (D)

3. Hydroxide ions in solution (A)
3 0
3 years ago
How far will u go if your wheel' diameter 34in?
Lapatulllka [165]
Technically you can go forever on and on, but maybe your question was like how many rotations in a certain distance?
4 0
3 years ago
A 34 kg bowling ball with a radius of 22 cm starts from rest at the top of an incline 4.4 m in height. Find the translational sp
liberstina [14]

Answer:

7.85 m/s.

Explanation:

Given,

Mass of the bowling ball, m = 34 Kg

radius, r = 0.22 cm

height of the inclination, h = 4.4 m

transnational velocity = ?

Moment of inertia of bowling ball,

I = \dfrac{2}{5}mr^2

Using conservation of energy

mgh = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2

We know that v = r\omega

mgh = \dfrac{1}{2}( \dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2

m gh = 0.7 m v^2

v =\sqrt{\dfrac{gh}{0.7}}

v =\sqrt{\dfrac{9.81\times 4.4}{0.7}}

v = 7.85\ m/s

Speed of the bowling ball is equal to 7.85 m/s.

3 0
3 years ago
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