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Mkey [24]
3 years ago
15

WILL UPVOTE EVERY ANSWER! MULTIPLE CHOICE!

Physics
1 answer:
pav-90 [236]3 years ago
6 0
The pointer of an analog meter is connected to a "<span>coil suspended by bearings"

Hope this helps!</span>
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Phyics again , pls and thank you!
Alika [10]

Answer: accelerating/ speeding up

Explanation: since it is rising constantly and keeps getting higher it is increasing

3 0
3 years ago
Read 2 more answers
Which one of the following experiments is the justification for the wave theory of light? a. Newton's rings experiment b. Frank-
Lunna [17]

Answer:

c

Explanation:

double alit experiment is for wave theory for light

4 0
3 years ago
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Using the midpoint and the distance formulas, calculate he coordinate of the midpoint and the length of the segment.
arsen [322]

Answer:

Explanation:

Formula to calculate the distance between two points (x_1,y_1) and (x_2,y_2) is,

d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Therefore, length of CD having coordinates C(3,1) and D(3, 3),

CD = \sqrt{(3-3)^2+(1-3)^2}

     = 2 units

Formula to find the midpoint of CD is,

M = (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

   = (\frac{3+3}{2},\frac{1+3}{2})

   = (3, 2)

Therefore, length of CD = 2 units and (3, 2) are the coordinates of the midpoint of CD.

7 0
3 years ago
This is a net gain or loss of electrons. *
krok68 [10]
A net gain of electrons.
8 0
3 years ago
Read 2 more answers
Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitu
ANEK [815]

Answer:

The electric field will be decreased by 29%

Explanation:

The distance between point P from the distance z = 2.0 R

Inner radius = R/2

Outer raidus = R

Thus;

The electrical field due to disk is:

\hat {K_a} = \dfrac{\sigma}{2 \varepsilon _o} \Big( 1 - \dfrac{z}{\sqrt{z^2+R_i^2}} \Big))

\implies \dfrac{\sigma}{2 \vaepsilon _o} \Big ( 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0\ R)^2+(R)^2}} \Big)

Similarly;

\hat {K_b} = \hat {k_a} - \dfrac{\sigma}{2 \varepsilon_o} \Big( 1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ r)^2 + (\dfrac{R}{2}^2)}}\Big)

However; the relative difference is: \dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{E_a -E_a + \dfrac{\sigma}{2 \varepsilon_o  \Big[1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ R)^2 + (\dfrac{R}{2})^2}} \Big] } } { \dfrac{\sigma}{2 \varepsilon_o \Big [ 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0 \ R)^2 + (R)^2}} \Big] }}

\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{1 - \dfrac{2.0}{\sqrt{(2.0)^2 + \dfrac{1}{4}}} }{1 - \dfrac{2.0 }{\sqrt{(2.0)^2 + 1}}}

= 0.2828 \\ \\ \mathbf{\simeq  29\%}

3 0
3 years ago
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