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Yakvenalex [24]
3 years ago
15

A test charge of +3 µC is at a point P where the electric field due to the other charges is directed to the right and has a magn

itude of 4 106 N/C. If the test charge is replaced with a charge of -3 µC, what happens to the electric field at P?
Physics
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

Electric field will remain same.

Explanation:

Given that

At initial condition ,charge q = +3μ C

Electric field E = 4106 N/c

As we know that

Electric field due to charge q

E=\dfrac{kq}{r^2}  

Now when charge is replaced by new charge q'= -3μ C

From the expression of electric field we can say that electric field will remain same from same quantity of electric charge.

So we can say that electric field will remain same.

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Which best explains why changes to the atomic theory were necessary?. A] differing ideas. .B] experimental evidence. . C] better
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3 years ago
Read 2 more answers
A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f
hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2    .... (2)

By equation the equation (1) and (2), we get

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u = 31.75 m/s

5 0
3 years ago
In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo
Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

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Suppose we find  the magnitude of the acceleration of the 602.0-g block

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Using equation of motion

s=ut+\dfrac{1}{2}at^2

Where, s = distance

t = time

a = acceleration

Put the value into the formula

60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2

a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}

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5 0
3 years ago
Find the average speed of a car that travels 45km in 2hours 15minutes​
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Answer:

20 km/h

Explanation:

45 km ÷ 2.25 hours (15 mins is 0.25 hours)

= 20

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3 0
2 years ago
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