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enyata [817]
3 years ago
15

Two identical charges, 2 m apart, exert forces of magnitude 4 N on each other. The value of each charge is: 1. 9 × 105 C 2. 4.2

× 10−5 C 3. 3.8 × 105 C 4. 2.1 × 10−5 C 5. 1.8 × 10−9 C
Physics
1 answer:
lesya692 [45]3 years ago
7 0

Answer:

The value of each charge is 4.22 x 10⁻⁵ C

Explanation:

Given;

distance between the two identical charges, d = 2 m

the force of repulsion between these two charges, F = 4N

Apply Coulomb's law;

F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C

Therefore, the value of each charge is 4.22 x 10⁻⁵ C

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W = 1225 N x 10 m = 12250

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3 years ago
An egg is thrown upward with a velocity of 4.5 m/s. How long will it take to reach it's maximum height?
Ray Of Light [21]

Answer:

0.45 seconds

Explanation:

Letting the value of g = 10 m/s/s

final velocity (v) = 0 m/s (since the egg will come to rest at the maximum height)

initial velocity(u) = 4.5 m/s

acceleration = -10 m/s/s (since the gravity is acting against the egg)

time = t seconds

From the first equation of motion:

<em>v = u + at</em>

<em>0 = 4.5 + (-10)t</em>

<em>t = -4.5 / -10</em>

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3 years ago
A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan
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Answer:

25 seconds

Explanation:

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Let she takes t seconds to cover the distance, s=115 m.

As acceleration, a=\frac{v-u}{t}=\frac{5-4.2}{t}

\Rightarrow at=0.8\cdots(i)

Now, from the equation of motion

s=ut+\frac 12 at^2

\Rightarrow s=ut+\frac 12 at(t)

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\Rightarrow 115=(4.2+0.4)t

\Rightarrow t= 115/4.6 = 25 seconds.

Hence, she takes 25 seconds to walk the distance.

5 0
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Answer:

  λ = 5940 Angstroms

Explanation:

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Where the speed in between the strr and the observer is positive if they move away

Let's use the relationship

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We replace

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Let's calculate

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