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Ganezh [65]
3 years ago
9

Two small, positively charged spheres have a combined charge of 5.0 x 10 -5 C. If each sphere is repelled from the other by an e

lectrostatic force of 1.0 N when the spheres are 2.0 m apart, what is the charge, in micro-coulomb, on the sphere with the smaller charge?
Physics
1 answer:
geniusboy [140]3 years ago
4 0

Explanation:

It is given that,

Let q₁ and q₂ are two small positively charged spheres such that,

q_1+q_2=5\times 10^{-5}\ C.............(1)

Force of repulsion between the spheres, F = 1 N

Distance between spheres, d = 2 m

We need to find the charge on the sphere with the smaller charge. The force is given by :

F=k\dfrac{q_1q_2}{d^2}

q_1q_2=\dfrac{F.d^2}{k}

q_1q_2=\dfrac{1\ N\times (2\ m)^2}{9\times 10^9}

q_1q_2=4.45\times 10^{-10}\ C............(2)

On solving the system of equation (1) and (2) using graph we get,

q_1=0.0000384\ C=38.4\ \mu C

q_2=0.0000116\ C=11.6\ \mu C

So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.

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while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity o
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Complete question:

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency​ reflected off the wall to the bat?

Answer:

The frequency reflected by the stationary wall to the bat is 41 kHz

Explanation:

Given;

frequency emitted by the bat, = 39 kHz

velocity of the bat, v_b = 8.32 m/s

speed of sound in air, v = 340 m/s

The apparent frequency of sound striking the wall is calculated as;

f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz

The frequency reflected by the stationary wall to the bat is calculated as;

f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz

f_s\approx 41 \ kHz

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3 years ago
A 25 kg mass is hanging from two cables, each with their own tension. Cable 1 is connected to the
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Answer:

a. one line down one line to the right one live to the northwest from the object

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3 years ago
¡¡¡AYUDA CON ESTOS EJERCICIOS DE FÍSICA!!!
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Answer:

(a) 8 V, (b) 144000 V, (c) 2 x 10^(-8) C

Explanation:

(a) charge, q = 5 μC , Work, W = 40 x 10-^(-6) J

The electric potential is given by

W = q V

40\times10^{-6}=5 \times10^{-6}\times V\\\\V = 8 V

(b)

charge, q = 8 x 10^(-6) C, distance, r = 50 cm = 0.5 m

Let the potential is V.

V =\frac{k q}{r}\\\\V =\frac{9\times 10^{9}\times 8\times 10^{-6}}{0.5}\\\\V =144000 V

(c)

Work, W = 8 x 10^(-5) J, Potential difference, V = 4000 V

Let the charge is q.

W= q V

8\times10^{-5}= q\times 4000\\\\q =2\times 10^{-8} C

3 0
3 years ago
A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that
astraxan [27]

Answer:

A) k=34867.3384\ N.m^{-1}

B) \omega'\approx84\ Hz

Explanation:

Given:

mass of car, m=1380\ kg

A)

frequency of spring oscillation, f=1.6\ Hz

We knkow the formula for spring oscillation frequency:

\omega=2\pi.f

\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f

\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6

k_{eq}=139469.3537\ N.m^{-1}

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

k=\frac{k_{eq}}{4}

k=\frac{139469.3537}{4}

k=34867.3384\ N.m^{-1}

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }

\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }

\omega'\approx84\ Hz

7 0
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