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mafiozo [28]
3 years ago
10

g Two identical metal spheres (small enough to be treated as particles) each with a mass of 3.1 × 10−3 kg are hung from separate

identical strings. The strings are 0.70 m long and are attached to the same nail in the ceiling. Surplus electrons are added to each sphere, and then the spheres are brought in contact with each other and released. Their equilibrium position is such that each string makes an angle of 15.0° with the vertical. How many surplus electrons are on each sphere?
Physics
1 answer:
gogolik [260]3 years ago
8 0

Answer:

Explanation:

Let q be the charges on each spheres , 2d be the   distance between them in equilibrium , T be tension in the string and F be the force of repulsion between them

F = k q² /4d²

For equilibrium

T sin15 = F

T cos 15 = mg

tan15 = F / mg

F = mg tan15

k q² /4d² = mg tan15

k q² = 4d² x mg tan15

= 4 x( .7 sin15)² x 3.1 x 10⁻³ x 9.8 x .2679

= 1.068 x 10⁻³

q = .3444 x 10⁻⁶ C

no of electrons

=  .3444 x 10⁻⁶  / 1.6 x 10⁻¹⁹

= 215.25 x 10¹⁰

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Explanation:

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Both are aquatic animals and are hunters

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Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = m_1=6.50 kg- 1.55 kg=4.95 kg

Velocity achieved by squid = v_1=2.60 m/s

Momentum gained by squid = P=m_1v_1

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Velocity by which water was released by squid = v_2

Momentum gained by water but in opposite direction = P'=m_2v_2

P = P'

m_1v_1=m_2v_2

v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E  =\frac{1}{2}m_1v_1^{2}

Kinetic energy of water = K.E' = \frac{1}{2}m_2v_2^{2}

Total kinetic energy created by this maneuver:

K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}

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4 0
3 years ago
Calculate the speed for a car that went a distance of 125 meters in 2 seconds time.
stiv31 [10]

Answer:

<h2>62.5 m/s</h2>

Explanation:

The speed of the car can be found by using the formula

s =  \frac{d}{t} \\

d is the distance

t is the time

From the question we have

s =  \frac{125}{2}  = 62.5 \\

We have the final answer as

<h3>62.5 m/s</h3>

Hope this helps you

5 0
1 year ago
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

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Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0
3 years ago
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