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Alenkasestr [34]
3 years ago
14

Two students are running down the hallway towards each other. Student 1 has a mass of 107kg and is running to the right at 4.1m/

s. Student 2 has a mass of 97kg and is running to the left at 7.1m/s. If they strike in an elastic collision, what is the velocity of student 2 if student 1 moves at 6.55m/s to the left after the collision?
Physics
1 answer:
antiseptic1488 [7]3 years ago
3 0

Answer:

Hi

Explanation:

You are awesome

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Waning Gibbous would be the phase?
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A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p
ella [17]

Answer:

r_B = 1.88 m

Explanation:

As we know that work done by electric force is given as

W_e = -q\Delta V

so here we know that charge is moving from

r_A = 2.97 m

to another position

so we will have

W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}

-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})

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r_B = 1.88 m

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Synonym of an applied force
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Explanation:

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Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det
chubhunter [2.5K]

Complete Question

The complete quetion is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

       The mass of the fly is  m_f =  11 mg =  11*10^{-3} g =  1.1*10^{-5} \ kg

        The extension of the web is  e=  4.00 \ mm = 0.004 \ m

       

The spring constant is mathematically evaluated as

          k = \frac{mg}{e}

substituting values

        k = \frac{1.1 *10^{-5} *9.8}{0.004}

         k = 0.027 \ N/m

The frequency of vibration is

         f =  \frac{1}{2 \pi} \sqrt{\frac{k}{m} }

substituting values

       f =  \frac{1}{2 * 3.142 } \sqrt{\frac{0.027}{1.1*10^{-5}} }

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8 0
3 years ago
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

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Again, centripetal force of the neutron star is given as

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Recall that v = wR(ns), so that

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For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

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Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

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w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0
3 years ago
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