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frez [133]
2 years ago
12

On an ice skating rink, a girl of mass 50 kg stands stationary, face to face with a boy of mass 80 kg. The children push off of

one another, and the boy moves away with a velocity of +3 m/s.
What is the final velocity of the girl?

–1.9 m/s
+1.9 m/s
–4.8 m/s
+4.8 m/s
Physics
2 answers:
andrey2020 [161]2 years ago
5 0

Answer:

Pretty sure its -4.8%

Explanation:

Others asked this question and got this as an answer.

Charra [1.4K]2 years ago
5 0

Answer:

C. –4.8 m/s

Explanation:

I just took the exam.

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Inertia is ______ to mass.
UkoKoshka [18]

Answer:

Inertia is directly proportional to mass of an object. Therefore, when the force of inertia increases the mass also increases, and when it decreases the mass also decreases.

Explanation:

6 0
3 years ago
Read 2 more answers
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
3 years ago
What is the definition of evolution
Anna35 [415]
<span>the process by which different kinds of living organisms are thought to have developed and diversified from earlier forms during the history of the earth.

Hope this helps.

</span>
7 0
2 years ago
Read 2 more answers
A circular coil of 185 turns has a radius of 1.60 cm. (a) Calculate the current that results in a magnetic dipole moment of magn
s344n2d4d5 [400]

Answer:

(a) the current in the coil is 12.03 A

(b) the maximum magnitude of the torque is 0.0854 N.m

Explanation:

Given;

number of turns, N = 185

radius of the coil, r = 1.6 cm = 0.016 m

magnetic dipole moment, μ = 1.79 A·m²

Part (a) current in the coil

μ = NIA

Where;

I is the current in the coil

A is the of the coil = πr² = π(0.016)² = 0.000804 m²

I = μ / (NA)

I = 1.79 / (185 x 0.000804)

I = 1.79 / 0.14874

I = 12.03 A

Part (b) the maximum magnitude of the torque

τ = μB

Where;

τ is the maximum magnitude of the torque

B is the magnetic field strength = 47.7 mT

τ = 1.79 x 0.0477 = 0.0854 N.m

4 0
3 years ago
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Which of Newton’s Laws involves mass and acceleration?
Temka [501]
The answer is <span>Newton’s</span> 2nd law. Thank you! 
4 0
3 years ago
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