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Elan Coil [88]
3 years ago
13

Two astronauts are floating together with zero speed in a gravity-free region of space. The mass of astronaut A is 110 kg and th

at of astronaut B is 74 kg. Astronaut A pushes B away, with B attaining a final speed of 0.4. The final recoil speed of astronaut A is:
Physics
1 answer:
Kobotan [32]3 years ago
4 0

Answer:

The recoil speed of Astronaut A is 0.26 m/s.            

Explanation:

Given that,

Mass of astronaut A, m_A=110\ kg

Mass of astronaut B, m_B=74\ kg

Astronaut A pushes B away, with B attaining a final speed of 0.4, v_B=0.4\ m/s

We need to find the recoil speed of astronaut A. The momentum remains conserved here. Using the law of conservation of linear momentum as :

m_Av_A=m_Bv_B\\\\v_A=\dfrac{m_Bv_B}{m_A}\\\\v_A=\dfrac{74\times 0.4}{110}\\\\v_A=0.26\ m/s

So, the recoil speed of Astronaut A is 0.26 m/s.                                

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2 years ago
A hard-boiled egg of mass 46.0 gg moves on the end of a spring with force constant 25.6 N/mN/m . The egg is released from rest a
soldi70 [24.7K]

Answer:

0.022kg/s

Explanation:

We are given that

Mass of boiled egg=46 g=\frac{46}{1000} kg=0.046 kg

1kg=1000 g

Constant force=F=25.6 N/m

Initial displacement=A_1=0.296 m

Final displacement=A_2=0.12 m

Time=t=4.55 s

Damping force=F_x=-bv_x

We have to find the  magnitude of damping constant b.

We know that the displacement of the oscillator under damping motion is given by

x=Ae^{-\frac{b}{2m}t}cos(w't+\phi)

For maximum displacement cos(w't+\phi)=1

Therefore , x=A_2

Substitute the values

A_2=A_1e^{-\frac{-b}{2m}t}

e^{-\frac{b}{2m}t}=\frac{A_2}{A_1}

-\frac{b}{2m}t=ln\frac{A_2}{A_1}

lnx=y\implies x=e^y

Substitute the values

-\frac{b}{2\times 0.046}\times 4.55=ln\frac{0.12}{0.296}

\frac{2\times 0.046}{4.55b}=ln\frac{0.296}{0.12}

\frac{2\times 0.046}{4.55}=0.9b

b=\frac{2\times 0.46}{4.55\times 0.9}=0.022kg/s

Hence,the  magnitude of damping constant b=0.022kg/s

3 0
3 years ago
Need Help !
Bad White [126]

Answer: 5.96m/s

Explanation:

Given the following :

Mass of car (m) = 1500kg

Velocity (V) = 5.25m/s

Forward force of engine = 1250N

Diatance moved = 4.8m

Final Velocity =?

Final kinetic energy = Initial kinetic energy + work done by engine

Initial kinetic energy = 0.5 × mass × velocity^2

Initial kinetic energy = 0.5 × 1500 × 5.25^2

Initial kinetic energy = 20671.875 J

Work done by engine = Force × distance

Work done by engine = 1250 × 4.8 = 6000J

Final kinetic energy = (20671.875 + 6000) J

= 26671.875 J

From kinetic energy = 0.5mv^2

26671.875 = 1/2 × 1500 × v^2

53343.75 = 1500v^2

v^2 = 35.5625

v = sqrt(35.5625)

v = 5.96m/s

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3 years ago
Which equation can be used to calculate the normal force on an object if you know the speed of the object, the coefficient of ki
tino4ka555 [31]

Answer:

D

Explanation:

The friction force is the weight force times the coefficient of friction. So diving by the coefficient gives you the weight force which is equivalent to the normal force.

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2 years ago
How do noise and vibration affect you when operating a boat?
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3 years ago
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