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BabaBlast [244]

3 years ago

10

A 62 kg astronaut is space-walking outside the space capsule and is stationary when the tether line breaks. As a means of return

ing to the capsule, he throws his 3.3 kg wrench at a speed of 20.5 m/s away from the capsule. At what speed does the astronaut move toward the capsule

2 answers:

DanielleElmas [232]3 years ago

8 0

To solve this problem we must apply the concepts related to the conservation of momentum. We have the two masses of the bodies and the velocity of one of those bodies. Therefore, the initial momentum is equivalent to the final momentum. Mathematically this can be expressed as

$P_i = P_f$

$m_1_v_1 = m_2 v_2$

Replacing with our values,

$3.3kg(20.5m/s) = (62kg)v_2$

Solving for $v_2$ ,

$v_2= 1.09m/s$

Therefore the speed that the astronaut move toward the capsule is 1.09m/s

Leokris [45]3 years ago

3 0

Answer:

Explanation:

Given that,

Mass of astronaut

M = 62kg

Mass of wrench

m = 3.3kg

Speed at which wrench move away from the capsule

v = 20.5m/s

Speed of astronaut?

Let speed of astronaut be V

Since there is no external force, we would apply conservation of momentum

Momentum of the astronaut is equal to the momentum of the wrench

Momentum is given as

P = Mass × Velocity

P(astronaut) = P(wrench)

M×V = m×v

V = m×v/M

V = 3.3×20.5/62

V = 1.09 m/s

The astronaut speed is 1.09m/s

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3 years ago

Calculate the energy absorbed when 13 kg of liquid water raises from 18°C to 100°C and then boils at 100°C.

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Answer:

the energy absorbed is 4.477 x 10⁶ J

Explanation:

mass of the liquid, m = 13 kg

initial temperature of the liquid, t₁ = 18 ⁰C

final temperature of the liquid, t₂ = 100 ⁰C

specific heat capacity of water, c = 4,200 J/kg⁰C

The energy absorbed is calculated as;

H = mcΔt

H = mc(t₂ - t₁)

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2 years ago

A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time el

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B is the answer to this

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2 years ago

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

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2 years ago

A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa

NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

$mV_1+3mV_2=(m+3m)V_f=4mV_f$

$V_1+3V_2=4V_f$ and making $V_f$ the subject then

$V_f=\frac {V_1+3V_2}{4}$ and since $V_1$ is initial velocity of car, value given as 4 m/s, $V_2$ is the initial velocity of the three cars stuck together, value given as 2 m/s and $v_f$ is the final velocity which is unknown. By substitution

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(b)

Initial kinetic energy is given by

$\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ$

Final kinetic energy is given by

$\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ$

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

6 0

3 years ago

Read 2 more answers