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photoshop1234 [79]
3 years ago
15

An astronaut and his space suit have a combined mass of 157 kg. The

Physics
1 answer:
alexgriva [62]3 years ago
6 0

Answer:

v₃ = 9.62[m/s]

Explanation:

To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.

We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.

Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.

In this way, we can construct the following equation.

-(m_{1}+m_{2})*v_{1}=(m_{1}*v_{2})-(m_{2}*v_{3})

where:

m₁ = mass of the astronaut = 157 [kg]

m₂ = mass of the toolkit = 5 [kg]

v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]

v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]

v₃ = velocity at which the toolkit should be thrown [m/s]

Now replacing:

-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]

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Which force is greater the earth’s pull on the moon or the moon’s pull on the earth
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Answer:

The earth's pull on the moon

Explanation:

Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth.

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A book sits on a bookshelf without moving until a student picks it up. Which law best explains why the book remains at rest unti
svetoff [14.1K]

Answer:

Newtons first law

Explanation:

object in rest stays at rest

object in motion stays in motion

8 0
3 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
2 years ago
A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

f' = (\frac{v}{v\pm v_s})f

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

v_s is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz

When the train has passed the platform, we have

v_s = +40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz

Therefore the overall shift in frequency is

\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

v=\lambda f

where

v is the speed of the wave

\lambda is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

5 0
3 years ago
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