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3 years ago

An astronaut at rest in space with mass 84 kg fires a thruster that expels 35 g of hot

Physics

2 answers:

storchak [24]3 years ago

4 0

Answer:

-0.364m/s

Explanation:

Mass of Astronaut (M₁) = 84kg

mass of gas(m₂) = 35g = 0.035kg

velocity of gas (v₂) = 875m/s

velocity of astronaut (v₁) =

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(84 * 0) + (0.035 * 0) = (84 * v₁) + (0.035 * 875)

0 = 84v₁ + 30.625

84v₁ = -30.625

v₁ = -0.364m/s

IRINA_888 [86]3 years ago

3 0

Answer:

The velocity of the astronaut is, v = - 0.36 m/s

Explanation:

Given data,

The mass of the astronaut, M = 84 kg

The mass of the gas expelled by the thruster, m = 35 g

= 0.035 g

The velocity of the gas expelled, v = 875 m/s

According to the conservation of momentum,

MV + mv = 0

V = - mv / M

Substituting the values,

V = - 0.035 x 875 / 84

= - 0.36 m/s

The negative sign in the velocity indicates the astronaut moves opposite to the direction of velocity of gas.

Hence, the velocity of the astronaut is, v = - 0.36 m/s

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A coffee filter of mass 1.5 grams dropped from a height of 3 m reaches the ground with a speed of 0.7 m/s. How much kinetic ener

Mademuasel [1]

The kinetic energy gained by the air molecules is 0.0437 J

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Given:

Mass of a coffee filter, m = 1.5 g

Height from which it is dropped, h = 3 m

Speed at ground, v = 0.7 m/s

Initially, the coffee filter has potential energy. It is given by :

$P =mgh$

P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m

P = 0.0441 J

Finally, it will have kinetic energy. It is given by :

$E= \frac{1}{2} mv^{2}$

$E= \frac{1}{2}$ × $1.4$ × 10⁻³ × (0.7)²

E = 0.000343 J

The kinetic energy Kair did the air molecules gain from the falling coffee filter is :

E = 0.000343 - 0.0441

= 0.0437 J

So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J

Learn more about kinetic energy here:

brainly.com/question/8101588

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8 0

2 years ago

An object is moving at 2.50 m/s [E]. At a time 3.00 seconds later the object is traveling at 1.50 m/s

babunello [35]

Given parameters:

First velocity = 2.50m/s

Time of travel = 3s

Second velocity = 1.50m/s

Unknown:

The displacement during the first interval = ?

Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.

Velocity = $\frac{Displacement}{Time taken}$

So;

Displacement = Velocity x Time taken

Now input the parameter for the first velocity and time of travel;

Displacement = 2.5 x 3 = 7.5m

The displacement id 7.5m

7 0

3 years ago

A jeweler working with a heated 47 g gold ring must lower the ring's temperature to make it safe to handle. If the ring is initi

Gelneren [198K]

Mass of gold m₁ = 47 g

Initial temperature of gold T₁ = 99 C

Specific heat of gold C₁ = 0.129 J/gC

final temperature T₂ = 38 C

Heat needed by the gold to cool down

Q =m₁ * C₁* ( T₁ - T₂)

Q = (47)(0.129)(99-38)

Q = 369.843 J

This heat will be given by the water

we need to find out mass of water m₂

and initial temperature of water is T₃ = 25 C

Specific heat of water C₂ = 4.184 J/gC

Q = m₂*C₂*(T₂ - T₃)

369.843 = m₂(4.184)(38-25)

m₂ = 6.8 g

6 0

3 years ago

Read 2 more answers

Absolute zero (K=0 or -273.15°C) is what temperature on the Farenheit scale? a) 459.4°F b) -301.43 °F c) 233.05°F d) -40.15°F e

ANEK [815]

Answer:

e) None of these is true

Explanation:

Given that

Temperature = 0 K

We know that relationship between kelvin and Farenheit scale

$\dfrac{K-273}{100}=\dfrac{F-32}{180}$

Now by putting the values

$\dfrac{K-273}{100}=\dfrac{F-32}{180}$

$\dfrac{0-273}{100}=\dfrac{F-32}{180}$

So F= - 459.67°F

So we can say that 0 K is equal to - 459.67°F.

So the our option e is correct.

3 0

4 years ago

An electron and a proton are each placed at rest in a uniform electric field of magnitude 560 N/C. Calculate the speed of each p

Butoxors [25]

Answer:

The speed of electron is $v=4.52\times 10^6\ m/s$ and the speed of proton is 2468.02 m/s.

Explanation:

Given that,

Electric field, E = 560 N/C

To find,

The speed of each particle (electrons and proton) 46.0 ns after being released.

Solution,

For electron,

The electric force is given by :

$F=qE$

$F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N$

Let v is the speed of electron. It can be calculated using first equation of motion as :

$v=u+at$

u = 0 (at rest)

$v=\dfrac{F}{m}t$

$v=\dfrac{8.96\times 10^{-17}}{9.1\times 10^{-31}}\times 46\times 10^{-9}$

$v=4.52\times 10^6\ m/s$

For proton,

The electric force is given by :

$F=qE$

$F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N$

Let v is the speed of electron. It can be calculated using first equation of motion as :

$v=u+at$

u = 0 (at rest)

$v=\dfrac{F}{m}t$

$v=\dfrac{8.96\times 10^{-17}}{1.67\times 10^{-27}}\times 46\times 10^{-9}$

$v=2468.02\ m/s$

So, the speed of electron is $v=4.52\times 10^6\ m/s$ and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.

6 0

4 years ago