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JulijaS [17]
3 years ago
13

A 60.0 kg astronaut is on a space walk away from the shuttle when her tether line breaks! She is able to throw her 10.0 kg oxyge

n tank away from the shuttle with a velocity of 12.0 m/s to propel herself back to the shuttle. Assuming that she starts from rest (relative to the shuttle), determine the maximum distance she can be from the craft when the line breaks and still return with 60.0 s ( the amount of time she can hold her breath).
Physics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

120 m

Explanation:

m_1 = Mass of Astronaut

m_1 = Mass of tank

v_1 = Velocity of Astronaut

v_2 = Velocity of tank

From conservation of linear momentum

m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{10\times 12}{60}\\\Rightarrow v_1=2\ m/s

Distance = Speed × Time

Distance=2\times 60=120\ m

So, the maximum distance she can travel is 120 m

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dmitriy555 [2]
<span>This is because astronomers have discovered that the Crab Nebula is 6500 light years from earth. This means it takes 6500 earth years for light to get from the Crab Nebula to earth. Therefore, if we are able to observe the Crab Nebula and beyond it then it means that earth has been in existence for 6500 years and over.  </span>




5 0
3 years ago
Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t
n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

y=\frac{\lambda (m+\frac{1}{2})D}{d}

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

\lambda is the wavelength of the light used

In this problem we have:

\lambda=683 nm = 683\cdot 10^{-9} m is the wavelength of the light

d=1.1 mm = 0.0011 m is the width of the slit

m = 13 is the order of the minimum

y=8.57 cm = 0.0857 m is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m

6 0
3 years ago
Which of the following is true? A. If an equal force is applied to two objects, the object with greater mass will accelerate mor
docker41 [41]
B


I hope this helps and have a wonderful day filled with joy!!


<3
4 0
3 years ago
A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result
Virty [35]

Hello!

A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result? A) law of differential mass B) law of conservation of momentum C) law of unequal forces D) law of accelerated collision

We have the following data¹:

ΔP (momentum before impact) = ?  

mA (mass) = 10 kg

vA (velocity) = 5 m/s

mB (mass) = 5 kg

vB (velocity) = 0 m/s

Solving:

ΔP = mA*vA + mB*vB

ΔP = 10 kg*5 m/s + 5 kg*0 m/s

ΔP = 50 kg*m/s + 0 kg*m/s

Δp = 50 kg*m/s ← (momentum before impact)

We have the following data²:

ΔP (momentum after impact) = ?  

mA (mass) = 10 kg

vA (velocity) = 0 m/s

mB (mass) = 5 kg

vB (velocity) = 10 m/s

Solving:

Δp = mA*vA + mB*vB

Δp = 10 kg*0 m/s + 5 kg*10 m/s

Δp = 0 kg*m/s + 50 kg*m/s

Δp = 50 kg*m/s ← (momentum after impact)

*** Then, which principle explains the result ?

Law of conservation of momentum, <u>since the total momentum of body A and B before impact is equal to the total momentum of body A and B after impact.</u>

Note:  Bodies of different masses and velocities may have the same kinetic energy, if proportionality between the units is maintained it can occur that they have the same kinetic energy.

Answer:

B) law of conservation of momentum

_______________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

5 0
3 years ago
A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What
kati45 [8]

Explanation:

The given data is as follows.

           radius (r) = 3.25 cm,    \alpha = 11.6 rad/s^{2}

Now, we will calculate the tangential acceleration as follows.

          a_{tangential} = \alpha \times r

Putting the given values into the above formula as follows.

         a_{tangential} = \alpha \times r

                      = 11.6 rad/s^{2} \times 3.25 cm

                      = 37.7 rad cm/s^{2}

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 rad cm/s^{2}.

6 0
3 years ago
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