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CaHeK987 [17]
3 years ago
11

An astronaut working with many tools some distance away from a spacecraft is stranded when the "maneuvering unit" malfunctions.

How can the astronaut return to the spacecraft by sacrificing some of the tools? (Note: the maneuvering unit is connected to the astronaut's spacesuit and is not removable)
Physics
1 answer:
N76 [4]3 years ago
6 0

Answer:

He can return to the spacecraft by sacrificing some of the tools employing the principle of conservation of momentum.

Explanation:

By carefully evaluating his direction back to the ship, the astronaut can throw some of his tools in the opposite direction to that. On throwing those tools of a certain mass, they travel at a certain velocity giving him velocity in the form of recoil in the opposite direction of the velocity of the tools. This is same as a gun and bullet recoil momentum conservation. It is also the principle on which the operational principles of their maneuvering unit is designed.

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Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s

8 0
3 years ago
The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How
Vlada [557]

Answer:

The value is  w =  7.54 \  m        

Explanation:

From the question we are told that

     The length of the crack is  a =  0.3 \  m

     The  frequency is  f =  30.0 \ kHz =  30 *10^{3} \  Hz

      The distance outside the cave that is being consider is  D =  100 \  m

      The speed of sound is v_s =  340 \  m/s

Generally the wavelength of the wave is mathematically represented as

        \lambda =  \frac{v}f}

=>     \lambda =  \frac{340 }{30*10^{3}}

=>     \lambda = 0.0113 \ m/s

Generally for a  single slit the path difference between the interference patterns of the sound wave and the center  is mathematically represented as  

          y =  \frac{ n *  \lambda * D}{a}

=>     y =  \frac{ 1  *  0.0113 * 100}{0.3}

=>     y = 3.77 \  m

Generally the width of the sound beam is mathematically represented as

         w =  2 *  y

=>      w =  2 *  3.77

=>      w =  7.54 \  m        

4 0
3 years ago
An object is accelerating at a constant rate due to a force being applied.what can be done to the force to decrease the objects
Sliva [168]

Answer:

y

Explanation:

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6 0
3 years ago
The space shuttle orbits 310 km above the surface of the Earth.
MAVERICK [17]

Answer:

44.7 N

Explanation:

The gravitational force between the objects is given by:

F=G\frac{mM}{r^2}

where

G is the gravitational constant

m and M are the masses of the two objects

r is the distance between the centres of the two objects

In this problem, we have:

m=5.0 kg is the mass of the sphere

M=5.98\cdot 10^{24} kg is the Earth's mass

R=6370 km is the Earth's radius, while h=310 km is the altitude of the sphere, so the distance of the sphere from Earth's centre is

r=6370 km+310 km=6680 km=6.68\cdot 10^6 m

Substituting into the equation, we find

F=(6.67\cdot 10^{-11})\frac{(5.0 kg)(5.98\cdot 10^{24} kg)}{(6.68\cdot 10^6 m)^2}=44.7 N

8 0
3 years ago
A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w
barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

Moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}

Moles\ of\ AgBr= 0.1225\ mol

The reaction taking place is:

MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4

From the reaction,

4 moles of AgBr is produced when 1 mole of MBr_4 undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of MBr_4 undergoes reaction

0.1225 mole of AgBr is produced when \frac {1}{4}\times 0.1225 mole of MBr_4 undergoes reaction

Moles of MBr_4 got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of MBr_4 = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

0.030625 = \frac{12.5843}{319.616 + x}

Solve for x,

we get, x = 91.2999 g/mol

<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

8 0
3 years ago
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