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damaskus [11]
3 years ago
7

A Space Walk An astronaut is taking a space walk near the shuttle when her safety tether breaks. What should the astronaut do to

get back to the shuttle? View Available Hint(s) Attempt to "swim" toward the shuttle Take slow steps toward the shuttle Take a toal from her tool belt and throw it away from the shuttle. Take the portion of the safety tether still attached to her belt and throw it toward the shuttle Previous Answers Completed Part B Assuming that the astronaut can throw a tool of any inertia with the same acceleration, what tool should be thrown to get back to the shuttle as quickly as possible? View Available Hint(s) Hint 1. How to approach the problem Recall that during an isolated interaction between two objects, the changes in momentum for the two objects are equal and magnitude and opposite in direction. Furthermore, since the time interval in which they interact is the same, the time rates of change in momentum are also equal in magnitude and opposite in direction, which can be expressed as follows (written in terms of x components): 1a-22 Therefore, how will changing the inertia of the tool affect the astronaut's acceleration (such that it is larger)? The tool with the smallest inertia The tool with the largest inertia. Any tool, since the inertia of the tool would make no difference
Physics
1 answer:
KiRa [710]3 years ago
8 0

Answer:

There is no short answer.

Explanation:

Part A

Examining the given situation in the question where an astronaut is trying to get back to the shuttle and trying to figure out a strategy to do so, the best solution in this situation would be to throw an object in the opposite direction of the shuttle since the throwing action will results in the same force being exerted on the astronaut but in the opposite direction and thus pushing him/her towards the shuttle.

Part B

Throwing the object with the biggest mass is the best option in this situation, the force will not be affected by this since the exerted force on the object will be the same, so will the force in the opposite direction but since the lighter the astronauts total weight, the larger the accelaretion will be and therefore throwing the heaviest tool is the best option.

I hope this answer helps.

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3 years ago
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde
boyakko [2]

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

V = V_{0}(1-e^{\frac{-t}{RC} }  )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} })  \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }

Taking natural log on both sides,

e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }  \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })

t = -RC ln (1-\frac{V}{V_{0} })        (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

  = \frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }

  = 18×10⁻¹²F

Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

t = -RC ln (1-\frac{V}{V_{0} })

 = -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

 = 15×10⁻⁹s

= 15 ns

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Which statement is true about two isotopes of the same element? (1 point)
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A

The scientific study of weather is called meteorology.
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The Schwarzschild radius RBH for an object of mass M is defined as: Rbh = (2GM)/(c^2)
Agata [3.3K]

Answers:

1) 2951.39 m

2) 0.000952 m

3) 8.765(10)^{16}kg

Explanation:

1) We know the Schwarzschild radius RBH is given by the following equation:

RBH=\frac{2GM}{c^{2}}   (1)

Where:

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Universal Gravitational Constant

M the mass of the black hole

c=3(10)^{8}m/s is the speed of light

Now, if we have a black hole with the mass of the Sun (M_{Sun}=1.99(10)^{30}kg), its radius will be:

RBH_{Sun}=\frac{2GM_{Sun}}{c^{2}}   (2)

RBH_{Sun}=\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1.99(10)^{30}kg)}{{(3(10)^{8}m/s)}^{2}}   (3)

RBH_{Sun}=2951.39m   (4) This is the radius of the black hole with the mass of the Sun

2) On the other hand, if the black hole has the mass of Mars (M_{Mars}=6.42(10)^{23}kg), its radius will be:

RBH_{Mars}=\frac{2GM_{Mars}}{c^{2}}   (5)

RBH_{Mars}=\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.42(10)^{23}kg)}{{(3(10)^{8}m/s)}^{2}}   (6)

RBH_{Mars}=0.000952m   (7) This is the radius of the black hole with the mass of Mars

3) In this case, we have to isolate M from (1):

M=\frac{RBH c^{2}}{2G}   (8)

Where RBH=1.30(10)^{-10}m

Solving (8) with the known values:

M=\frac{(1.30(10)^{-10}m)(3(10)^{8}m/s)^{2}}{2(6.674(10)^{-11}m^{3}/kgs^{2}}   (9)

M=8.765(10)^{16}kg   (10) This is the mass of the black hole

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3 years ago
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