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sweet [91]
3 years ago
5

Physics help please ;-; . An astronaut has a mass of 82.0 kg. What is the astronaut's stationary weight at a position 4230 kilom

eters above Earth's surface? Note: Earth's radius is 6380 km and the Earth's mass is 5.972 x 1024 kg.
Two rocks, each of mass 72 kg, are positioned 95 m away from each other in deep space. What is the magnitude of the gravitational attraction between them?

if at the least explain how id work it out.​
Physics
1 answer:
sweet [91]3 years ago
3 0

Answer:

F = G*\frac{m_{1} * m_{2} }{r^2}

Explanation:

Just find the force of gravity.

1) m1 is the mass of earth, m2 is mass of astronaut. r is the earth's radius + astronaut's position in meters. G is the universal constant of gravity.

2) Same as above. m1 is 72 kg, m2 is 72 kg, r is 95 m and G is gravitational constant.

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A heavier car is always safer in a crash than a lighter car.
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not true because the mass from the heavy car will cause it to damage more

Explanation:

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<h2>1. Friction is A. a force</h2>

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2 years ago
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A 900 kilogram compact car was stopped at a traffic light when an 1800 kilogram luxury sedan struck it from behind the two cars
SOVA2 [1]
The answer is B. 22.7 kg-m/s
7 0
3 years ago
A truck with 28-in.-diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min, *hint convert mile
solong [7]

Answer:

Angular speed ω=3771.4 rad/min

Revolution=5921 rpm

Explanation:

Given data

d=28in\\r=d/2=28/2=14in\\v=50mi/hr

To find

Angular speed ω

Revolution per minute N

Solution

First we need to convert the speed of truck to inches per mile

as

1 mile=63360 inches

1 hour=60 minutes

so

v=(50*\frac{63360}{60} )\\v=52800in/min

Now to solve for angular speed ω by substituting the speed v and radius r in below equation

w=\frac{v}{r}\\ w=\frac{52800in/min}{14in}\\ w=3771.4rad/min

To solve for N(revolutions per minute) by substituting the angular speed ω in the following equation

N=\frac{w}{2\pi }\\ N=\frac{3771.4rad/min}{2\pi }\\ N=5921RPM  

3 0
3 years ago
A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predete
zmey [24]

Answer:

The diameter of wire should be 4.04 \times 10^{-4} m

Explanation:

Given:

Current density J = 500 \times 10^{4} \frac{A}{m^{2} }

Current I = 0.64 A

From the formula of current density,

  J = \frac{I}{A}

Where A = area of cylindrical wire = \pi r^{2}

  \pi r^{2} = \frac{I}{J}

  r^{2} = \frac{I}{\pi J }

   r = \sqrt{\frac{0.64}{3.14 \times 500 \times 10^{4} } }

   r = 2.02 \times 10^{-4}m

For finding the diameter of wire,

   d = 2r

   d = 4.04 \times 10^{-4}m

Therefore, the diameter of wire should be 4.04 \times 10^{-4} m

8 0
3 years ago
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