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gogolik [260]
3 years ago
10

An 88.0 kg spacewalking astronaut pushes off a 645 kg satellite, exerting a 110 N force for the 0.450 s it takes him to straight

en his arms. How far apart are the astronaut and the satellite after 1.40 min?
Physics
1 answer:
arlik [135]3 years ago
7 0

Answer:

The astronaut and the satellite are 53.718 m apart.

Explanation:

Given;

mass of spacewalking astronaut, = 88 kg

mass of satellite, = 645 kg

force exerts by the satellite, F = 110N

time for this action, t = 0.45 s

Determine the acceleration of the satellite after the push

F = ma

a = F / m

a = 110 / 645

a = 0.171 m/s²

Determine the final velocity of the satellite;

v = u + at

where;

u is the initial velocity of the satellite = 0

v = 0 + 0.171 x 0.45

v = 0.077 m/s

Determine the displacement of the satellite after 1.4 m

d₁ = vt

d₁ = 0.077 x (1.4 x 60)

d₁ = 6.468 m

According to Newton's third law of motion, action and reaction are equal and opposite;

Determine the backward acceleration of the astronaut after the push;

F = ma

a = F / m

a = 110 / 88

a = 1.25 m/s²

Determine the final velocity of the astronaut

v = u + at

The initial velocity of the astronaut = 0

v = 1.25 x 0.45

v = 0.5625 m/s

Determine the displacement of the astronaut after 1.4 min

d₂ = vt

d₂ = 0.5625 x (1.4 x 60)

d₂ = 47.25 m

Finally, determine the total separation between the astronaut and the satellite;

total separation = d₁ + d₂

total separation = 6.468 m + 47.25 m

total separation = 53.718 m

Therefore, the astronaut and the satellite are 53.718 m apart.

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Answer:

= 1000 hours

Explanation:

Earth's circumference is 10⁴ mile

speed of a sailboat is 10¹ mile/hour

distance = speed × time

10⁴ = 10¹  × t

t = 10⁴ / 10¹

t = 10³

= 1000 hours

4 0
3 years ago
Which waves have wavelengths longer than those of visible light? Give an example of how each kind of wave is used.
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1. Radio Waves
ex. Wi-Fi
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ex. Mobile Phones
3. Infrared Radiation
ex. Heat Lamps
6 0
3 years ago
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What are radioactive isotopes?
melisa1 [442]

Answer:

<em><u>R</u></em><em><u>adioactive isotope</u></em> , also called radioisotope,

radionuclide, or radioactive nuclide, any of

several species of the same chemical element

with different masses whose nuclei are unstable

and dissipate excess energy by spontaneously

emitting radiation in the form of alpha , beta , and

gamma rays .

A brief treatment of radioactive isotopes follows.

For full treatment, see isotope: <u>Radioactive</u>

<u>Radioactiveisotopes.</u>

Every chemical element has one or more

radioactive isotopes. For example, hydrogen, the

lightest element, has three isotopes with mass

numbers 1, 2, and 3. Only hydrogen-3 (tritium ),

however, is a radioactive isotope , the other two

being stable. More than 1,000 radioactive

isotopes of the various elements are known.

Approximately 50 of these are found in nature;

the rest are produced artificially as the direct

products of nuclear reactions or indirectly as the

radioactive descendants of these products.

Radioactive isotopes have many useful

applications. In medicine , for example, cobalt -60

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arrest the development of cancer. Other

radioactive isotopes are used as tracers for

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Explanation:

#learnwithbrainly

8 0
3 years ago
How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J
algol13

Answer:

Work done by the frictional force is 3.41\times 10^5\ J

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

W=k_f-k_i

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)

W = −340605 J

or

W=3.41\times 10^5\ J

Hence, the correct option is (a).

6 0
3 years ago
A box is pulled to the right with a force of 65 N at an angle of 58 degrees to the horizontal. The surface is frictionless. The
Citrus2011 [14]
The free-body diagram is missing, but I assume the only forces acting on the box are the force F pushing the box, the weight of the object and the normal reaction of the surface.

Since the weight and the normal reaction acts in the vertical (y) direction, the only force acting on the box in the horizontal (x) direction is the horizontal component of the force F, which is given by
F_x = F \cos 58^{\circ} = (65 N)(\cos 58^{\circ} )=34.4 N
And so this is the net force in the x-direction.
5 0
3 years ago
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