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11Alexandr11 [23.1K]
3 years ago
11

A space-walking astronaut has become detached from her spaceship. She's floating in space with her handy tool belt attached to h

er waist, thinking about how she might get back to the ship, which she can see 50 meters from her current location.
How can she get back to the ship? Use language from the Laws of Motion in your answer.
Physics
1 answer:
ziro4ka [17]3 years ago
8 0

Astronauts on a spacewalk use the Newton's third law for navigation. Newton's third law states that every action has an equal and opposite reaction. Action and reaction forces act on different bodies.If  a person applies a force on an object, the object applies an equal and opposite force on the person.

The astronaut who has been detached from her spaceship, has a handy tool belt attached to her waist. If she throws a tool in a direction opposite to the direction of the location of the spaceship, the tool exerts an equal force on the astronaut, propelling her in the direction of the ship. To reach the ship, she can keep throwing tools, taking care to throw it  exactly in the opposite direction in which she needs to travel.

Every time a tool is thrown backwards, the astronaut will experience a force in the forward direction in accordance with Newton's third law, ultimately enabling her to get back to the ship.

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Is Environmental Nature or Nurture?
erma4kov [3.2K]
ANSWER: NATURE




EXPLAINTION:
4 0
2 years ago
A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration
Tpy6a [65]

Answer:

7.5 m/s

Explanation:

We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:

v^2 = u^2 + 2as

where v = final velocity

u = initial velocity

a = acceleration

s = distance traveled

From the question:

u = 28 m/s

a = -4 m/s^2

s = 91 m

Therefore:

v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s

The velocity of the boat when it reaches the buoy is 7.5 m/s.

6 0
2 years ago
Solving:
STatiana [176]

Answer:

1. 31,536,000 seconds

2. Car B traveled a longer distance

3. Volume of box = 0.235887 cubic meters

Explanation:

Q1. Age in seconds

1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds

Therefore 1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

In scientific notation this would be 3.1536\times  10^7 \text{ seconds}

Q2. Comparing km and miles

Given:
1 km = 1000m and 1 m = 3.3ft,   I km = 1000 x 3.3 = 3300 ft.

Convert Car A distance of 25.7km to feet :
25.7 km. = 25. 7 x 3300 ft. = 84,810 ft.

For Car B  that traveled 20 miles,
20 miles = 20 x 5280 = 105,600 ft.

Since 105,600 > 84,810, car B traveled a longer distance

Q3. Volume of wooden box

The wooden box is in the shape of a rectangular prism
It volume is L x W x H
Volume = 1.525 x 0.30 x 0.5156 = 0.235887 cubic meters

5 0
9 months ago
From 0 seconds to 4 seconds is the acceleration positive or negative?
Alenkinab [10]

Answer:

From 0 -4 seconds the acceleration is positive. (The graph is going upwards.)

From 6-10 seconds the acceleration is negative. (The graph is going downwards.)

7 0
3 years ago
A sound from a source has an intensity of 270 dB when it is 1 m from the source.
Rufina [12.5K]
Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: \frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2
where
I_{2} is the intensity at distance 2
I_{1} is the intensity at distance 1
d_{2} is distance 2
d_{1} is distance 1

We can infer for our problem that I_{1}=270, d_{1}=1, and d_{2}=3. Lets replace those values in our formula to find I_{2}:
\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2
\frac{I_{2} }{270} =( \frac{1}{3} )^2
\frac{I_{2} }{270} = \frac{1}{9}
I_{2}= \frac{270}{9}
I_{2}=30 dB

We can conclude that the intensity of the sound when is <span>3 m from the source is 30 dB.</span>
8 0
3 years ago
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