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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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Pls I need help for the problem solving part.
Bogdan [553]

i don't know the anss , sorry.

3 0
3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
3 years ago
Know the monosaccharides, the disaccharides and the polysaccharides. Know the hormones involved in glucose regulation.
BabaBlast [244]

Answer:

Explanation:

Monosaccharides are the simplest form of carbohydrates which are called as single sugars. These are the building blocks of bigger carbohydrates.

Disaccharides are the sugars that are formed when two monosaccharides combine together by glycosidic bonds.

Polysaccharides are the long chains of carbohydrate molecules. These are formed by the monosaccharide units bonded by the glycosidic linkages.

The insulin and glucagon are the two hormones secreted by the pancreas that regulate the blood glucose levels. Insulin is secreted by the beta cells of pancreas. It is secreted when the blood glucose level is high. Glucagon is secreted by the beta cells of pancreas when the blood glucose level is low.  

7 0
3 years ago
A 58 kg skier is going down a 35 degree slope. The areaof each
maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

N = mgcos\theta

Since the skier is standing on two skis, his weight will be divide by two

N' = \frac{mgcos\theta}{2}

Pressure is given as the force applied in a given area, that is

P = \frac{F}{A}

Replacing F with N'

P = \frac{N'}{A}

P = \frac{\frac{mgcos\theta}{2}}{A}

Our values are given as,

m = 58kg

g = 9.8m/s^2

\theta = 35\°

A = 0.3m^2

Replacing we have that

P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}

P = 776.01Pa

Therefore the pressure exerted by each ski on the snow is 776.01Pa

6 0
3 years ago
The speed of the wave depends on what?
kolbaska11 [484]

The speed of a mechanical wave depends on the mechanical properties
of the medium through which the wave is traveling.

The speed of an electromagnetic wave depends on the electrical properties
of the medium through which the wave is traveling.

It's pretty well unanimous for choice <em>A</em> .


4 0
3 years ago
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