1. An astronaut in a spacesuit has a mass of 80 kilograms. What is the weight of this astronaut on the surface of the Moon where the strength of gravity is approximately 1/6 that of Earth? (2 Points)

1. An Astronaut In A Spacesuit Has A Mass Of 80 Kilograms. What Is The Weight Of This Astronaut On The

Answers

Answer 1
Weight = 80 x (9.8/6) = .... N

Related Questions

A system of 18 electrons and 11 neutrons has a net charge of

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Answer:

-28.8 × 10-19 C

describe three events you cannot explain about matter and energy

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Answer:

describe three events you cannot explain about matter and energy

Explanation:

as anything that has mass and takes up space. Matter is found in 3 major states; solid, liquid and gas. ... Atoms are the smallest particle of matter. They are so small that you cannot see them with your eyes or even with a standard microscope.

Hope that helped.

Answer:

1. Different kinds of balls bounce to different heights when dropped on the same floor.

2. Sugar dissolves faster in hot milk than in cold milk.

3. Plants grow more slowly when they are not near a window.

Explanation:

PLATO/edmentum

Please help me with this.

Answers

It is not velocity because velocity includes a direction and this does not.
velocity is direction

What is the Speed of child falling from tree 5.0 meters high

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What are the answers givin?

Please help willing to give brainliest

Electromagnetic waves do not require a medium to travel. Which one
below is an example of an electromagnetic wave?

1.ultraviolet waves
2.sound waves
3.water waves

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Answer:

sound water because sound is the most important electronic wave for water

Predict the magnitude of force applied on you if you push against a tree with a force of 50 and directed to the

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Answer: -50N

Explanation:

Newton's 3rd law says that for every force applied, there is an opposite force with equal magnitude. The person pushing on the tree will experience a force of -50N

Black and splinter cleavage barely scratches glass​

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Answer:

oh I know that sounds good to me

1)
What is the velocity in meters per second of a runner who runs exactly 110 m toward the beach in 72 seconds

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Answer:

1.53 m/s toward the beach

Explanation:

1.53 m/s toward the beach

Explanation:

The magnitude of the velocity of the runner is given by:

where

d is the displacement of the runner

t is the time taken

In this case, d=110 m and t=72 s, so the velocity of the runner is

Velocity is a vector, so it consists of both magnitude and direction: we already calculate the magnitude, while the direction is given by the problem, toward the beach.

I need help with these 2 questions pls help

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the measure of how many a wave passes on a certain point in a certain time is

frequency

wheel and axle is also called a continuous lever​

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Explanation:

When load is connected to axle and effort is on the wheel it acts as first class lever with in fulcrum in the middle ..So wheel and axle is continuous lever

Wheel and axle is called continuous lever, why? When load is connected to axle and effort is on the wheel, it acts as a first class lever with fulcrum in the middle. ... So, it is called a continuous lever. Hopes that helped.


how long will a speed boat cross a 2.o km lake with a speed of 40 km/hr​

Answers

Explanation:

s=d/t

t= d/s

t= 2/40

t= 0.05 hr

100 POINTS. PLEASE PROVIDE EXPLANATION FOR PART D

Answers

Answer:

0.298

Explanation:

Work equals change in energy.  So the work done by friction equals the final kinetic energy minus the initial potential energy.

W = ΔE

Fd = ½ mv² − mgh

Nμ (-d) = 46.0 J − (25 N) (3.30 m)

-Nμ d = -36.5 J

Using a free body diagram, the normal force is:

N = mg cos θ

N = (25 N) (4.90 m / d)

N = 122.5 J / d

Nd = 122.5 J

Therefore:

-(122.5 J) μ = -36.5 J

μ = 0.298

What is the net force on a 1200-kg car that is accelerating at 8.5 m/s2?

Answers

Answer:

10,200 N

Explanation:

Given:

Mass of the car m = 1200 kg

Acceleration of the car a = 8.5 m/s^2

Force F=?

[tex]\because F = ma\\

\therefore F = 1200 \times 8.5\\

\therefore F = 10,200 N \\ [/tex]

In the International Space Station which orbits Earth, astronauts experience apparent weightlessness because:_________
a) the astronauts and the station are in free fall towards the center of the Earth.
b) there is no gravity in space. the station is kept in orbit by a centrifugal force that counteracts the Earth's gravity.
c) the station's high speed nullifies the effects of gravity.
d) the station is so far away from the center of the Earth.

Answers

Answer:

a) the astronauts and the station are in free fall towards the center of the Earth

Explanation:

Weightlessness is only a sensation of zero weight, a body experiences when it is in free fall towards the center of the Earth, caused by lack of contact force on such body.

Weightlessness doesn't mean the object has zero actual weight, is just a sensation of no weight due to lack of contact force to produce upward reaction on the object which gives the real sense of ones weight.

Thus, the astronauts experience apparent weightlessness because:

a) the astronauts and the station are in free fall towards the center of the Earth.

If you want to make a strong battery, should you pair two metals with high electron affinities, low electron affinities, or a mix? Explain your answer.

Answers

A battery has two electrodes, at one end it has the anode and the other end has the cathode. Electrons travel through the circuit from the anode (negative) to the cathode (positive), and this is the driving force that provides electricity to flow through circuits.
The anode needs to have a low electron affinity because it needs to readily release electrons, and the cathode needs to have a high electron affinity because it needs to readily accept electrons.

What does Weber's Law about 'just noticeable differences' predict about how much someone has to change the brightness of a light before we can notice the difference? a. It depends on how bright the light was in the first place - the brighter it was, the less change is needed before we realize it. b. It depends on how long we have been looking at the light - the longer we have been looking, the more change is needed. c. It is always the same amount - 7 lux. d. It depends on how bright the light was in the first place - the brighter it was, the more change is needed before we realize it.

Answers

Answer:

answer A is the correct one

Explanation:

Weber's law states that  "the smallest discernible change of a stimulus and proportional to the stimulus".

Applying this law to cases of optical intensity, the ratio must be

          k = cte = ΔI / I

where ΔI is the variation of the intensity and I is the value of the intensity

In general, for humans, the constant is 0.15 for the rods and 0.015 for the cones of the retina.

When reviewing the answers, answer A is the correct one, since in order for the previous relationship to be maintained, the magnitudes must rise proportionally

If an object is on a 27º frictionless incline, what will be the acceleration of the object on the incline?

Answers

The acceleration is g Sin (27)
= 9.8 x Sin (27) = ..... m/s^2

Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.
What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?

Answers

Answer:

a)    F = -1.82 10⁻¹⁵ N,  b) K = 9.1 10⁻¹⁶ J

Explanation:

a) To calculate the force between the nucleus and the electrons, let's use the Coulomb equation

           F = k q Q / r²

as the nucleus occupies a very small volume compared to electrons, we can suppose it as punctual

let's calculate

          F = 9 10⁹ (-1.6 10⁻¹⁹) (79 1.6 10⁻¹⁹) / (10⁻¹⁰)²

          F = -1.82 10⁻¹⁵ N

b) they ask us for kinetic energy

let's use Newton's second law

         F = m a

acceleration is centripetal

         a = v² / r

we substitute

         F = m v² / r

         v = √ (F r / m)

         v = √ (1.82 10⁻¹⁵ 10⁻¹⁰ / 9.1 10⁻³¹)

         v = √ (0.2 10⁻¹⁶)

         v = 0.447 10⁸ m / s

kinetic energy is

          K = ½ m v²

          K = ½ 9.1 10⁻³¹ (0.447 10⁸)²

          K = 0.91 10⁻¹⁵ J

          K = 9.1 10⁻¹⁶ J

if a bust starts to move and its velocity becomes 90 km after 8 seconds . calculate its acceleration answer it quick please

Answers

Answer:

a = 3.125 [m/s^2]

Explanation:

In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.

[tex]90\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s} \\= 25 \frac{m}{s}[/tex]

[tex]v_{f} =v_{i} + (a*t)[/tex]

where:

Vf = final velocity = 25 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t  = time = 8 [s]

The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.

25 = 0 + a*8

a = 3.125 [m/s^2]

As the chief design engineer for a major toy company, you are in charge of designing a loop-the-loop toy for youngsters. The idea is that a ball of mass m and radius r will roll down an inclined track and around the loop without slipping. The ball starts from rest at a height h above the tabletop that supports the whole track. The loop radius is R. Determine the minimum height h, in terms of R and r, for which the ball will remain in contact with the track during the whole of its loop-the-loop journey

Answers

Answer: h = 2.7 R - 1.7 r

Explanation:    

normally, force is 0 at the top ;

-N - mg = - m v^2 / ( R - r )

-0 - mg = (- mv^2) / ( R-r )

mg = (mv^2) / (R - r)

g = v^2 / ( R - r ) ; ----------------equation 1

conservation of energy ;

ΔK + ΔP = 0 ;

1/2 I ω^2 + 1/2 m v^2 + mg ( h2 - h1 ) = 0 ;  

0.5 * ( 2/5 ) m r^2 * ( v / r )^2 + 0.5 m v^2 + mg ( ( 2R - r ) -h ) = 0 ;

0.5 * ( 2/5 ) m r^2 * ( v / r )^2 + 0.5 m v^2 = mg ( - 2R + r + h ) ;

0.5 * ( 2/5 )r^2 * ( v / r )^2 + 0.5 v^2 = g ( - 2R + r + h ) ;

0.5 * ( 2/5 ) v^2 + 0.5 v^2 = g ( - 2R + r + h ) ;

[ 0.5 * ( 2/5 ) + 0.5 ] v^2 = g ( - 2R + r + h ) ;-------------equation 2

from equation 1 ,  v^2 = g ( R - r ), input in equation 2

[ 0.5 * ( 2/5 ) + 0.5 ] [ g ( R - r ) ] = g ( - 2R + r + h )

[ 0.5 * ( 2/5 ) + 0.5 ] [ ( R - r ) ] = ( - 2R + r + h )

0.7 ( R - r ) = h - 2R + r

0.7R - 0.7r = h - 2R + r

solve for h

h = 0.7R + 2R - 0.7r - r

h = 2.7 R - 1.7 r

a negatively charged particle is attracted to: A) all particles that are located close by B)negatively charged particles C) positively charged particles D)only particles that are large

Answers

Answer:

The answer is C) Positively charged particles

Explanation:

Particles with the same charge are repelled by one another, while particles with different charges are attracted to one another.

Answer:

positively charged particles (c)

Explanation:

I got it right on Study Island

The day time temperature of Mercury is 800 F . What is Mercury’s temperature in Celsius and kelvin?

Answers

Answer:

In Celsius it is 426.667

In Kevin it is 699.817

Explanation:

If the efficiency of a machine is equal to 85% and the work input is 10.0 , what is the work output?

Answers

Answer:

8.5

Explanation:

to find efficiency take the ratio between the output and input

85/100=output/input

What determines the radiation that an electromagnetic wave emits?

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The question is a little confusing. An electromagnetic wave is radiation. One doesn’t emit the other. Take another look at the question and write it again please.

Pretty Easy question please answer only 20 minutes left:
A summary of the results of a scientific investigation is called a:
observation
research
hypothesis
conclusion

Answers

Answer:

D. Conclusion.

Explanation:

9.
Un Pldle UI TUL!
What acceleration is produced on a mass of 200 g, when a force of 10 N is
exerted on it ?​

Answers

Given parameters:

Mass of the body  = 200g

Force on the body  = 10N

Unknown parameters:

Acceleration produced by the force  = ?

To solve this problem we must first define force in terms of mass and acceleration. This is possible due to the Newton's first law of motion.

  Force  = mass x acceleration

Here the unknown is acceleration and we can easily solve for it.

But we must take the mass to kilogram in order for it to cancel out.

        1000g  = 1 kg

        200g  = x kg =   [tex]\frac{200}{1000}[/tex]   = 0.2kg

Now input the parameters and solve;

         10  = 0.2 x acceleration

   Acceleration  = [tex]\frac{10}{0.2}[/tex]   = 50m/s²

The acceleration produced by the body is 50m/s²

HELP!!! HURRY!!!!! 100 POINTS!!!!!!!!!

Answers

what happens at Point C is sublimation. the increase in temperature affects the Vapour pressure soon as you can see the curve is increasing with increasing pressure there is increase in temperature that is the sublimation Curve

Answer:

what happens at Point C is sublimation. the increase in temperature affects the Vapour pressure soon as you can see the curve is increasing with increasing pressure there is increase in temperature that is the sublimation Curve.

re testing a new amusement park roller coaster with an empty car with a mass of 100 kg. One part of the track is a vertical loop with a radius of 12.0 m. At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of 8.00 m/s . You may want to review (Pages 203 - 212) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of A vertical circle with friction. Part A As the car rolls from point A to point B, how much work is done by friction?

Answers

Answer:

-4530 J

Explanation:

Given that

Mass of the car, m = 100 kg

Speed of the car at point A, v1 = 25 m/s

Speed at point B, v2 = 8 m/s

Radius of the track, r = 12 m and with respect to the origin of the center of the track, we say that y1 = -12 at point A and y2 = 12 at point B

We also know that

W(total) = W(grav) + W(other) = K₂ - K₁

Work done by the gravitational force, W(grav) = -U(grav) = mgy1 - mgy2

Kinetic Energy, K = ½mv²

Adding all together, we have

½mv₁² + mgy1 + W(other) = ½mv₂² + mgy2

½ * 100 * 25² + 100 * 9.8 * -12 + W = ½ * 100 * 8² + 100 * 9.8 * 12

50 * 625 + 980 * -12 + W = 50 * 64 + 980 * 12

31250 - 11760 + W = 3200 + 11760

19490 + W = 14960

W = 14960 - 19490

W = -4530 J

An object starts from rest and accelerates at a rate of 4 m/s2 for 3 seconds. What is it's displacement from the start position?

Answers

Answer:

36m

Explanation:

a = 4 m/s^2

u = 0 ( it's starting from rest )

t = 3

a=v-u/t ....... 4=(v-0)/3, v = 12m/s

Using s=d/t,

12=d/3,

d= 36m

If an object starts from rest and accelerates at a rate of 4 m/s2 for 3 seconds, then its displacement from the start position would be 44.1 meters.

What are the three equations of motion?

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2 × a × t²

v² - u² = 2 × a × s

As given in the problem If an object starts from rest and accelerates at a rate of 4 m/s² for 3 seconds,

S = ut + 1/2 × a × t²

S = 0 + 0.5 × 9.8 × 3²

S = 44.1 meters

Thus, If an object starts from rest and accelerates at a rate of 4 m/s2 for 3 seconds, then its displacement from the start position would be 44.1 meters.,

To learn more about equations of motion here, refer to the link given below ;

brainly.com/question/5955789

#SPJ2

Assume you push a 20 kg child in a little red wagon with a force of 100 N. However, the dirt and rocks below the wheels
create a frictional force of 25 N. What is the acceleration on the wagon and child?
A. 1.25 m/s2
B.3.75 m/s2
C. 5 m/s2
D. 0.2 m/s2

Answers

Answer:

a

Explanation:

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