All categories

3 years ago

29. If you traveled to Mars, your mass would be than your mass on Earth. *

Physics

2 answers:

marin [14]3 years ago

8 0

Yes it would be the same your weight would change:)

Bingel [31]3 years ago

8 0

If you traveled to Mars, your mass would be exactly, precisely equal to your mass on Earth, PROVIDED you did not relieve the boredom of space travel by eating too many astronaut steaks along the way.

The only things that can affect your mass are your personal diet, exercise, sleep, and stress habits.

Your mass has nothing to do with gravity. Your WEIGHT has.

You might be interested in

What is robot??????????

S_A_V [24]

Answer:

a machine capable of carrying out a complex series of actions automatically, especially one programmable by a computer.

6 0

3 years ago

When a crown of mass 14.7 kg is submerged in water, an accurate scale reads only 13.4 kg. What material is the crown made of?

Bogdan [553]

Answer:

ρ = 1.13 10⁴ km/m³

Explanation:

For this exercise we use Newton's equilibrium equation

B –W + W_scale = 0

Where B is the thrust and W_scale is the balance reading

The push is given by Archimedes' law

B = ρ_water g V

B = W- W_scale

B = m g - m_scale g

Let's calculate

B = 14.7 9.8 - 13.4 9.8

B = 12.74 N

ρ_water g V = 12.74

V = 12.74 / ρ_water g

V = 12.74 / 1000 9.8

V = 0.0013 m³

Let's use density

ρ = m / V

We replace

ρ = 14.7 / 0.0013

ρ = 1.13 10⁴ km/m³

3 0

3 years ago

What is the difference between deviation and incidence

yKpoI14uk [10]

Explanation:

Deviation :  

It means the difference between a expected value of a measurement or an observation vs the actual value.

Incidence :

A straight line, ray of light, etc.., hits a surface at a point.

3 0

3 years ago

Read 2 more answers

Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha

Ksivusya [100]

Answer:

$A_c=87.73*10^{21}m/s$

Explanation:

From the question we are told that

$r=5\times 10^{-11}$

$T=1.5 \times 10^{-16}$

Generally the equation for velocity is mathematically given as

$Velocity (v)=\frac{2 \pi r}{t}$

$V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}$

Generally the equation for Centripetal acceleration is mathematically given as

$A_c=\frac{V^2}{r}$

$A_c=(\frac{20.944*10^5)}{r5*10^{-11}}$

$A_c=87.73*10^{21}m/s$

8 0

3 years ago

A train moving with a velocity of 42.9 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s^2 North until

ankoles [38]

Answer:

3658.24m

Explanation:

Hello!

the first thing that we must be clear about is that the train moves with constant acceleration

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

$\frac {Vf^{2}-Vo^2}{2.a} =X$

Vf = final speed =160km/h=44.4m/s

Vo = Initial speed =42.9km/h=11.92m/s

A = acceleration =0.25m/s^2

X = displacement

solving

$\frac {44.4^{2}-11.92^2}{2.(0.25)} =X\\X=3658.24m$

the distance traveled by the train is 3658.24m

5 0

3 years ago