Ask question

Ask question

All categories

2 years ago

7

Suppose an astronaut were to visit a planet where the force of gravity is half that of Earth. His mass on that planet would be:

1 answer:

wolverine [178]2 years ago

8 0

"Mass" is what YOU ARE. It doesn't matter whether you're on Earth, on the moon, on an asteroid, on a meteoroid, on the sun, swimming in chicken soup on Saturn, or in a space capsule falling from one solar system body to another one. Your mass DOESN'T change.

Whatever mass the astronaut had after he ate that steak dinner just before he was launched, THAT's the mass he has while he's on the way to another planet, and THAT's the same mass he'll have when he gets there. It makes no difference what the gravity is like on the other planet. His mass will not change.<em> (choice-B)</em>

Only his WEIGHT changes when he's in different places.

You might be interested in

Matt, Erin, and Lauren were having a challenge to see who could dissolve a one-pound cube of sugar the quickest.

AnnZ [28]

It should be letter a

7 0

3 years ago

A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As th

Umnica [9.8K]

Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

We need to calculate the difference between each of the lengths

Using formula for length

$\Delta L=L_{2}-L_{1}$

$\Delta L=74.7-58.6$

$\Delta L=16.1\ m$

For an open-open tube,

We need to calculate the fundamental wavelength

Using formula of wavelength

$\lambda=2\Delta L$

Put the value into the formula

$\lambda=2\times16.1$

$\lambda=32.2\ cm$

We need to calculate the frequency of the tuning

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{343}{32.2\times10^{-2}}$

$f=1065.2\ Hz$

$f=1.065\ kHz$

Hence, The frequency of the tuning is 1.065 kHz

3 0

2 years ago

What means 10² what we call this

damaskus [11]

we call it as well as 100 . so easey

7 0

3 years ago

Read 2 more answers

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

antoniya [11.8K]

Answer:

$r=5.278\times 10^{-4}\ m$

Explanation:

Given that:

magnetic field intensity, $B=0.07\ T$

kinetic energy of electron, $KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J$

we have mass of electron, $m=9.1\times 10^{-31}\ kg$

<em>Now, form the mathematical expression of Kinetic Energy:</em>

$KE= \frac{1}{2} m.v^2$

$1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2$

$v^2=4.2198\times 10^{11}$

$v=6.496\times 10^6\ m.s^{-1}$

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

$r=\frac{m.v}{q.B}$

$r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}$

$r=5.278\times 10^{-4}\ m$

6 0

3 years ago

I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2

olga nikolaevna [1]

Answer:

Explanation:

$a = \frac{v_f-v_0}{t}$ which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator. For us:

$a=\frac{2.7-1.6}{1.2}$ so

a = .92 m/s/s (NOT negative because you're speeding up)

5 0

2 years ago