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3 years ago

If an astronaut is 66 kg, what is their weight in N on the moon?

2 answers:

6 0

Mass isn’t dependent on gravity, so the man would still be 66
66
k
g
. His weight, however, is calculated using the formula:

=
w
=
m
g

So, weight, equal mass times the acceleration due to gravity. On the moon, acceleration due to gravity is 1.62/²
1.62
m
/
s
²
. If we plug these “givens” in, we get:

=(66)(1.62/²)
w
=
(
66
k
g
)
(
1.62
m
/
s
²
)

Therefore, weight equals 106.92
106.92
N
e
w
t
o
n
s
(the unit of weight)

Blababa [14]3 years ago

5 0

The mass of the person would be 66 kg on both Earth and the Moon.

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Physics B 2020 Unit 3 Test

weqwewe [10]

Answer:

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

$F=qvB sin \theta$

where here:

For the proton in this problem:

$q=1.602\cdot 10^{-19}C$ is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

$\theta=65^{\circ}$ is the angle between the directions of v and B

So the force is

$F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N$

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

The right hand rule gives the direction of the force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

The radius of a particle moving in a magnetic field is given by:

$r=\frac{mv}{qB}$

where here we have:

$m=6.64\cdot 10^{-22} kg$ is the mass of the alpha particle

$v=2155 m/s$ is the speed of the alpha particle

$q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C$ is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

$r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m$

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

$f=\frac{qB}{2\pi m}$

where here:

$q=1.602\cdot 10^{-19}C$ is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

$m=9.31\cdot 10^{-31} kg$ is the mass of the electron

Substituting, we find:

$f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz$

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if $\Delta \Phi \neq 0$ , which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.

The flux is calculated as

$\Phi = BA sin \theta$

where

B = 5.5 T is the strength of the magnetic field

A is the area of the coil

$\theta=18^{\circ}$ is the angle between the direction of the field and the plane of the loop

Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is

$A=(0.15 m)(0.08 m)=0.012 m^2$

So the flux is

$\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb$

See the last 7 answers in the attached document.

Download docx

5 0

3 years ago

Scientists need to know how to make measurements

Verizon [17]

Part of the scientific process involves sharing your results with other scientists. To do this, we all need to use the same measurement system, which you'll learn about in this lesson.

Imagine you're trying to find out how much an elephant weighs. You're pretty sure it weighs a lot, but you don't know the exact number. So you ask your teacher, and she tells you an elephant weighs the same as three hippos.

Well that's nice to know, but how much does a hippopotamus weigh? Again, you ask your teacher, and she tells you a hippopotamus weighs the same as five alligators. That's a cool fact to know, but you still don't understand how much an elephant weighs because comparing elephants to alligators can be confusing.

plz mark me as brainliest :)

4 0

3 years ago

Given two positively charged particles, of equal magnitude, separated by a distance “d”. What will happen to the force field bet

Ulleksa [173]

The force between the two particles will quadruple

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{d^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

d is the separation between the two charges

In this problem, let's call F the initial force between the two charges, when they are at a distance of d.

Later, the distance between the two particles is halved, so the new distance is:

$d'=\frac{d}{2}$

This means that the new electrostatic force will be:

$F'=k\frac{q_1 q_2}{(d/2)^2}=4(k\frac{q_1 q_2}{d^2})=4F$

Therefore, the force between the two particles will quadruple.

Learn more about electrostatic force:

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5 0

3 years ago

Two small pith balls, each of mass m = 14.2 g, are suspended from the ceiling of the physics lab by 0.5 m long fine strings and

vichka [17]

Answer:

1.424 μC

Explanation:

I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),

tension T (acting along the string - to the pivot point), and

F (electric force – acting along the line connecting the charges).

We then have something like this

x: T•sin α = F,

y: T•cosα = mg.

Dividing the first one by the second one we have

T•sin α/ T•cosα = F/mg, ultimately,

tan α = F/mg.

Since we already know that

q1=q2=q, and

r=2•L•sinα,

k=9•10^9 N•m²/C²

Remember,

F =k•q1•q2/r², if we substitute for r, we have

F = k•q²/(2•L•sinα)².

tan α = F/mg =

= k•q²/(2•L•sinα)² •mg.

q = (2•L•sinα) • √(m•g•tanα/k)=

=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =

q = 0.486 • √(8.61•10^-12)

q = 0.486 • 2.93•10^-6

q = 1.424•10^-6 C

q = 1.424 μC.

8 0

3 years ago

Select all that apply. The resistance to an object being pulled over a surface is called _____. starting friction, sliding frict

jok3333 [9.3K]

That is kinetic friction.

8 0

3 years ago

Read 2 more answers