Answer:
a) the plastic tube need to be 24.7 m long
b) the kilometer of copper wire required is 15.7
c) the resistance of this solenoid is 5538.8 2 ohms
Explanation:
Given the data in the question;
we determine the length of the plastic tube. assuming the solenoid is long.
the self inductance of a long solenoid is;
L = μ₀n²πr²l
μ₀ = 4π × 10⁻⁷ T-m/A
where
n = number of turns per unit length
r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)
l = length of the solenoid or the length of the plastic tube
we find n = number of turns per unit length
given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter
number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter
So from our previous formula, we find l
L = μ₀n²πr²l
we substitute
1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)
1 = 0.04048 × l
l = 1 / 0.04048
l = 24.7 m
Therefore, the plastic tube need to be 24.7 m long
b)
n = number of turns per unit length = 1265.8 turns/metre
so, the length of the plastic tube over which the copper wire is to be wound,
number of turns of copper wire required = n × l
= 1265.8 turns/meter × 24.7 m
= 31,265.26 turns
Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.
so for each turn length of copper wire required = 2π × r
= 2π × 0.08 m
= 0.5026548 m
So copper wire required for 31,265.26 turns will be;
⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km
Therefore, the kilometer of copper wire required is 15.7
c)
p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m
Resistance = pl/a
where l is length of copper wire, a is cross sectional area;
diameter of copper wire is 0.79-mm
radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m
area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²
Resistance = pl/a
we substitute
Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]
Resistance = 5538.8 2 ohms
Therefore, the resistance of this solenoid is 5538.8 2 ohms
5.
2075 Set A Q.No. 20 2070 Supp. Set B Q.No. 2 B What
happens to the kinetic energy of photo electrons when
intensity of light is doubled?
[2]
Answer:
The energy of each photon can be transformed into kinetic energy and as this energy does not change, the energy of both photoelectrons remains constant,
Explanation:
The photoelectric effect was explained by Einstein, who assumed that the lz is made up of particles called photons each of a given energy, therefore the photoelectric effect can be explained as a collision of particles.
From this explanation we see that the intensity is proportional to the number of existing particles, when we double the intensity we double the number of particles, but the energy of each particle does not change, therefore if we use the conservation of energy.
The energy of each photon can be transformed into kinetic energy and as this energy does not change, the energy of both photoelectrons remains constant, only the number of electrons expelled changes.
On the Moon's surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. The Earth's atmosphere slows down light. Assume the distance to the Moon is precisely 3.84×108 m, and Earth's atmosphere (which varies in density with altitude) is equivalent to a layer 30.0 km thick with a constant index of refraction n=1.000293. What is the difference in travel time for light that travels only through space to the moon and back and light that travels through the atmosphere and space?
Answer:
a) space only t = 1.28 s
b) space+ atmosphere t_ {total} = 1.28000003 s
Explanation:
The speed of light in each material medium is constant, which is why we can use the uniform motion relations
v= x / t
a) let's look for time when it only travels through space
t = x / c
t = 3.84 10⁸/3 10⁸
t = 1.28 s
b) we look for time when it travels part in space and part in the atmosphere
space
as it indicates that the atmosphere has a thickness of e = 30 10³ m
t₁ = (D-e) / c
t₁ = (3.84 10⁸ - 30.0 10³) / 3 10⁸
t₁ = 1.2799 s
atmosphere
we use the refractive index
n = c / v
v = c / n
we substitute in the equation of time
t₂ = e n / c
t₂ = 30 10³ 1,000293 /3 10⁸
t₂ = 1.000293 10⁻⁴ s
therefore the total travel time is
t_ {total} = t₁ + t₂
t_ {total} = 1.2799+ 1.000293 10⁻⁴
t_ {total} = 1.28000003 s
we can see that the time increase due to the atmosphere is very small
Two blocks (with masses of 2.0 kg and 4.0 kg) are on a bench tied together with string. They are being pulled to the right with a force of 30N. They each experience a friction force between the block and the bench.
(Refer to image)
The 2 kg block experiences a friction force with a friction coefficient of 0.30 and the 4 kg experiences a friction with a friction coefficient of 0.20.
Assume that g (the acceleration due to gravity) is 10.0 m/s/s.
Find the magnitude of the friction forces. Find the magnitude of the acceleration of the blocks. Use these answers to help you find the answer to the question:
What is tension in the string connecting the two blocks? (Submit just this answer in Newtons)
Answer:
T = 34/3 N
Explanation:
Magnitude of the friction force on 2kg block = 0.3x10x2 = 6N
Magnitude of the friction force on 4kg block = 0.2x10x4 = 8N
Magnitude of the acceleration of the blocks
F = ma
30 - 8 - 6 = (2+4)a
a = 8/3 m s^-2
Tension in the string connecting the two blocks
Consider the 2kg block,
T - f = ma
T - 6 = 2(8/3)
T = 34/3 N
A students walks at a rate of 4 miles per hour to school. If she leaves her
house at 7:40am how long will it take her to travel 2 miles?
Answer:
30 minutes or 1/2 hour
she'll get there at 8:10am but that's not important
Explanation:
u can divide 4mph by two to find how long it would take her to travel 2 miles
she travels at 2 miles per 1/2 hour
hope this helps chu <3
A large wooden block of weight 30n, is observed to push down a ramp. The block is pushed down an inclined plane as shown below and the applied force is parallel to the plane and has a magnitude of 5 n as shown. The force of kinetic friction is 17n the block is pushed 6m down along the inclination of the ramp is 30 degrees with the horizontal
Answer:
OMG IM ON THE SAME QUESTION
Explanation:
a boy throws a ball straight up into the air it reaches the highest point of its flight after 4 seconds how fast was the ball going when it left the boy's hand
Answer:
Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative. The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.
uniform solid sphere has a mass of 1.765 kg and a radius of 0.854 m.a. Find the torque required to bring the sphere from rest to an angular velocity of 317 rad/s, clockwise, in 15.5 s.b. What magnitude force applied tangentially at the equator would provide the needed torque
Answer:
a) the torque required is 10.53 N-m
b) The magnitude force applied tangentially is 12.33 N
Explanation:
Given the data in the question;
mass m = 1.765 kg
radius r = 0.854 m
first we calculate the moment of inertia;
[tex]I[/tex] = [tex]\frac{2}{5}[/tex]mr²
we substitute
[tex]I[/tex] = [tex]\frac{2}{5}[/tex] × 1.765 × (0.854)²
[tex]I[/tex] = 0.514897 kg.m²
a)
Find the torque required to bring the sphere from rest to an angular velocity of 317 rad/s, clockwise, in 15.5 s
ω[tex]_{initial[/tex] = 0
ω[tex]_{final[/tex] = 317 rad/s
t = 15.5 s
we know that; ω[tex]_{final[/tex] = ω[tex]_{initial[/tex] + ∝t
so we substitute
317 = 0 + ∝(15.5)
∝ = 317 / 15.5
∝ = 20.4514 rad/s²
so
ζ = [tex]I[/tex] × ∝
we substitute
ζ = 0.514897 × 20.4514
ζ = 10.53 N-m
Therefore, the torque required is 10.53 N-m
b)
What magnitude force applied tangentially at the equator would provide the needed torque.
ζ = F × r
we substitute
10.53 = F × 0.854
F = 10.53 / 0.854
F = 12.33 N
Therefore, magnitude force applied tangentially is 12.33 N
Calculate the first and second order angles for light of wavelength 400. nm and 700. nm of the grating contains 1.00 x 104 lines/cm.
Answer:
[tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex]
[tex]44.43^{\circ}[/tex], second order does not exist
Explanation:
n = Number of lines grating = [tex]1\times10^4\ \text{Lines/cm}[/tex]
[tex]\lambda[/tex] = Wavelength
m = Order
Distance between slits is given by
[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{1\times 10^4}\\\Rightarrow d=10^{-6}\ \text{m}[/tex]
[tex]\lambda=400\ \text{nm}[/tex]
m = 1
We have the relation
[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}\\\Rightarrow \theta=\sin^{-1}\dfrac{1\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=23.58^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=53.13^{\circ}[/tex]
The first and second order angles for light of wavelength 400 nm are [tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex].
[tex]\lambda=700\ \text{nm}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 700\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=44.43^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 700\times 10^{-9}}{10^{-6}}[/tex]
Here [tex]\dfrac{2\times 700\times 10^{-9}}{10^{-6}}=1.4>1[/tex] so there is no second order angle for this case.
The first order angle for light of wavelength 700 nm are [tex]44.43^{\circ}[/tex].
Second order angle does not exist.
an iron Tyre of diameter 50cm at 288k is to be shrank on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm.Linear expansivity of iron is 0.000012k-1
Answer:
The answer should be D
Explanation:
The gravitational force between two objects with masses 1kg and 28kg separated by a distance 7m is ____________10-11 N.
a.
3.81
b.
26.68
c.
9151.24
d.
1307.32
Hhhhhellllppp fastt
Answer:
a. 3.81
Explanation:
F = GMm/r^2
F = (6.67 x 10^-11 x 28 x 1) / 7^2
F = 3.81 x 10^-11 N
The speed limmit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.
Answer: The amount of light that enters the pupil is controlled by the Iris
Explanation:
At position A within a tube containing fluid that is moving with steady laminar flow, the speed of the fluid is 12.0 m/s and the tube has a diameter 12.00 cm. At position B, the speed of the fluid is 18.0 m/s and the tube has a diameter 6.00 cm. What is the ratio of the density of the fluid at position A to the density of the fluid at position B
Answer:
0.375
Explanation:
For incompressible flow, we know that;
ρ1•v1•A1 = ρ2•v2•A2
Where;
ρ1 = density of fluid at position A
v1 = speed of fluid at position A
A1 = area of tube
ρ2 = density of fluid at position B
v2 = speed of fluid at position B
A2 = area of tube
We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.
Thus;
ρ1/ρ2 = (v2•A2)/(v1•A1)
Now, the tube will have the same height.
But we are given;
diameter of A = 12.00 cm = 0.12 m
diameter of B = 6 cm = 0.06 m
Thus;
A1 = π(d²/4)h = πh(0.12²/4)
A2 = πh(0.06²/4)
We are also given;
v1 = 12 m/s
v2 = 18 m/s
Thus;
ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))
πh/4 will cancel out to give;
ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)
ρ1/ρ2 = 0.375
The rods, which number over 100 million, can only be activated by a certain range of wavelengths, but they do not pass any color information to the brain. In other words, they note differences in shades of grey (from black to white) and are responsible for a person's ability to see in dim light. Cones, which number around 6 million, give us color vision. Cones come in three different kinds: 64%% of cones are sensitive to long wavelengths of visible light (toward the red end of the spectrum), 32%% are sensitive to medium wavelengths, and the remaining 2%% are sensitive to short wavelengths (toward the blue end of the spectrum). Colors are differentiated on the basis of the extent to which visible light stimulates each kind of cone.
Do rods have their peak sensitivity at a higher or lower frequency than cones?
a) Higher
b) Lower
Answer:
The correct answer is a
peak sensitivity is much higher for cones
Explanation:
After reading this interesting problem, where it gives a good description of the types of photoreceptor cells that exist in the eyes
The Cone has its name because of the shape of a cone that has this shape that allows to perceive very small amounts of intensity
The Canes have the shape of a cane and are filled with a substance that is sensitive to color colors, but they need a greater intensity of light to be activated, for which reason they work in the daytime, when it gets dark the intensity of the light is insufficient to activate these cells and the only ones that send information to the brains are the cones.
With this explanation it is clear that cones have high sensitivity at all frequencies and rods have low to medium sensitivity at specific frequencies.
Therefore peak sensitivity is much higher for cones
The correct answer is a
why food cook faster with salt water than cook with pure water
Answer:
oil heats faster
Explanation:
How does the REE change as a person ages?
Explain
I go to k12 too and i am writing this test now too. I am like pretty sure my answer is correct. I guess we can kinda help eachother with this assignment cuz i am sorta stuck too.
The older a person gets, the greater REE he has.I dont know i hope this helped a bit I know its not much i need help with this too.
The average mean distance of Saturn from the sun is
Answer:
From an average distance of 886 million miles (1.4 billion kilometers), Saturn is 9.5 astronomical units away from the Sun. One astronomical unit (abbreviated as AU), is the distance from the Sun to Earth. From this distance, it takes sunlight 80 minutes to travel from the Sun to Saturn.
we have that from the Question"The average mean distance of Saturn from the sun is" it can be said that Tthe average mean distance of Saturn from the sun is
A distance of 1427 x 10^6 km or 886 696 691 milesFrom the Question we are told
The average mean distance of Saturn from the sun is
Generally
The Sun is the star of the milky way galaxy and its distance from every planet in the milky way determines in one way or another its properties and in-habitability
Saturn being a Planet of the milky way we see that Saturn is a significant distance away from sun
A distance of 1427 x 10^6 km or 886 696 691 miles
Therefore
The average mean distance of Saturn from the sun is
A distance of 1427 x 10^6 km or 886 696 691 miles
For more information on this visit
https://brainly.com/question/19007362?referrer=searchResults
Use the drop-down menus to complete each statement about tornado safety.
Before a tornado warning occurs, you should
✔ have a disaster plan ready.
If you’re indoors during a tornado, you should
✔ go to the basement.
If you’re outdoors during a tornado, you should
✔ lie in a ditch or low-lying area.
Answer:
correct tysmm
Explanation:
23
In order for a 12 Volt power source
to produce a current of 0.085 amps,
a resistance of...
[?] Ohms is needed.
Enter
Haven't learned this yet.
Answer:
141.18 ohms
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12
Current (I) = 0.085 A
Resistance (R) =?
The resistance needed can be obtained as follow:
V = IR
12 = 0.085 × R
Divide both side by 0.085
R = 12 / 0.085
R = 141.18 ohms
Therefore, a resistor of resistance 141.18 ohms is needed.
A small dog is trained to jump straight up a distance of 1.1 m. How much kinetic energy does the 7.7 kg dog need to jump this high?
Answer:
83.09 J
Explanation:
The potential energy at the point of the top of the jump is represented by the equation
[tex]mgdeltah[/tex]
when the dog jumps, all the potential energy converts to kinetic energy (1/2mv^2). Plugging in the values:
(7.7)(4.184)(1.1) = 83.0907 J
(5 Points)
a) At ground level, the pressure of the helium in a balloon is 1x105
Pa. The volume occupied by the helium is 9.6m The balloon is
released and it rises quickly through the atmosphere. Calculate
the pressure of the helium when it occupies a volume of 12m3.
(3 Marks)
b) A box is 15m below the surface of the sea. The density of sea-
water is 1020 kg/m.
Calculate the pressure on the box due to the sea-water.
(2 Marks)
Answer:
1. [tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
Explanation:
1. From Boyles' law;
[tex]P_{1}[/tex][tex]V_{1}[/tex] = [tex]P_{2}[/tex][tex]V_{2}[/tex]
[tex]P_{1}[/tex] = 1 x [tex]10^{5}[/tex] Pa
[tex]V_{1}[/tex] = 9.6 [tex]m^{3}[/tex]
[tex]V_{2}[/tex] = 12 [tex]m^{3}[/tex]
Thus,
1 x [tex]10^{5}[/tex] x 9.6 = [tex]P_{2}[/tex] x 12
[tex]P_{2}[/tex] = [tex]\frac{100000 x 9.6}{12}[/tex]
= 80000
[tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. Pressure, P = ρhg
where: ρ is the density of the fluid, h is the height/ depth and g is the acceleration due to gravity (9.8 m/[tex]s^{2}[/tex]).
Thus,
P = 1020 x 15 x 9.8
= 149940
P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine:
This question is incomplete, the complete question is;
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).
Answer:
the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Explanation:
Given the data in the question and image below and as illustrated in the second image;
distance S = 40 m
V[tex]_B[/tex] = 54 km/hr
V[tex]_A[/tex] = 72 km/hr
α = 100 m
now, angular velocity of Bxy will be;
ω[tex]_B[/tex] = V[tex]_B[/tex] / α
so, we substitute
ω[tex]_B[/tex] = ( 54 × 1000/3600) / 100
ω[tex]_B[/tex] = 15 / 100
ω[tex]_B[/tex] = 0.15 rad/s
Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
ASAP Even though the force exerted on each object in a collision is the same strength, if the objects have different masses, their will be different. * O changes in velocity O amount of force O speed and direction
Answer:
it should be changes in velocity
Explanation:
I hope this helps!
A light year is the amount of time it takes for light from the Sun to reach the Earth.
True
False
A positive charge Q2 is uniformly distributed over a nonconducting disc of radius a which has a concentric circular hole of radius b. At the center of the hole there is another nonconducting disc of radius d where a charge Q1 is uniformly distributed.
a) Find the surface charge density of the disc with the hole σ2.
b) Find the surface charge density 01 of the disc of radius d.
c) Find the total charge enclosed by the circle of radius
Answer:
a) σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] , b) σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂, σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
Explanation:
a) The very useful concept of charge density is defined by
σ = Q / A
In this case we have a circular disk
The are of a circle is
A = π r²
in this case we have a hole in the center of radius r = b, so
A_net = π r² - π r_ {hollow} ²
A_ {net} = π (a² - b²)
whereby the density is
σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]
b) The density of the other disk is
σ = Q₂ / A₂
σ = [tex]\frac{Q_2}{d^2}[/tex]
c) The total waxed load is requested by the larger circle
Q_ {total} = Q₁ + Q₂
the net charge density, in the whole system is
σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]
the area is
A_{total} = π a²
since the other circle is inside, we are ignoring the space between the two circles
σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
An atom of tin has an atomic number of 50 and a mass number of 119. How many protons, electrons, and neutrons are found in one neutral atom of tin?
O 50 protons, 69 electrons, 50 neutrons
O 50 protons, 50 electrons, 69 neutrons
69 protons, 50 electrons, 69 neutrons
69 protons, 69 electrons, 50 neutrons
Answer:
50 protons 50 electrons and 69 neutrons...
Explanation:
the number of protons is equal to number of electrons. then mass number is the sum of protons and neutrons in a nucleus so for we to get the number of neutrons we take the mass number subtract the protons number.
What are some possible factors that can be the X
and Y axis of a motion graph?
Answer:
x-Speed/velocity
y-time.
Explanation:
because Speed is a rate of change of distance while time how long it takes a a car to move to a specific point
What factors affect the speed of a wave? Check all that apply.
the amplitude of the wave
the energy of the wave
the temperature of the medium
the type of wave
the type of medium
Answer:
the amplitude of the wave
the energy of the wave
the type of wave
the type of medium
PLEASE CLICK ON THIS IMAGE I NEED HELP
Answer:
Explanation:
you can say the law of superpoition can tell us that each rock layer is older than the one above it. So, the relative age of the rock or fossil in the rock or fossil in the rock is older if it is farther down in the rock layers. hope it helps
If you could help me please.
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
2) How much does a 3-kg object weigh (on earth) in newtons?
3) How much does a 20-kg object weigh (on earth) in newton?
4) What must happen for the mass of an object to change?
5) What are 2 ways the weight of an object can change?
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
No. 1kg of mass does not weigh 9.8N on the moon.
Weight = (mass) x (gravity).
Gravity is 9.8 m/s² on Earth, but gravity is only 1.62 m/s² on the moon.
2) How much does a 3-kg object weigh (on earth) in newtons?
Weight = (mass) x (gravity)
Gravity = 9.8 m/s² on Earth.
Weight = (3 kg) x (9.8 m/s² )
Weight = 29.4 N
3) How much does a 20-kg object weigh (on earth) in newton?
Weight = (mass) x (gravity)
Gravity = 9.8 m/s² on Earth.
Weight = (20 kg) x (9.8 m/s² )
Weight = 196 N
4) What must happen for the mass of an object to change?
When an object moves, its mass increases. The faster it moves, the greater its mass gets. But this is all part of Einstein's "Relativity". The object has to move at a significant fraction of the speed of light before any change can be noticed or measured. So as far as we are concerned, in everyday life, the mass of an object doesn't change, no matter where it is, or what you do to it.
5) What are 2 ways the weight of an object can change?
First, remember that the mass of an object doesn't change, no matter where it is, what you do to it, or what else is around it.
But its weight can change, because its weight depends on the strength of gravity in the place where the object is, and that gravity is the result of what else is around it in the neighborhood. So the weight can change even though the mass doesn't.
The weight of an object changes if you take it to a place where gravity is stronger or weaker.
Let's say we have an object whose mass is 90.72 kilograms. Like me !
As long as I stay on earth, where gravity is 9.8 m/s² , I weigh 889 Newtons (200 pounds).
. . . Fly me to the moon. Gravity = 1.62 m/s² Weight = 147 Newtons (33 lbs)
. . . Drag me to Jupiter. Gravity = 24.8 m/s² Weight = 2,249 N (506 pounds)
My mass never changed, but my weight sure did.
