A cylindrical bar that us well insulated around its sides connects hot and cold reservoirs and conducts heat at a rate of 10.0 J/s under steady state conditions. If all of its linear dimensions (diameter and length) are reduced by half, the rate at which it will now conduct heat between the same reservoirs is closest to
Answer: the at which the bar conducts now is 5 Js⁻¹
Explanation:
Given the data in the question;
we know that; Heat transfer by conduction is given by;
Q ∝ [tex]A / l[/tex]
such that,
[tex]Q_{1}/Q_{2}[/tex] = [tex]A_{1}l_{2} / A_{2}l_{1}[/tex]
[tex]Q_{2} = Q_{1} A_{2}l_{1}/A_{1}l_{2}[/tex]
so
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r}{2}[/tex])² × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r^{2} }{4}[/tex]) × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × πr² × 1/4 × [tex]l[/tex] ) / (πr² × 1/2 × [tex]l[/tex] )
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × 1/4) / ( 1/2)
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × 1/4) / ( 1/2)
[tex]Q_{2} =[/tex] 5 Js⁻¹
Therefore, the at which the bar conducts now is 5 Js⁻¹
The Short Answer:
Earth's tilted axis causes the seasons. Throughout the year, different
parts of Earth receive the Sun's most direct rays. So, when the North
Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And
when the South Pole tilts toward the Sun, it's qvinter in the Northern
Hemisphere.
1. What direction is the Earth tilted in the summer here in
Buffalo?
a) towards the Sun
b) away from the Sun
23.5
Answer:
I cannot fully see the picture, but I'm going to have to tell you to go with towards the sun because it says summer.
An object with an initial horizontal velocity of 20 ft/s experiences a constant horizontal acceleration due to the action of a resultant force applied for 10 s. The work of the resultant force is 10 Btu. The mass of the object is 55 lb. Determine the constant horizontal acceleration, in ft/s2.
Answer:
a = 7.749 ft/s²
Explanation:
First to all, we need to convert all units, so we can work better in the calculations.
The horizontal acceleration is asked in ft/s² so the units of speed will be the same. The Work is in BTU and we need to convert it in ft.lbf in order to get the acceleration and final speed in ft/s:
W = 10 BTU * 778.15 Lbf.ft / BTU = 7781.5 lbf.ft
Now, to get the acceleration we need to get the final speed of the object first. This can be done, by using the following expression:
W = ΔKe (1)
And Ke = 1/2mV²
So Work would be:
W = 1/2 mV₂² - 1/2mV₁²
W = 1/2m(V₂² - V₁²) (2)
Finally, we need to convert the mass in lbf too, because Work is in lbf, so:
m = 55 lb * 1 lbf.s²/ft / 32.174 lb = 1.7095 lbf.s²/ft
Now, we can calculate the final speed by solving V₂ from (2):
7781.5 = (1/2) * (1.7095) * (V₂² - 20²)
7781.5 = 0.85475 * (V₂² - 441)
7781.5/0.85475 = (V₂² - 400)
9103.83 + 400 = V₂²
V₂ = √9503.83
V₂ = 97.49 ft/s
Now that we have the speed we can calculate the acceleration:
a = V₂ - V₁ / t
Replacing we have:
a = 97.49 - 20 / 10
a = 7.749 ft/s²Hope this helps
What will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of the d'Arsonval movement is negligible
Answer:
hello your question is incomplete attached below is the complete question
answer :
20.16 v
Explanation:
The reading of the voltmeter at the instant the switch returns to position a
L = 5H
i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo
= 1/5 ∫ 3*10^-3 d(t) + 0 = 0.6 * 10^-3 t
iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA
Rm ( resistance ) = 21 * 1000 = 21 kΩ
The reading of the voltmeter ( V )
V = IR
= 0.96 mA * 21 k Ω = 20.16 v
A ball is projected at an angle of 53º. If the initial velocity is 48 meters/second, what is the vertical component of the velocity with which it was
launched?
OA. 31 meters/second
OB. 38 meters/second
OC
44 meters/second
OD
55 meters/second
Answer: B
Explanation:
The vertical component of a vector such as velocity is the magnitude of the vector multiplied by the sine of the angle.
[tex]V_y=48*sin(53)=38.3m/s[/tex]
You push a box across the floor at a constant speed of 1 m/s, applying a horizontal constant force of magnitude 20 N. Your friend pushes the same box across the same floor at a constant speed of 2 m/s, applying a horizontal force. What is the magnitude of the force that your friend applies to the box
Answer:
the force your friend applied on the box is 40 N.
Explanation:
Given;
speed of the box, v₁ = 1 m/s
force applied to the box, F₁ = 20 N
the speed of the box when your friend pushes it, v₂ = 2 m/s
then your friends applied force, F₂ = ?
Assuming the time, t, through which both forces were applied and mass of the box, m, to be constant;
[tex]F_1 = \frac{mv_1}{t} \\\\\frac{m}{t} = \frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
[tex]F_2 = \frac{F_1v_2}{v_1} \\\\F_2 = \frac{20\times 2}{1} \\\\F_2 = 40 \ N[/tex]
Therefore, the force your friend applied on the box is 40 N.
A square plate is produced by welding together four smaller square plates,
each of side
a. The weight of each of the four plates is
shown in the figure.
Find the x-coordinate of the center of gravity (as a multiple of a).
Answer in units of a.
(PICTURED)
PART TWO
Find the y-coordinate of the center of gravity
(as a multiple of a).
Answer in units of a
Answer:
eExplanation:
Daffy Duck is standing 6.8 m away from Minnie Duck. The attractive gravitational force between them is 5.4x10-8 N. If Daffy Duck has a mass of 86.5 kg, What is Minnie Duck's mass?'
Answer:
432.78 Kg
Explanation:
From the question given above, the following data were obtained:
Distance apart (r) = 6.8 m
Force of attraction (F) = 5.4×10¯⁸ N
Mass of Daffy Duck (M₁) = 86.5 kg
Mass of Minnie Duck (M₂) =?
NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
The mass of Minnie Duck can be obtained as follow:
F = GM₁M₂ / r²
5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²
5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24
Cross multiply
6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24
Divide both side by 6.67×10¯¹¹ × 86.5
M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5
M₂ = 432.78 Kg
Therefore, the mass of Minnie Duck is 432.78 Kg
Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, as if they had been migrating. For years curiosity mounted about why the stones moved. One explanation was that strong winds during the occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force is needed on a 25 kg stone (a typical mass) to maintain the stone's motion once a gust has started it moving
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
You've been hired to design the hardware for an ink jet printer. You know that these printers use a deflecting electrode to cause charged ink drops to form letters on a page. The basic mechanism is that uniform ink drops of about 30 microns radius are charged to varying amounts after being sprayed out towards the page at a speed of about 20 m/s. Along the way to the page, they pass into a region between two deflecting plates that are 1.6 cm long. The deflecting plates are 1.0 mm apart and charged to 1500 volts. You measure the distance from the edge of the plates to the paper and find that it is one-half inch. Assuming an uncharged droplet forms the bottom of the letter, how much charge is needed on the droplet to form the top of a letter 3 mm high (11 pt. type)
Answer:
the required charged is 7.06 × 10⁻¹³ C
Explanation:
Given that;
Radius = 30 microns = 30 × 10⁻⁶
Speed v = 20 m/s
length x = 1.6 cm = 0.016 m
spacing d = 1.0 mm = 0.001 m
Voltage V = 1500 V
from the question, the electric field between the plates is uniform and equal to Voltage divided by the distance between the plates.
Electric field E = V/d
E = 1500 V / 0.001 m
E = 1.5 × 10⁶ V/m
Mass of ink drop m = pv
m = 10³ kg/m³ × [tex]\frac{4}{3}[/tex]πr³
m = 1000 kg/m³ × [tex]\frac{4}{3}[/tex]π × (30 × 10⁻⁶)³
m = 1.131 × 10⁻¹⁰ Kg
Time taken to travel t = x / sped
t = 0.016 m / 20 m/s
t = 0.0008 s
From the kinematic equation
to form the top of a letter 3 mm ( 0.003 m )high
y = [tex]\frac{1}{2}[/tex]at²
2y = at²
a = 2y/t²
we substitute
a = (2 × 0.003 m) / (0.0008 s)²
a = 9375 m/s²
Now Force F = Eq = ma
so
q = ma / E
we substitute
q = ( 1.131 × 10⁻¹⁰ Kg × 9375 m/s² ) / ( 1.5 × 10⁶ V/m )
q = 7.06 × 10⁻¹³ C
Therefore, the required charged is 7.06 × 10⁻¹³ C
How would you compare the acceleration between the unbalanced net force of 100 N and of 50 N
Answer:
The acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Explanation:
Given;
first net force, F₁ = 100 N
second net force, F₂ = 50 N
If we consider equal mass for the two net forces, and apply Newton's second law of motion, the acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Let a₁ be the acceleration produced by the first net force
then, a₂ be the acceleration produced by the second net force
Thus, a₁ = 2a₂
A person stands on the ball of one foot. The normal force due to the ground pushing up on the ball of the foot has magnitude 750 N. Ignore the weight of the foot itself. The other significant forces acting on the foot are the tension in the Achilles tendon pulling up and the force of the tibia pushing down on the ankle joint. If the tension in the Achilles tendon is 2225 N, what is the force exerted on the foot by the tibia
Answer:
the force exerted on the foot by the tibia would be 2975 N
Explanation:
Given the data in the question;
To maintain equilibrium between the foot and the ball vertically, the addition normal normal force [tex]N^>[/tex] (750 N) and the tension in the Achilles tendon [tex]F^>_{Achilles}[/tex] (2225 N) must be equal to the force exerted on the foot by the tibia;
so
| [tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] | = | [tex]F^>_{Tibia}[/tex] |
so force exerted on the foot by the tibia will be;
| [tex]F^>_{Tibia}[/tex] | = |[tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] |
so we substitute IN OUR VALUES
| [tex]F^>_{Tibia}[/tex] | = 750 N + 2225 N
| [tex]F^>_{Tibia}[/tex] | = 2975 N
Therefore, the force exerted on the foot by the tibia would be 2975 N
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antinode is 2.20 mm and the speed of propagation of transverse waves on the string is 260 m/s. The string extends along the x-axis, with one of the fixed ends at x= 0, so that there is a node at x =0. The smallest value of x where there is an antinode is x= 0.150m.
Required:
a. What is the maximum transverse speed of a point on the string at an antinode?
b. What is the maximum transverse speed of a point on the string at x = 0.075 m?
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so [tex]V_{max1}[/tex] = A × ω
[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s
[tex]V_{max1}[/tex] = 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
[tex]V_{max2}[/tex] = A' × ω
[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69
[tex]V_{max2}[/tex] = 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.05 m/sm/s . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 40.8 mm above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 7.00 ss after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.
Required:
a. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
b. Where is Henrietta when she catches the bagels?
Answer:
v = 10.46 m/s
x = 30.134 m from house
Explanation:
given data
speed = 3.05 m/s
time = 7 s
height = 40.8 mm
solution
we get here first time required to fall that is t
t = [tex]\sqrt{\frac{2\times 40.8}{9.8}}[/tex] ..................1
t = 2.88 s
now we take here initial speed that is v
so v × t = 3.05 × ( t+ 7)
v = [tex]\frac{3.05\times (2.88 + 7)}{2.88}[/tex]
v = 10.46 m/s
and
when she catch the bagel henrietta was at
x = 3.05 × ( 2.88 + 7)
x = 30.134 m from house
Define position
i am not sure?
Which mode of kinetic energy contributes to temperature?
Answer:
Kinetic energy Temperature
Explanation:
What does m/s/s mean?
Explanation:
There are two answers
m/(s/s)=m
or
(m/s)/s=m/s²
g Galileo's telescopes were not of high quality by modern standards. He was able to see the moons of Jupiter, but he never reported seeing features on Mars. Use the small-angle formula to find the angular diameter of Mars when it is closest to Earth. How does that compare with the maximum angular diameter of Jupiter
Answer:
θ₂/ θ₁= 2.58
Explanation:
In this exercise you are asked to compare the angular diameters of Mars and Jupiter. The angular diameter or angle in radians is
θ = D / R
where D is the diameter of the body, the distance from Earth to the body of interest and θ is angle in radians
The different distances are tabulated with respect to the Sun
Sun -Earth 1,496 10¹¹ m
Sun- Mars 2.28 10¹¹ m
Sun - Jupiter 7.78 10 m
The Radii of the planets are
Mars 3.37 10⁶ m
Jupiter 6.99 10⁷ m
let's calculate the angles for each body
a) Mars
θ₁ = 2r / R'
the distance from the ground is
R ’= D_planet - D_earth
R ’= 2.28 10¹¹ - 1.496 10¹¹
R ’= 0.784 10¹¹ m
let's calculate
θ₁ = [tex]\frac{2 \ 3.37 \ 10^6 }{0.784 \ 10^{11}}[/tex]
θ₁ = 8.6 10⁻⁵ radians
b) Jupiter
R ’= 7.78 10¹¹ - 1.496 10¹¹
R ’= 6.284 10¹¹ m
let's calculate
θ₂ = [tex]\frac{2 \ 6.99 \ 10^7}{6.284 \ 10^{11}}[/tex]
θ₂ = 2.22 10⁻⁴ radians
the ratio of the angular diameters is
θ₂/ θ₁ = [tex]\frac{2.22 \ 10^{-4}}{8.6 \ 10^{-5}}[/tex]
θ₂/ θ₁= 2.58
I need help please will mark brainliest
Answer: IT B TRUST ME MANN OK
Explanation: OK BYE TRUST
a car travels 10 miles east in 30 minutes. what is its velocity in miles per hour. what is its velocity in miles per hour?
Answer:
popu
Explanation:
2u2uwju2i2je82jei
For this assignment, you should mathematically solve and record a video testing your solution for the following prompt: Two rolls of toilet paper, of equal mass and radius, are dropped from different heights so that they hit the ground at the same time. One roll of toilet paper is dropped normally while the other is dropped while a person holds onto a sheet of toilet paper such that the roll unravels as it descends. Determine the ratio of heights h1/h2, where h1 represents the height of the toilet paper dropped normally and h2 represents the height of the toilet paper that unravels, so that both rolls hit the ground at the same time.
Answer:
h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
Explanation:
Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )
Determine the ratio of heights h1/h2
mass of tissues = same
radius of tissues = same
h1 = height of tissue 1
h2 = height of tissue 2
For the first tissue ( Tissue that dropped manually )
potential energy = kinetic energy
mgh = 1/2 mv^2
therefore the final velocity ( v^2 ) = 2gH ----- ( 1 )
second tissue ( Tissue that dropped while rotating )
gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )
To determine the ratio of heights we will equate equations 1 and 2
hence :
gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )
∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
I need help please will mark brainliest
Answer: 30 to 40 s
Explanation:
If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
364N
185N
173N
73N
ANSWER AND I WILL GIVE YOU BRAINILIEST
Answer:
73N
Explanation:Just multiply 1.2^2 by 50
What is the speed of a cyclist that rides west 88 km in 32 minutes?
Answer:
The speed of the cyclist is 2.75 km/min.
Explanation:
Given
The distance d = 88 km Time t = 32 minutesTo determine
We need to find the speed of a cyclist.
In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.
Using the formula involving speed, time, and distance
[tex]s=\frac{d}{t}[/tex]
where
s = speed d = distance covered t = time takensubstitute d = 88, and t = 32 in the formula
[tex]s=\frac{d}{t}[/tex]
[tex]s=\frac{88}{32}[/tex]
Cancel the common factor 8
[tex]s=\frac{11}{4}[/tex]
[tex]s=2.75[/tex] km/min
Therefore, the speed of the cyclist is 2.75 km/min.
A spring with a spring constant of 22 N/m is stretched from equilibrium to 2.9 m. How much work is done in the process?
O A. 186 )
OB. 47 J
O C. 933
OD. 121 )
what is the answer ?
Answer:
using W=1/2kW2
k=22N/m w=2.9
w=1/2×22×2.9×2.9
w=92.51Joules
Approximately 93J answer is C
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field towards the other end. If they move fast enough when they strike the positive electrode at the other end, they will give up their energy as X-Rays
(a) Through what potential difference should electrons be accelerated so that their speed is 1% of the speed of light?
(b) What potential difference would be needed to give the protons same kinetic energy as electrons?
(c) What speed would this potential difference give to the protons, both in m/s and as a % of the speed of light.
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = [tex]\frac{1}{2}[/tex] mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
[tex]K_e = K_p[/tex]
K_p = ½ m v_e²
K_p = [tex]\frac{1}{2}[/tex] 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]
v/c= 2.33 10⁻⁴
3. A stone is thrown vertically upwards from the top of a building 50 m tall with an initial
velocity of 20.0 ms. If the stone just misses the edge of the roof on its return, determine
(a) The time is taken the stone to get to its maximum height.
(b) The maximum height reached by the stone
(c) The time at which the stone return to the point where it was thrown
(d) The velocity of the stone at this instance
(e) The velocity and position of the stone at t = 5 s.
Answer:
13.4436
Explanation:
a device that spreads light into different wavelengths is a what?
maybe a spectrograph ?
A skier of mass 72 kg is pulled up a slope by a motor-driven cable. a. how much work is required to pull him 75 m up a 30 degree slope (assumed frictionless) at a constant speed of 3.4 m/s
Answer: [tex]26.460\times 10^3\ J[/tex]
Explanation:
Given the mass of skier m=72 kg
distance traveled d=75 m
constant speed v=3.4 m/s
If speed is constant then there must no force acting in the direction of motion
i.e. tension force must be equal to the component of weight
[tex]T=mg\sin 30^{\circ}[/tex]
Work done is given by
[tex]\Rightarrow W=F\cdot d=Td\cos 0^{\circ}\\\Rightarrow W=mg\sin 30^{\circ} d=72\times 9.8\times 0.5\times 75\\\Rightarrow W=26.460\times 10^3\ J[/tex]
Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.65 m/s. a. Calculate the magnitude of the initial angular momentum of the system by treating the astronauts as particles.
Answer:
L = 5076.5 kg m² / s
Explanation:
The angular momentum of a particle is given by
L = r xp
L = r m v sin θ
the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1
as we have two bodies
L = 2 r m v
let's find the distance from the center of mass, let's place a reference frame on one of the masses
[tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_{i} m_{i}[/tex]i
x_{cm} = [tex]\frac{1}{m+m} ( 0 + l m)[/tex]
x_{cm} = [tex]\frac{1}{2m} lm[/tex]
x_{cm} = [tex]\frac{1}{2}[/tex]
x_{cm} = 13.1 / 2 = 6.05 m
let's calculate
L = 2 6.05 74.3 5.65
L = 5076.5 kg m² / s
Answer:
5076.5
Explanation: