Answer:
F12 = G M1 M2 / R12^2
F12' = G M1 M2 / R12'^2
F12' / F12 = R12'^2 / R12^2 = (1/3)^2
F12' = 1/9 F12
The new force is 1/9 the of the old force
Please help, I keep trying a bunch of things but keep getting them wrong. I don't know where I am going wrong here.
1. Boyle's Law states the volume and pressure of a gas are inversely proportional.
Name the three units of the constant of proportionality between pressure and volume in alphabetical order. (**I have the first two)
2. The ideal gas law can be written as (PV/nT=R). Name the units for R.
The units of the constant of proportionality between pressure and volume in alphabetical order are
1. Celsius (°C)
2. Fahrenheit (°F)
3. Kelvin (K)
The units for R, that is, the ideal gas constant are
1. J K⁻¹ mol⁻¹
2. L atm K⁻¹ mol⁻¹
We will start by completing the Boyle's Law stated
Boyle's Law states the volume and pressure of a gas are inversely proportional, provided that the temperature remains constant.
This means temperature is the constant of proportionality.
Now, we will name the three units of the constant of proportionality, that is, temperature. The units are
1. Degree Celsius (°C)
2. Degree Fahrenheit (°F)
3. Kelvin (K)
2. In the ideal gas equation (PV/nT=R), R represents the ideal gas constant.
The units for R, that is, the ideal gas constant are
1. J K⁻¹ mol⁻¹
2. L atm K⁻¹ mol⁻¹
Hence,
The units of the constant of proportionality between pressure and volume in alphabetical order are
1. Celsius (°C)
2. Fahrenheit (°F)
3. Kelvin (K)
The units for R, that is, the ideal gas constant are
1. J K⁻¹ mol⁻¹
2. L atm K⁻¹ mol⁻¹
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Find the first three harmonics of a string of linear mass density 2.00 g/m and length 0.600 m when it is subjected to tension of 50.0 N.
Hi there!
We can use the following equation to find the frequency of each harmonic:
[tex]f_n = \frac{n}{2L} \sqrt{\frac{T}{\lambda}}[/tex]
n = nth harmonic
L = length of string (m)
T = Tension of string (N)
λ = linear density (kg/m)
Begin by converting the linear mass density to kg:
2.00g /m · 1 kg / 1000g = 0.002 kg/m
Now, we can use the equation to find the first three harmonics.
First harmonic:
[tex]f_1 = \frac{1}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{131.76 Hz}[/tex]
Second harmonic:
[tex]f_2 = \frac{2}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{263.52Hz}[/tex]
Third harmonic:
[tex]f_3 = \frac{3}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{395.28Hz}[/tex]
Two blocks are set in a pully system as shown in fig below. Block A sits on the frictionless table while block B hags freely. The pully is light and frictionless towards the light string that runs over it. If the Block A has mass of 3.4 kg and Block has 3.5 kg, what would be the magnitude of the acceleration (in ms-2) of the blocks? [g = 9.8 ms=2]
Answer:
Explanation:
F = ma
a = F/m
a = mBg / (mB + mA)
a = 3.5(9.8)/(3.5 + 3.4)
a = 4.971014...
a = 5.0 m/s²
If you want to use individual Free Body Diagrams
mass A will have downward weight and upward normal forces equal at mAg
and a horizontal force of string tension T
F = ma
T = mAa
mass B will have a downward force of mBg and an upward force of T
mBg - T = mBa
substitute for T
mBg - mAa = mBa
mBg = a(mB + mA)
a = mBg / (mB + mA) which is identical to the above answer.
the c component of vector a is 5.3 units, and it’s y component is -2.3 units. the angle that vector a makes with the +x axis is closest to
110
160
23
340
250
Answer:
340
Explanation:
Sorry I don't know how to do this one yet, I just found the answer in a textbook.
The angle that vector a makes with the +x axis is closest to 23.
What is direction of a vector?The direction of a vector is represented tangent of angle equal to the ratio of the y component and the x component of the vector quantity.
tangent of angle = y/x
angle = tan⁻¹ (-2.3/5.3)
angle = 23.46°
Thus, the angle that vector makes with +x is 23.
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73 ml of water is followed by 25 ml of juice. What is the percent strength of juice?
Total liquid: 73 + 25 = 98 ml
Percent juice = (25/98) x 100 = 25.5 %
What are the 7 different states of matter in Chemistry?How many states of matter are there?
Answer:
The 7 states of matter are solid, loquid, gas, fermionoc condensate, quark gluton plasm, bose einetein condensate amd ionised plasm but its usually only 3 they teach you
Answer:
7
Explanation:
solid, liquid,gas,fermionoc condensate,quark glutton plasm,bose einetein condensate amd ionised plasm.
2) A rolling disk, mass m and radius R, approaches a step of height R/2 with velocity v. (i) Taking the corner of the step as the pivot point, what is the initial angular momentum of the disk
The rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
Using the law of conservation of energy, the initial mechanical energy E of the disk equals its final mechanical energy E' as it climbs the step.
So, E = E'
1/2Iω + 1/2mv² + mgh = 1/2Iω' + 1/2mv'² + mgh'
where I = rotational inertia of disk = 1/2mR² where m = mass of disk and R = radius of disk, ω = initial angular speed of disk, v = initial velocity of disk, h = initial height of disk = 0 m, ω' = final angular speed of disk = 0 rad/s (assumung it stops at the top of the step), v' = final velocity of disk = 0 m/s (assumung it stops at the top of the step), and h' = final height of disk = R/2.
Substituting the values of the variables into the equation, we have
1/2Iω² + 1/2mv² + mgh = 1/2Iω'² + 1/2mv'² + mgh'
1/2(1/2mR² )ω² + 1/2mv² + mg(0) = 1/2I(0)² + 1/2m(0)² + mgR/2
mR²ω²/4 + 1/2mv² + 0 = 0 + 0 + mgR/2
mR²ω²/4 + 1/2mv² = mgR/2
R²ω²/4 = gR/2 + 1/2v²
R²ω²/4 = (gR + v²)/2
ω² = 2(gR + v²)/R²
ω² = √[2(gR + v²)/R²]
ω = √[2(gR + v²)]/R
Since angular momentum L = Iω, the rolling disk's initial angular momentum is
L = 1/2mR² ×√[2(gR + v²)]/R
L = mR√[2(gR + v²)]/2
the rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
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A skater is spinning with his arms outstretched. He has a 2 lb weight in each hand. In an attempt to change his angular velocity he lets go of both weights (by just opening his grip). Does he succeed in changing his angular velocity
Answer:
No
Explanation:
Changing momentum of any kind requires work. Work is a force acting over a distance. While holding the weights at arms length and spinning will create a force (centripetal), there is no radial distance change incurred. Releasing the weights will reduce the force to zero, still no work done and no change in angular momentum.
If he was holding the weights at arms length while spinning and he pull his hands to his chest, there now exists both the centripetal force and a distance in the direction of that force (inward radial) this work will result in an increase in angular velocity as moment of inertia has decreased with the work done.
No, the skater doesn't succeed in changing his angular velocity.
Conservation of angular momentumThe final angular velocity of the skater is determined by applying the principle of conservation of angular momentum as shown below;
Li = Lf
[tex]Ii\omega _i = I_f \omega _f[/tex]
where;
Ii is the initial moment of inertia of the skaterIf is the final moment of inertia of the skaterωi is the initial angular speed of the skaterωf is the final angular speed of the skaterWhen the skater holds the weight, the momnet of inertia of both arms is the same. Also when the skater drops the weight, the moment of inertia of both arms is still the same. Thus, at any instant, the moment of inertia of the two arms is the same.
To change the angular speed, the initial and final moment of inertia of the two arms must be different. Thus, the skater doesn't succeed in changing his angular velocity.
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1. Explain who is doing more work and why: a bricklayer carrying bricks and placing them on the wallof a building being
constructed, or a project supervisor observing and recording the progress of the
workers from an observation booth.
Look at the simple machine shown below to determine the mechanical advantage
Answer:
A
Explanation:
how do all organisms begin life
Answer:
All organisms begin their lives as single cells.Overtime,these organisms grow and take on the characteristics of their species...All organisms grow,and different parts of organisms may grow at different rates.Organisims made out of only one cell
may change little during their lives, but they do grow
Explanation:
brainlest me please
a Answer the following questions 1. On a cold wintery day, you burn firewood to keep yourself warm. The firewood undergoes a change in state. a. Name the change in state of matter that you see
Answer:
heating prosedure takes place the opposite of condensation
Anita Knapp needs to get hay to cows in a frozen field using an airplane flying
80.0 m/s, at a height of 300,m. If at the last minute, how far from the cow would
she have to release the hay in order to hit the cow?*
756 m
626m
700m
575 m
Other:
Answer:
626m
Explanation:
A mass vibrates back and forth from the free end of an ideal spring of spring constant 20 N/m with an amplitude of 0.30 m. What is the kinetic energy of this vibrating mass when it is 0.30 m from its equilibrium position?
Hi there!
We can begin by using the work-energy theorem in regards to an oscillating spring system.
Total Mechanical Energy = Kinetic Energy + Potential Energy
For a spring:
[tex]\text{Total ME} = \frac{1}{2}kA^2\\\\\text{KE} = \frac{1}{2}mv^2\\\\PE = \frac{1}{2}kx^2[/tex]
A = amplitude (m)
k = Spring constant (N/m)
x = displacement from equilibrium (m)
m = mass (kg)
We aren't given the mass, so we can solve for kinetic energy by rearranging the equation:
ME = KE + PE
ME - PE = KE
Thus:
[tex]KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2\\\\[/tex]
Plug in the given values:
[tex]KE = \frac{1}{2}(20)(0.3^2) - \frac{1}{2}(20)(0.3^2) = \boxed{0 \text{ J}}[/tex]
We can also justify this because when the mass is at the amplitude, the acceleration is at its maximum, but its instantaneous velocity is 0 m/s.
Thus, the object would have no kinetic energy since KE = 1/2mv².
Conservation of Energy Roller Coaster A roller coaster cart of mass 100kg travels on a track with one loop. Fill in blanks A-H. А. KE=OJ PE=120000J h= А. V= B B KE=___CE PE=60000J h= _D V= E KE=__F PE=40000J h=__G_ V= KE= PE= h=Om v= K D E F G H K
(a) The height of the roller coaster at 120,000 potential energy is 122.45 m.
(b) The velocity of the roller coaster at 0 J kinetic energy is 0.
(c) The height of the roller coaster at 60,000 potential energy is 61.23 m.
(d) The velocity of the roller coaster at 60,000 J kinetic energy is 34.64 m/s.
(e) The height of the roller coaster at 40,000 potential energy is 40.82 m.
(f) The velocity of the roller coaster at 80,000 J kinetic energy is 40 m/s.
The given parameters:
mass of the roller coaster, m = 100 kgWhen the kinetic energy = 0 and potential energy = 120,000 J
The height of the roller coaster is calculated as follows;
P.E = mgh
[tex]h = \frac{P.E}{mg}\\\\h = \frac{120,000}{100 \times 9.8} \\\\h = 122.45 \ m[/tex]
Since the kinetic energy = 0, the velocity of the roller coaster = 0
When the potential energy, P.E = 60,000 J, the kinetic energy, K.E is calculated as;
P.E + K.E = M.A
P.E + K.E = 120,000
60,000 + K.E = 120,000
K.E = 120,000 - 60,000
K.E = 60,000 J
The height of the roller coaster at 60,000 potential energy is calculated as follows;
[tex]h = \frac{P.E}{mg} \\\\h = \frac{60,000}{100 \times 9.8} \\\\h =61.23 \ m[/tex]
The velocity of the roller coaster at 60,000 J kinetic energy is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2\\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{ \frac{2K.E}{m}} \\\\v = \sqrt{ \frac{2\times 60,000}{100}}\\\\v = 34.64 \ m/s[/tex]
When the potential energy, P.E = 40,000 J, the kinetic energy, K.E is calculated as;
P.E + K.E = M.A
40,000 + K.E = 120,000
K.E = 120,000 - 40,000
K.E = 80,000
The height of the roller coaster at 40,000 potential energy is calculated as follows;
[tex]h = \frac{P.E}{mg} \\\\h = \frac{40,000}{100 \times 9.8} \\\\h = 40.82 \ m[/tex]
The velocity of the roller coaster at 80,000 J kinetic energy is calculated as follows;
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 80,000}{100} } \\\\v = 40 \ m/s[/tex]
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A 2 kg ball is rolling down a hill at a constant speed of 4 m/s. How much kinetic energy does the ball have?
What is most likely the amount of energy available at a trophic level of primary consumers if the amount of energy available to secondary consumers in that food web is 200 kilocalories?
0 kilocalories
20 kilocalories
200 kilocalories
2,000 kilocalories
Answer:
200 kilocalories
Explanation:
When 587.9 nm passes through a single slit 0.73 mm wide, it creates a diffraction pattern. (a) What distance away is the wall if the first minimum is 0.86 mm from the central maximum
From Young's single slit experiment, the distance away from the wall will be 1.068 m
Given that 587.9 nm of wavelength of light passes through a single slit 0.73 mm wide, it creates a diffraction pattern.
From the question, the following parameters are given:
The wavelength of the light λ = 587.9 nm
The width of the slit a = 0.73 mm
Fringe width X = 0.86 mm
The distance away from the wall D = ?
The fringe width is related to the wavelength of the light source by the equation:
X = Dλ ÷ a
Substitute all the parameters into the formula
0.83 × [tex]10^{-3}[/tex] = 587.9 × [tex]10^{-9}[/tex] D ÷ 0.73 ×
Cross multiply
587.9 × [tex]10^{-9}[/tex] D = 6.278 × [tex]10^{-7}[/tex]
make D the subject of the formula
D = 6.278 × [tex]10^{-7}[/tex] ÷ 587.9 × [tex]10^{-9}[/tex]
D = 1.068 m
Therefore, the distance away from the wall is 1.068 m
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If the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.
Given the data in the question;
Wavelength; [tex]\lambda = 587.9nm = 5.879*10^{-7}m[/tex]Width of slit; [tex]a = 0.73mm = 0.00073m[/tex] First minimum; [tex]y = 0.86mm = 0.00086m[/tex]Since its first, order number; [tex]m = 1[/tex]Distance; [tex]L = \ ?[/tex]From Thomas Young's single slit experiment:
[tex]\frac{a*y}{L} = m * \lambda[/tex]
Where a is the width of the slit, y is first minimum, L is the distance, m is the order number and λ is the wavelength.
We substitute our values into the equation
[tex]\frac{0.00073m\ *\ 0.00086m}{L} = 1\ *\ ( 5.879*10^{-7}m)\\\\\frac{0.0000006278m^2}{L} = 5.879*10^{-7}m\\\\L = \frac{0.0000006278m^2}{5.879*10^{-7}m} \\\\L = 1.07m[/tex]
Therefore, if the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.
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A stomp rocket takes 3.1 seconds to reach its maximum height.
- What is its initial velocity? (Do not use units. If the answer is negative, please put a
negative sign in front of the answer.)
- What is its maximum height? (Do not use units. If the answer is negative, please put a
negative sign in front of the answer.)
Answer:
Explanation:
Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.
v = at
v = 9.8(3.1)
v = 30.38
v = 30 m/s
max height can be found knowing the velocity is zero at the top of its flight.
v² = u² + 2as
s = (v² - u²) / 2a
s = (0.00² - 30.38²) / (2(-9.8))
s = 47.089
s = 47 m
A stomp rocket takes 3.1 seconds to reach its maximum height then the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.
What is Velocity?Velocity is defined as rate of change of position with respect to time.
SI unit of velocity is m/sec. Velocity is a vector quantity.
Given that in the question time taken by rocket to reach maximum height is 3.1 sec. Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.
v = at
v = 9.8(3.1)
v = 30.38
v = 30 m/s
Max height can be found knowing the velocity is zero at the top of its flight.
v² = u² + 2as
s = (v² - u²) / 2a
s = (0.00² - 30.38²) / (2(-9.8))
s = 47.089
s = 47 m
So, the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.
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F=mass x what does this equal?
Answer:
Force = mass × acceleration.
Which performs a function that is most like the function of a retina?
Answer:
The answer is ciliary body and focus the pupil. In addition, the ciliary body is a portion of the eye that contains the ciliary muscle that reins the shape of the lens and the ciliary epithelium that yields the aqueous humor. The ciliary body is a share of the uvea which is the layer of tissue that transports oxygen and nutrients to the eye nerves while the pupil is a hole positioned in the midpoint of the iris of the eye that permits light to foray the retina. It looks black since light rays incoming the pupil are moreover engrossed by the tissues in the eye openly or engrossed after diffuse reflections in the eye that typically miss leaving the fine pupil
Explanation:
Answer:the eye has many parts that must work together in order to produce clear vision
Explanation: correct on my test
EXAM ENDS IN 30 MINS
PLSSS HELPPP ILL MAKE U BRAINLIEST
Explanation:
F = Icurrent×length×Bfieldstrength×sin(angle field to wire)
in our case
Icurrent = 10 A
length = 0.02km = 20 meters
B = 10^-6 T
angle = 30 degrees.
F = (20 A)(20m)(10^-6 T)×sin(30) = 400× 10^‐6 ×0.5 N =
= 200 × 10^-6 = 2 × 10^‐4 N
What do alcohol, drugs, and tobacco all have in common?
All have some medicinal value.
All are harmful to the body.
All are depressants.
All are stimulants.
Answer:
all are harmful to the body
How do humans obtain the carbon and energy they use in their bodies?
A. by breathing in carbon dioxide from the atmosphere
B. by consuming plants or other animals
C. by absorbing energy from sunlight
D. by absorbing carbon found in the soil
Answer:
B. . by consuming plants or other animals
Someone with a force of 900 N does not float in a freshwater pool. To prevent them from sinking, 20 N must be applied in an upward fashion. What is their volume and average density?
Explanation:
The buoyant force [tex]F_B[/tex] is defined as
[tex]F_B = \rho_wgV[/tex]
where [tex]\rho_w[/tex] is the density of the displaced fluid (freshwater), g is the acceleration due to gravity and V is the volume of the submerged object. In the case of freshwater, its density is [tex]997\:\text{kg/m}^3.[/tex] Since the buoyant force is 20 N, we can solve for the volume of the displaced fluid:
[tex]F_B = \rho_wgV \Rightarrow V = \dfrac{F_B}{\rho_wg}[/tex]
Plugging in the values, we get
[tex]V = \dfrac{20\:\text{N}}{(997\:\text{kg/m}^3)(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:= 2.05×10^{-3}\:\text{m}^3[/tex]
Recall that the weight of an object in terms of its density and volume is given by
[tex]W = \rho gV[/tex]
Using the value for the volume above, we can solve for the density of the object as follows:
[tex]\rho = \dfrac{W}{gV} = \dfrac{900\:\text{N}}{(9.8\:\text{m/s}^2)(2.05×10^{-3}\:\text{m}^3)}[/tex]
[tex]\:\:\:\:\:= 44,798\:\text{kg/m}^3[/tex]
2. Two equipotential lines are separated by a distance of 2.17 cm. If the potential values of the lines are 5.9 volts and 8.6 volts, what would the strength of the electric field between the lines be
An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]
a.
The object's speed at 2.20 m below balcony level is 8.74 m/s
Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.
Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and
So, E = E'
U + K + f = U' + K' + f'
where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).
So,
U + K + f = U' + K' + f'
mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh
mgh = mgh' + 1/2mv² + 0.10mgh
Dividing through by m, we have
gh = gh' + 1/2v² + 0.10gh
So, gh - 0.10gh = gh' + 1/2v²
0.90gh = gh' + 1/2v²
1/2v² = 0.90gh - gh'
1/2v² = g(0.90h - h')
v² = 2g(0.90h - h')
Taking square-root of both sides, we have
v = √[2g(0.90h - h')]
where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the values of the variables into the equation, we have
v = √[2g(0.90h - h')]
v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]
v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]
v = √[2 × 9.8 m/s²(3.901 m)]
v = √[76.4596 m²/s²]
v = 8.74 m/s
So, the object's speed at 2.20 m below balcony level is 8.74 m/s
b.
Yes it does matter when we apply 10% loss before V calculations
We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.
So, yes it does matter when we apply 10% loss before V calculations
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How do organisms use communication to survive?
Answer: Im not entirly sure but I think It's D all the above. I think all but B because I never really heard of that but if you look in our history I think that happen im not sure I would wait untill you know that somone knows for sure.
Explanation:
-
9 Two bodies of 6 kg and 4 kg masses have their
velocity 5i - 2j +10k and 10î – 2ġ +5ť,
respectively
. Then, the velocity of centre of me
(a) 5ỉ +23 - 8 (b) 7î+ 29 - 8
(2) 7î – 2į +8ỉ (d) 5î – 29 +8k
+
-
Answer:
I don't know he he.
just joking
An object of mass m is hanging by a string from the ceiling of an elevator. The elevator is moving down at constant speed. What is the tension of the string?
A. Zero
B. Equal to mg
C. Less than mg
D. Greater than mg
Answer:
D. Greater than mg
Explanation:
According to Newton’s second law of motion, the net force equals mass times acceleration. We are going to use a free body diagram (force diagram) to show that the equation of the motion is given by
T – mg = – ma
Thereby,
T = mg – ma
and the answer is: (d)
D. Greater than mg
_________________________________
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