Answer:
a) k = 95.54 N / m, c = 19.55 , b) m₃ = 0.9078 kg
Explanation:
In a simple harmonic movement with friction, we can assume that this is provided by the speed
fr = -c v
when solving the system the angular value remains
w² = w₀² + (c / 2m)²
They give two conditions
1) m₁ = 1 kg
f₁ = 1.1 Hz
the angular velocity is related to frequency
w = 2π f₁
Let's find the angular velocity without friction is
w₂ = k / m₁
we substitute
(2π f₁)² = k / m₁ + (c / 2m₁)²
2) m₂ = 2 kg
f₂ = 0.8 Hz
(2π f₂)² = k / m₂ + (c / 2m₂)²
we have a system of two equations with two unknowns, so we can solve it
we solve (c / 2m)² is we equalize the expression
(2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁
k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)
k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)
a) Let's calculate
k = 4 π² (0.8² -1.1²) / (½ -1/1)
k = 39.4784 (1.21) / (-0.5)
k = 95.54 N / m
now we can find the constant of friction
(2π f₁) 2 = k / m₁ + (c / 2m₁)²
c2 = ((2π f₁)² - k / m₁) 4m₁²
c2 = (4ππ² f₁² - k / m₁) 4 m₁²
let's calculate
c² = (4π² 1,1² - 95,54 / 1) 4 1²
c² = (47.768885 - 95.54) 8
c² = -382.1689
c = 19.55
b) f₃ = 0.2 Hz
m₃ =?
(2πf₃)² = k / m₃ + (c / 2m₃) 2
we substitute the values
(4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²
1.579 = 95.54 / m₃ + 95.542225 / m₃²
let's call
x = 1 / m₃
x² = 1 / m₃²
- 1.579 + 95.54 x + 95.542225 x² = 0
60.5080 x² + 60.5080 x -1 = 0
x² + x - 1.65 10⁻² = 0
x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2
x = [1 ± 1.03] / 2
x₁ = 1.015 kg
x₂ = -0.015 kg
Since the mass must be positive we eliminate the second results
x₁ = 1 / m₃
m₃ = 1 / x₁
m₃ = 1 / 1.1015
A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. If the mass is set in motion in a medium that imparts a damping force numerically equal to 16 times the velocity, the correct differential equation for the position x (t ), of the mass at a function of time, t is
Answer:
d²x/dt² = - 4dx/dt - 4x is the required differential equation.
Explanation:
Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,
F = mg
kx = mg
k = mg/x
= 4 kg × 9.8 m/s²/2.45 m
= 39.2 kgm/s²/2.45 m
= 16 N/m
Now the drag force f = 16v where v is the velocity of the mass.
We now write an equation of motion for the forces on the mass. So,
F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass
-kx - 16v = 4a
-16x - 16v = 4a
16x + 16v = -4a
4x + 4v = -a where v = dx/dt and a = d²x/dt²
4x + 4dx/dt = -d²x/dt²
d²x/dt² = - 4dx/dt - 4x which is the required differential equation
A stone is dropped from the upper observation deck of a tower, 50 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t. h(t) = (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) s (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) m/s (d) If the stone is thrown downward with a speed of 9 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)
Answer:
A. Using displacement =Ut + 1/2gt²
=> 0 + 1/2 (-9.8)t²
= -4.9t²
So
h(t) = 50+ displacement
= 50 - 4.9t²
B. To reach the ground
h(t) = 0
So
50-4.9t²= 0
t = √ (50/4.9)
= 3.2s
C. Using
V = u+ gt
U= 0
V= - 9.8(3.2)
= 31.4m/s
D. If u = -9m/s
Then s = ut + 1/2gt²
5t- 1/2gt²
But distance from the ground is
=.> 50-5t- 4.8t²= 0
So t solving the quadratic equation
t= 3.58s
(a) The distance of the stone above the ground level at time t is [tex]h(t) = 50 - 4.9t^2[/tex]
(b) The time taken for the stone to strike the ground is 3.19 s.
(c) The velocity of the stone when it strikes the ground is 31.4 m/s.
(d) The time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
The given parameters;
height above the ground, h₀ = 50 mThe distance of the stone above the ground level at time t is calculated as;
[tex]h(t) = h_0 - ut - \frac{1}{2} gt^2\\\\h(t) = 50 - 0 -0.5\times 9.8t^2\\\\h(t) = 50 - 4.9t^2[/tex]
The time taken for the stone to strike the ground is calculated as;
[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 50}{9.8} } \\\\t = 3.19 \ s[/tex]
The velocity of the stone when it strikes the ground is calculated as;
[tex]v =u + gt\\\\v = 0 + 3.2 \times 9.8\\\\v = 31.4 \ m/s[/tex]
The time taken for the stone to reach the ground when thrown at speed of 9 m/s is calculated as;
[tex]50 = 9t + \frac{1}{2} (9.8)t^2\\\\50 = 9t + 4.9t^2\\\\4.9t^2 + 9t - 50 = 0\\\\a = 4.9 \, \ b = 9, \ \ c = -50\\\\solve \ the \ quadratic \ equation\ using \ formula \ method\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-9 \ \ + /- \ \ \sqrt{(9)^2 - 4(4.9 \times -50)} }{2(4.9)} \\\\t = 2.41 \ s \ \ or \ \ - 4.24 \ s[/tex]
Thus, the time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
Learn more here:https://brainly.com/question/9527588
A cook preparing a meal for a group of people is an example of
O kinetic energy because he has the ability to make a meal
O potential energy because he has the ability to make a meal
O kinetic energy because he is making the meal
o potential energy because he is making the meal
A certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied. In approximately what distance would the car stop had it been going 66.0mph
Answer: 156.02 metre.
Explanation:
Give that a certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied.
Let us use third equation of motion,
V^2 = U^2 + 2as
Since the car is decelerating, V = 0
And acceleration a will be negative.
U = 33 mph
S = 39 m
Substitute both into the formula
0 = 33^2 - 2 × a × 39
0 = 1089 - 78a
78a = 1089
a = 1089 / 78
a = 13.96 m/h^2
If we assume that the car decelerate at the same rate.
the distance the car will stop had it been going 66.0mph will be achieved by using the same formula
V^2 = U^2 + 2as
0 = 66^2 - 2 × 13.96 × S
4356 = 27.92S
S = 4356 / 27.92
S = 156.02 m
Therefore, the car would stop at
156.02 m
As a skydiver falls, his potential energy ___ and his kinetic energy __
increases,increases
increases,decreases
decreases,increases
decreases, decreases
Answer:
Hey there!
PE=mgh, so as height decreases, so does the potential energy.
KE=mv^2, so as velocity increases, kinetic energy increases.
Thus, the correct answer would be Decreases, Increases.
Let me know if this helps :)
A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?
Answer:
Distance= 4 miles
Time = 36.3 seconds
Explanation:
80 mph = 178.95 m/s
90 mph = 201.32 m/s
V = u +at
201.32= 0+a(4.5)
201.32/4.5= a
44.738 m/s² = a
Acceleration of the cop car
= 44.738 m/s²
Distance traveled at 4.5seconds
For the cop car
S= ut + ½at²
S= 0(4.5) + ½*44.738*4.5
S= 100.66 meters
Distance traveled at 4.5seconds
For the speeding car
4.5*178.95=805.275
The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal
X/178.95= (704.675+x)/201.32
X-0.89x= 626.37
0.11x= 626.37
X= 5694.3 meters
The time = 5694.3/178.95
Time =31.8 seconds
So the distance they meet
= 5694.3+805.275
= 6499.575 meters
= 4.0 miles
The Time = 4.5+31.8
Time = 36.3 seconds
When the adjustable mirror on the Michelson interferometer is moved 20 wavelengths, how many fringe pattern shifts would be counted
Answer:
The number of fringe pattern shift is m = 40
Explanation:
From the question we are told that
The Michelson interferometer is moved 20 wavelengths i.e [tex]20 \lambda[/tex]
Generally the distance which the Michelson interferometer is moved is mathematically represented as
[tex]d = \frac{m * \lambda}{2}[/tex]
Here [tex]m[/tex] is the number of fringe pattern shift
So
[tex]20 \lambda = \frac{m * \lambda}{2}[/tex]
[tex]40 \lambda = m * \lambda[/tex]
m = 40
Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field exceeds 0.100 T. In the absence of any external magnetic field, determine the maximum current a 5.68-mm-diameter niobium wire can carry and remain superconducting.
Answer:
The current is [tex]I = 1420 \ A[/tex]
Explanation:
From the question we are told that
The diameter of the wire is [tex]d = 5.68 \ mm = 0.00568 \ m[/tex]
The magnetic field is [tex]B = 0.100 \ T[/tex]
Generally the radius of the wire is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{ 0.00568}{2}[/tex]
[tex]r = 0.00284 \ m[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * I}{ 2 \pi r }[/tex]
=> [tex]I =\frac{ B * 2 \pi r }{\mu_o}[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]
substituting values
=> [tex]I =\frac{ 0.100 * 2 * 3.142 * 0.00284 }{ 4 \pi * 10^{-7}}[/tex]
=> [tex]I = 1420 \ A[/tex]
A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. A parallel plate capacitor sets up an electric field E which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by 3 mm and the value of the magnetic field is 0.3 T, what voltage between the plates will allow particles of speed 5 x 105 m/s to pass straight through without deflection? A. 70 V B. 140 V C. 450 V D. 1,400 V E. 2,800 V
Answer:
C. 450v
Explanation:
Using
Voltage= B*distance of separation*velocity
3mm x 0.3T x 5E5m/s
= 450v
a person Travels along a straight road for half the distance with velocity V1 and the remaining half the distance with velocity V2 the average velocity is given by
Answer: (V1+V2)/2
Explanation: This is because basically with the question they are trying to say u(initial velocity) is V1 and v(final velocity) is V2 as the journey starts off with V1 and ends with V2 so therefore we know an equation where average velocity=(u+v)/2. So here it’s (V1+V2)/2
A diffraction grating 19.2 mm wide has 6010 rulings. Light of wavelength 337 nm is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of θ at which maxima appear on a distant viewing screen?
Answer:
(a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
Explanation:
Given that,
Width of diffraction grating [tex]w= 19.2\ mm[/tex]
Number of rulings[tex]N=6010[/tex]
Wavelength = 337 nm
We need to calculate the distance between adjacent rulings
Using formula of distance
[tex]d=\dfrac{w}{N}[/tex]
Put the value into the formula
[tex]d=\dfrac{19.2\times10^{-3}}{6010}[/tex]
[tex]d=3.19\times10^{-6}\ m[/tex]
We need to calculate the value of m
Using formula of constructive interference
[tex]d \sin\theta=m\lambda[/tex]
[tex]\sin\theta=\dfrac{m\lambda}{d}[/tex]
Here, m = 0,1,2,3,4......
[tex]\lambda[/tex]=wavelength
For largest value of θ
[tex]\dfrac{m\lambda}{d}>1[/tex]
[tex]m>\dfrac{d}{\lambda}[/tex]
Put the value into the formula
[tex]m>\dfrac{3.19\times10^{-6}}{337\times10^{-9}}[/tex]
[tex]m>9.46[/tex]
[tex]m = 9[/tex]
(a). We need to calculate the largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{9\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=71.9^{\circ}[/tex]
(b). We need to calculate the second largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{8\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=57.7^{\circ}[/tex]
(c). We need to calculate the third largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{7\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=47.7^{\circ}[/tex]
Hence, (a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
Find the focal length of contact lenses that would allow a nearsighted person with a 130 cmcm far point to focus on the stars at night.
Answer:
130cmExplanation:
The lens equation is expressed as;
1/f = 1/u+1/v where;
f is the focal length of the lens
u is the object distance
v is the image distance
Since the near sighted person wants focus the starts at nigt, the stars at night are the images located that infinity. Hence the image distance v = ∞.
The object distance u = 130cm
Substituting the given parameters in the formula to get the focal length f
[tex]\frac{1}{f} = \frac{1}{\infty} + \frac{1}{130} \\\\As \ x \ tends \ to \ \infty, \, \frac{a}{x} \ tends \ to \ 0 \ where\ 'a' \ is \ a\ constant \\\\} \\\\[/tex]
[tex]\frac{1}{f} = 0+ \frac{1}{130}\\\\[/tex]
[tex]\frac{1}{f} =\frac{1}{130}\\cross\ multiply\\\\f = 130*1\\\\f = 130cm[/tex]
Hence the focal length of contact lenses that would allow a nearsighted person with a 130 cm far point to focus on the stars at night is 130cm
A hot cup of coffee is placed on a table. Which will happen because of conduction? Answer options with 4 options A. The temperature of the coffee will decrease while the temperature of the table decreases. B. The temperature of the coffee will increase while the temperature of the table increases. C. The temperature of the coffee will decrease while the temperature of the table increases. D. The temperature of the coffee will increase while the temperature of the table decreases.
Answer:
C.Explanation:
C. The temperature of the coffee will decrease while the temperature of the table increases.
How much time will elapse if a radioisotope with a half-life of 88 seconds decays to one-sixteenth of its original mass?
Answer:
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
Explanation:
The decay of radioisotopes are represented by the following ordinary differential equation:
[tex]\frac{dm}{dt} = -\frac{t}{\tau}[/tex]
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]\tau[/tex] - Time constant, measured in seconds.
[tex]m[/tex] - Mass of the radioisotope, measured in grams.
The solution of this expression is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]m_{o}[/tex] is the initial mass of the radioisotope, measured in kilograms.
The ratio of current mass to initial mass is:
[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
The time constant is now calculated in terms of half-life:
[tex]\tau = \frac{t_{1/2}}{\ln2}[/tex]
Where [tex]t_{1/2}[/tex] is the half-life of the radioisotope, measured in seconds.
Given that [tex]t_{1/2} = 88\,s[/tex], the time constant of the radioisotope is:
[tex]\tau = \frac{88\,s}{\ln 2}[/tex]
[tex]\tau \approx 126.957\,s[/tex]
Now, if [tex]\frac{m(t)}{m_{o}(t)} = \frac{1}{16}[/tex] and [tex]\tau \approx 126.957\,s[/tex], the time is:
[tex]t = -\tau \cdot \ln\frac{m(t)}{m_{o}}[/tex]
[tex]t = -(126.957\,s)\cdot \ln \frac{1}{16}[/tex]
[tex]t \approx 352\,s[/tex]
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
Astronomers think planets formed from interstellar dust and gases that clumped together in a process called? A. stellar evolution B. nebular aggregation C. planetary accretion D. nuclear fusion
Answer:
C. planetary accretion
Explanation:
Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.
Answer:
[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]
Explanation:
Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.
Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.
A collector that has better efficiency in cold weather is the:
flat-plate collector due to reduced heat loss
evacuated tube collector due to its larger size
flat-plate collector due to the dark-colored coating
O evacuated tube collector due to reduced heat loss
Question 23 (1 point) Saved
One of the following is not found in Thermosyphon systems
o
Answer:
D. evacuated tube collector due to reduced heat loss
Explanation:
Evacuated tube collectors has vacuum which reduces the loss of heat and increase the efficiency of the collector. It has a major application in solar collector, and converts solar energy to heat energy. It can also be used for heating of a definite volume of water majorly for domestic purpose.
During cold weather, the conservation and efficient use of heat is required. Therefore, evacuated tube collector is preferred so as to reduce heat loss and ensure the maximum use of heat energy.
A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes from 17.0 cm to 6.0 cm in 0.53 s.
A) Determine the direction of the induced current.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?
Answer:
(A). The direction of the induced current will be clockwise.
(B). The magnitude of the average induced emf 16.87 mV.
(C). The induced current is 6.75 mA.
Explanation:
Given that,
Magnetic field = 0.45 T
The loop's diameter changes from 17.0 cm to 6.0 cm .
Time = 0.53 sec
(A). We need to find the direction of the induced current.
Using Lenz law
If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.
(B). We need to calculate the magnetic flux
Using formula of flux
[tex]\phi_{1}=BA\cos\theta[/tex]
Put the value into the formula
[tex]\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0[/tex]
[tex]\phi_{1}=0.01021\ Wb[/tex]
We need to calculate the magnetic flux
Using formula of flux
[tex]\phi_{2}=BA\cos\theta[/tex]
Put the value into the formula
[tex]\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0[/tex]
[tex]\phi_{2}=0.00127\ Wb[/tex]
We need to calculate the magnitude of the average induced emf
Using formula of emf
[tex]\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})[/tex]
Put the value into t5he formula
[tex]\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})[/tex]
[tex]\epsilon=0.016867\ V[/tex]
[tex]\epsilon=16.87\ mV[/tex]
(C). If the coil resistance is 2.5 Ω.
We need to calculate the induced current
Using formula of current
[tex]I=\dfrac{\epsilon}{R}[/tex]
Put the value into the formula
[tex]I=\dfrac{0.016867}{2.5}[/tex]
[tex]I=0.00675\ A[/tex]
[tex]I=6.75\ mA[/tex]
Hence, (A). The direction of the induced current will be clockwise.
(B). The magnitude of the average induced emf 16.87 mV.
(C). The induced current is 6.75 mA.
A wire carries current in the plane of this paper toward the top of the page. The wire experiences a ma netic force toward the right edge of the page. The direction of the magnetic field causing this force is:
A. in the plane of the page and toward the left edge
B. in the plane of the page and toward the bottom edge
C. upward out of the page
D. downward into the page
Answer:
D) True. In this case the thumb goes up the page, the fingers are extended out of the page and the palm points to the left
Explanation:
The magnetic force on a conductor is given by
F = i L x B
bold letters indicate vectors. We can write this expression in the form of magnitudes
F = i L B sin θ
The direction of the force can be found by the rule of the right hand, the thumb points in the direction of the current, the fingers extended in the direction of the magnetic field and the palm gives the direction of the force
Let's apply this expression to the case presented.
A) False. In this case the force is out of the page and is in contradiction with the real force
B) False. In this case the force is zero since the displacement of the current and the field would be parallel
C) False. In this case the force is to the left
D) True. In this case the thumb goes up the page, the fingers are extended out of the page and the palm points to the left
A loop of wire is at the edge of a region of space containing a uniform magnetic field B. The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. That is, dA/dt=−c, with c>0.Required:a. The induced emf in the loop is measuredto be V. What is the magnitude B of the magnetic field that the loop was in?b. For the case of a square loop of sidelength L being pulled out of the magneticfield with constant speed v, What is the rate of change of area c= -dA/dt
Answer:
The question is not clear enough. So i have attached a copy of the correct question.
A) B = V/c
B) c = Lv
Explanation:
A) we know that formula for magnetic flux is;
Φ = BA
Where B is magnetic field and A is area
Now,
Let's differentiate with B being a constant;
dΦ/dt = B•dA/dt
From faradays law, the EMF induced is given as;
E = -dΦ/dt
However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.
Thus, V = -dΦ/dt
Thus, we can now say that;
-V = B•dA/dt
Now from the question, we are told that dA/dt = - c
Thus;
-V = B•-c
So, V = Bc
Thus, B = V/c
B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:
Thus;
E =LvB or can be written as; V = LvB
Where;
V is EMF
L is length of bar
v is velocity
From the first solution, we saw that;
V = Bc
Thus, equating both of the equations, we have;
Bc = LvB
B will cancel out to give;
c = Lv
Explanation:
Select the situation for which the torque is the smallest.
a. A 200 kg piece of silver is placed at the end of a 2.5 m tree branch.
b. A 20 kg piece of marble is placed at the end of a 25 m construction crane arm.
c. A 8 kg quartz rock is placed at the end of a 62.5 m thin titanium rod.
d. The torque is the same for two cases.
e. The torque is the same for all cases.
Answer:
e. The torque is the same for all cases.
Explanation:
The formula for torque is:
τ = Fr
where,
τ = Torque
F = Force = Weight (in this case) = mg
r = perpendicular distance between force an axis of rotation
Therefore,
τ = mgr
a)
Here,
m = 200 kg
r = 2.5 m
Therefore,
τ = (200 kg)(9.8 m/s²)(2.5 m)
τ = 4900 N.m
b)
Here,
m = 20 kg
r = 25 m
Therefore,
τ = (20 kg)(9.8 m/s²)(25 m)
τ = 4900 N.m
c)
Here,
m = 8 kg
r = 62.5 m
Therefore,
τ = (8 kg)(9.8 m/s²)(62.5 m)
τ = 4900 N.m
Hence, the correct answer will be:
e. The torque is the same for all cases.
What is the separation in meters between two slits for which 594 nm orange light has its first maximum at an angle of 32.8°?
Answer:
1.1micro meter
Explanation:
Given that
Constructive interference is
ma = alpha x sin theta
Alpha = 1 x 594 x10^ -9/ sin 32.8°
= 1.1 x 10^ -6m
Explanation:
A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is the period if the amplitude of the motion is doubled
Answer:
The period of the motion will still be equal to T.
Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as
[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]
where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.
If a vacuum pump reduces the pressure of a gas to 1.0 x 10-6 atm, what is the pressure expressed in millimeters of mercury
Answer:
[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.
Explanation:
According to current SI unit conversions, 1 atmosphere is equal to 760 millimeters of mercury. The current pressure is determined by simple rule of three:
[tex]p = \frac{760\,mm\,Hg}{1\,atm} \times (1\times 10^{-6}\,atm)[/tex]
[tex]p = 7.6\times 10^{-4}\,mm\,Hg[/tex]
[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.
The A block, with negligible dimensions and weight P, is supported by the coordinate point (1.1/2) of the parabolic fixed grounded surface, from equation y = x^2/2 If the block is about to slide, what is the coefficient of friction between it and the surface; determine the force F tangent to the surface, which must be applied to the block to start the upward movement.
Answer:
μ = 1
F = P√2
Explanation:
The parabola equation is: y = ½ x².
The slope of the tangent is dy/dx = x.
The angle between the tangent and the x-axis is θ = tan⁻¹(x).
At x = 1, θ = 45°.
Draw a free body diagram of the block. There are three forces:
Weight force P pulling down,
Normal force N pushing perpendicular to the surface,
and friction force Nμ pushing up tangential to the surface.
Sum of forces in the perpendicular direction:
∑F = ma
N − P cos 45° = 0
N = P cos 45°
Sum of forces in the tangential direction:
∑F = ma
Nμ − P sin 45° = 0
Nμ = P sin 45°
μ = P sin 45° / N
μ = tan 45°
μ = 1
Draw a new free body diagram. This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.
Sum of forces in the tangential direction:
∑F = ma
F − Nμ − P sin 45° = 0
F = Nμ + P sin 45°
F = (P cos 45°) μ + P sin 45°
F = P√2
Light of wavelength 550 nm is incident on a slit having a width of 0.200 mm. The viewing screen is 1.90 m from the slit. Find the width of the central bright fringe
Answer:
The width of Center bright fringe is 10.2mm
Explanation:
Given that if
Y/ L << 1 then
Sin theta will be approx Y/L
So sin theta approx Y/L = lamda/a
Y= a x lambda/a
By substituting
1.9x 10^ -3m x 550*10^-9/ 0.2 x 10^-3m
= 5.2mm
But
Change in y = 2y = 10.4mm
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In the tug of war game, none of the teams won. What can you conclude about the forces of the two teams ? Write all the evidence to support your answer.
Answer:
Explanation:
We can conclude that the forces of the two teams are equal and opposite and hence they cancel each other. Therefore none of the teams won as the rope did not move.
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On a separate sheet of paper, tell why scientists in different countries can easily compare the amount of matter in similar objects in their countries
Answer: no u
Explanation: no u
In a physics lab, light with a wavelength of 490 nm travels in air from a laser to a photocell in a time of 17.5 ns . When a slab of glass with a thickness of 0.800 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 21.5 ns to travel from the laser to the photocell.What is the wavelength of the light in the glass? Use 3.00×108 m/s for the speed of light in a vacuum. Express your answer using two significant figures.
Answer:
196 nm
Explanation:
Given that
Value of wavelength, = 490 nm
Time spent in air, t(a) = 17.5 ns
Thickness of glass, th = 0.8 m
Time spent in glass, t(g) = 21.5 ns
Speed of light in a vacuum, c = 3*10^8 m/s
To start with, we find the difference between the two time spent
Time spent on glass - Time spent in air
21.5 - 17.5 = 4 ns
0.8/(c/n) - 0.8/c = 4 ns
Note, light travels with c/n speed in media that has index of refraction
(n - 1) * 0.8/c = 4 ns
n - 1 = (4 ns * c) / 0.8
n - 1 = (4*10^-9 * 3*10^8) / 0.8
n - 1 = 1.2/0.8
n - 1 = 1.5
n = 1.5 + 1
n = 2.5
As a result, the wavelength of light in a medium with index of refraction would then be
490 / 2.5 = 196 nm
Therefore, our answer is 196 nm
What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×10-34 Js.)
Answer:
9.82 × [tex]10^{-35}[/tex] Hz
Explanation:
De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:
λ = [tex]\frac{h}{mv}[/tex]
where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.
Given that: h = 6.63 ×[tex]10^{-34}[/tex] Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;
λ = [tex]\frac{h}{mv}[/tex]
= [tex]\frac{6.63*10^{-34} }{2.5*2.7}[/tex]
= [tex]\frac{6.63 * 10^{-34} }{6.75}[/tex]
= 9.8222 × [tex]10^{-35}[/tex]
The wavelength of the object is 9.82 × [tex]10^{-35}[/tex] Hz.
A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down.
In the 10.0 s period following the inital spin, the bike wheel undergoes 60.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ????s will it take the bike wheel to come to a complete stop?
The bike wheel has a mass of 0.625 kg0.625 kg and a radius of 0.315 m0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ????fτf that was acting on the spinning wheel.
Answer:
a) Δt = 24.96 s , b) τ = 0.078 N m
Explanation:
This is a rotational kinematics exercise
θ = w₀ t - ½ α t²
Let's reduce the magnitudes the SI system
θ = 60 rev (2π rad / 1 rev) = 376.99 rad
w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s
α = (w₀ t - θ) 2 / t²
let's calculate the annular acceleration
α = (43.98 10 - 376.99) 2/10²
α = 1,258 rad / s²
Let's find the time it takes to reach zero angular velocity (w = 0)
w = w₀ - alf t
t = (w₀ - 0) / α
t = 43.98 / 1.258
t = 34.96 s
this is the total time, the time remaining is
Δt = t-10
Δt = 24.96 s
To find the braking torque, we use Newton's law for angular motion
τ = I α
the moment of inertia of a circular ring is
I = M r²
we substitute
τ = M r² α
we calculate
τ = 0.625 0.315² 1.258
τ = 0.078 N m
The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
Given data:
The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex] (rps means rotation per second).
The time interval is, t' = 10.0 s.
The number of rotations made by wheel is, n = 60.0.
The mass of bike wheel is, m = 0.625 kg.
The radius of wheel is, r = 0.315 m.
The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,
[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]
Here, [tex]\theta[/tex] is the angular displacement, and its value is,
[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]
And, angular speed is,
[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]
Solving as,
[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]
Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.
[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]
Then total time is,
T = t - t'
T = 35.18 - 10
T = 25.18 s
Now, use the standard formula to obtain the value of braking torque as,
[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]
Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
Learn more about the rotational motion here:
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