Answer:
The torque exerted on the merry-go-round is 766.95 Nm
Explanation:
Given;
mass of the merry-go-round, m = 416 kg
radius of the disk, r = 1.7 m
angular speed of the merry-go-round, ω = 3.7 rad/s
time of motion, t = 2.9 s
The torque exerted on the merry-go-round is calculated as;
[tex]\tau = Fr= I\alpha\\\\\tau = (\frac{1}{2} m r^2)(\frac{\omega }{t} )\\\\\tau = (\frac{1}{2} \times 416 \times 1.7^2)( \frac{3.7}{2.9} )\\\\\tau = 766.95 \ Nm[/tex]
Therefore, the torque exerted on the merry-go-round is 766.95 Nm
12. Identify the Leader
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
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:
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the first light source has a wavelength of 640 nm. Two different interference patterns are observed. If the 10th order bright fringe from the first light source coincides with the 12th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source
Answer:
533.33 nm
Explanation:
Since dsinθ = mλ for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.
Since the fringes coincide,
m'λ = m"λ'
λ' = m'λ/m"
= 10 × 640 nm/12
= 6400 nm/12
= 533.33 nm
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 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
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Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight
The answer is Motion
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.
a cohesive force between the liquids molecules is responsible for the fluids is called
Answer:
static force
Explanation:
mark me brainliest
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
What is happening in the graph shown below?
A.
The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
B.
The object moves toward the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 2 m/s.
C.
The object moves toward the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 8 m/s.
D.
The object moves away from the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 8 m/s.
Answer:
D. The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
Explanation:
I just got it right lol
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
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
why food cook faster with salt water than cook with pure water
Answer:
oil heats faster
Explanation:
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
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.
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.
someone please help I can mark brainless
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]
Scientific work is currently under way to determine whether weak oscillating magnetic fields can affect human health. For example, one study found that driv- ers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long expo- sure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude 1.00 3 1023 T, oscillating sinusoidally at 60.0 Hz. If the diameter of a red blood cell is 8.00 mm, determine the maximum emf that can be generated around the perimeter of a cell in this field.
Answer:
fem = 7.58 10⁻⁵ V
Explanation:
For this exercise we use Faraday's law
fem = [tex]- \frac{d \Phi _B}{dt}[/tex]
the magnetic flux is
Ф_B = B. A = B A cos θ
Tje bold are vectros. Suppose the case where the normal to the surface of the red blood cell is parallel to the field, therefore the angle is zero and the cos 0 = 1
The red blood cell area is
A =π r²
indicate that the diameter is r = 8.00 mm = 8.00 10⁻³ m
the magnetic field has a frequency of f=60 Hz, and B₀ = 1.00 10⁻³T, therefore we can write it
B = B₀ sin (wt) = B₀ sin( 2π f t)
we substitute
fem = - A dB / dt
fem = - A B₀ [tex]\frac{ d (sin ( 2\pi f t)}{dt}[/tex]
fem = - π r² Bo (2πf cos 2πft)
the maximum electromotive force occurs when the function is ±1
fem = 2 π² r² B₀ f
let's calculate
fem = 2π² (8.00 10⁻³)² 1.00 10⁻³ 60
fem = 7.58 10⁻⁵ V
If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic heard if you blow much harder?
___Hz
Answer:
Generally, the lowest overtone for a pipe open at one end and closed would be at y / 4 where y represents lambda, the wavelength.
Since F (frequency) = c / y Speed/wavelength
F2 / F1 = y1 / y2 because c is the same in both cases
F2 = y1/y2 * F1
F2 = 3 F1 = 750 /sec
Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe
and y1 = 3 y2
The second harmonic will be three times the first harmonic. The answer is 750 Hz
VIBRATION OF WAVES IN PIPESClosed pipes have odd multiples of frequencies or harmonics. That is,
If [tex]F_{0}[/tex] = fundamental frequency = first harmonic
[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic
[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic
[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic
Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.
By using the formula above,
second harmonic will be 3 x 250 = 750Hz
Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz
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A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
the cat's displacement during this motion?
A. 6m
B. 2.5 m
C. 8.5 m
D. 4.5 m
20
А
B
D
Answer:
see that the correct one is B
Explanation:
To solve this exercise let us use the kinematic relations
v² = v₀² - 2 a x
as they indicate that the car stops, therefore the final speed is yield v = 0
x = v₀² / 2a
let's calculate
x = 2²/(2 0.8)
x = 2.5 m / s²
When reviewing the answers we see that the correct one is B
1. Define force and give its Sl unit
Answer:
Force is an external agency that changes or tends to change the state of body from rest to motion or motion to rest.
The SI unit of force is newton(N)
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:
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:
Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths from the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t[tex]_{min[/tex] = ( m + 1/2) × λ/2n
we substitute
t[tex]_{min[/tex] = ( 0 + 1/2) × 711 /2(1.21)
t[tex]_{min[/tex] = 0.5 × 711/2.42
t[tex]_{min[/tex] = 0.5 × 293.80165
t[tex]_{min[/tex] = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm
starting from rest, your bicycle can reach a speed of 4.0 m/s in 50 s. Assuming that your bicycle accelerates at a constant rate, what is its acceleration?
Answer:
0.08 ms^-2
Explanation:
by using v= u + at
initial velocity is zero as it is starting from rest
4= 0 + a x 50
4/50 = a = 0.08 ms^-2
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.
A light year is the amount of time it takes for light from the Sun to reach the Earth.
True
False
A hair dryer draws a current of 12.8 A.
(a)How many minutes does it take for
6.8 x 10° C of charge to pass through the
hair dryer? The fundamental charge is
1.602 x 10-19 C.
Answer in units of min.
(b)How many electrons does this amount of
charge represent?
Answer in units of electrons.
Answer:
(a) 8.85×10⁻³ minutes
(b) 4.24×10¹⁹ electrons
Explanation:
(a) Using,
Q = it............................. Equation 1
Where Q = quantity of charge, i = current, t = time.
Make t the subject of the equation
t = Q/i............................. Equation 2
Given: Q = 6.8×10⁰ C, i = 12.8 A
Substitute these values into equation 2
t = 6.8×10⁰/12.8
t = 8.85×10⁻³ minutes
(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3
Where n = number of electrons.
Given: Q = 6.8×10⁰ C
Substitute into equation 2
n = 6.8×10⁰/1.602×10⁻¹⁹
n = 4.24×10¹⁹ electrons
(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b) Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
What will be the time of the charge and number of the electrons in the charge ?As we know Q = IT
Where Q = quantity of charge, i = current, T = time.
From the above equation
T= Q/I.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A
Substitute these values
T= [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8
T = [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes
Now the number of the electrons present in the charge will be
n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])
Where n = number of electrons.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C
Substitute Value of Q
n = [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]
n = [tex]4.24\times\d10^{19}[/tex] electrons
Thus
(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b)Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
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