Answer:
θ₁ = 0.04º , θ₂ = 0.00118º
Explanation:
The equation that describes the diffraction pattern of a network is
d sin θ = m λ
where the diffraction order is, in this case they indicate that the order
m = 1
θ = sin⁻¹ (λ / d)
Trfuvsmod ls inrsd fr ll red s SI units
d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm
the distance between two lines we can look for it with a direct proportions rule
If there are 4724 lines in a centimeter, the distance for two hundred is
d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm
let's calculate the angles
λ = 300 10-9 m
θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)
θ₁ = sin⁻¹ (7.08 10-4)
θ₁ = 0.04º
λ = 5000
θ₂ = sin-1 (500 10-9 / 4,2337 10-4)
θ₂ = 0.00118º
Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and total work transfer
Answer:
Total heat transfer is positive
Total work transfer is positive
Explanation:
The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.
Heat transfer to a system is positive and that transferred from the system is negative.
Also, work done by a system is positive while the work done on the system is negative.
Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)
Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).
The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?
Answer:
The potential at the center of the sphere is -924 V
Explanation:
Given;
radius of the sphere, R = 0.22 m
electric field at the surface of the sphere, E = 4200 N/C
Since the electric field is directed towards the center of the sphere, the charge is negative.
The Potential is the same at every point in the sphere, and it is given as;
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)
The electric field on the sphere is also given as;
[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]
[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]
Substitute in the value of q in equation (1)
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]
Therefore, the potential at the center of the sphere is -924 V
find the value of k for which the given pair of vectors are not equal
2ki +3j and 8i + 4kj
Answer:
5
Explanation:
An amusement park ride has a vertical cylinder with an inner radius of 3.4 m, which rotates about its vertical axis. Riders stand inside against the carpeted surface and rotate with the cylinder while it accelerates to its full angular velocity. At that point the floor drops away and friction between the riders and the cylinder prevents them from sliding downward. The coefficient of static friction between the riders and the cylinder is 0.87. What minimum angular velocity in radians/second is necessary to assure that the riders will not slide down the wall?
Answer:
The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.
Explanation:
The riders will experience a centripetal force from the cylinder
[tex]F_{C}[/tex] = mrω^2 .... equ 1
where
m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of of the rider
For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as
[tex]F_{f}[/tex] = μR ....equ 2
where
μ = coefficient of friction = 0.87
R is the normal force from the rider = mg
where
m is the rider's mass
g is the acceleration due to gravity = 9.81 m/s
substitute mg for R in equ 2, we'll have
[tex]F_{f}[/tex] = μmg ....equ 3
Equating centripetal force of equ 1 and frictional force of equ 3, we'll get
mrω^2 = μmg
the mass of the rider cancels out, and we are left with
rω^2 = μg
ω^2 = μg/r
ω = [tex]\sqrt{\frac{ug}{r} }[/tex]
ω = [tex]\sqrt{\frac{0.87*9.81}{3.4} }[/tex]
ω = 1.58 rad/second
The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s
The riders will experience a centripetal force from the cylinder
[tex]F = mrw^2[/tex]
where m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of the rider
For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as
f = μN
where
μ = coefficient of friction = 0.87
N is the normal force = mg
f = μmg
Equating centripetal force of and frictional force of we'll get
[tex]mrw^2 = umg[/tex]
[tex]rw^2 = ug[/tex]
[tex]w^2 = ug/r[/tex]
[tex]w= \sqrt{ug/r}[/tex]
[tex]w= \sqrt{0.87*9.8/3.4}[/tex]
ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.
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A tightly wound toroid of inner radius 1.2 cm and outer radius 2.4 cm has 960 turns of wire and carries a current of 2.5 A.
Requried:
a. What is the magnetic field at a distance of 0.9 cm from the center?
b. What is the field 1.2 cm from the center?
Answer:
a
[tex]B = 0.0533 \ T[/tex]
b
[tex]B = 0.04 \ T[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r = 1.2 \ cm = 0.012 \ m[/tex]
The outer radius is [tex]r_o = 2.4 \ cm = \frac{2.4}{100} = 0.024 \ m[/tex]
The nu umber of turns is [tex]N = 960[/tex]
The current it is carrying is [tex]I = 2. 5 A[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * N* I }{2 * \pi * r }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value
[tex]\mu = 4\pi * 10^{-7} N/A^2[/tex]
And the given distance where the magnetic field is felt is r = 0.9 cm = 0.009 m
Now substituting values
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.009 }[/tex]
[tex]B = 0.0533 \ T[/tex]
Fro the second question the distance of the position considered from the center is r = 1.2 cm = 0.012 m
So the magnetic field is
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.012 }[/tex]
[tex]B = 0.04 \ T[/tex]
The magnetic field at a distance of 0.9 cm from the center of the toroid is 0.053 T.
The magnetic field at a distance of 1.2 cm from the center of the toroid is 0.04 T.
The given parameters;
radius of the toroid, r = 1.2 cm = 0.012 mouter radius of the toroid, R = 2.4 cm = 0.024 mnumber of turns, N = 960 turnscurrent in wire, I = 2.5 AThe magnetic field at a distance of 0.9 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.009} \\\\B = 0.053 \ T[/tex]
The magnetic field at a distance of 1.2 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.012} \\\\B = 0.04 \ T[/tex]
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Can an object travel at the speed of
light? Why or why nbt?
Answer:
no the only things that can travel at the speed of light are waves in the electromagnetic spectrum
A length of organ pipe is closed at one end. If the speed of sound is 344 m/s, what length of pipe (in cm) is needed to obtain a fundamental frequency of 50 Hz
Answer:
The length = 27.52m
Explanation:
v=f x wavelength
Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
Assuming that the skateboarder's acceleration is gsin 18.0 ∘, find his speed when he reaches the bottom of the ramp in 3.50 s .
Answer:
Explanation:
v= u + at
v is final velocity , u is initial velocity . a is acceleration and t is time
Initial velocity u = 0 . Putting the given values in the equation
v = 0 + g sin 18 x 3.5
= 10.6 m /s
For a skateboarder who starts from the rest, the speed when he reaches the bottom of the ramp will be 10.6 m/s.
What are Velocity and Acceleration?The term "velocity" refers to a vector measurement of the rate and direction of motion. Velocity is the rate of movement in a single direction, to put it simply. Velocity can be used to determine how fast a rocket is heading into space and how fast a car is moving north on a congested motorway.
There are several types of velocity :
Instantaneous velocityAverage VelocityUniform VelocityNon-Uniform VelocityThe pace at which a person's velocity changes is known as acceleration. This implies that an object is accelerating if its velocity is rising or falling. An object that is accelerating won't have a steady change in location every second like an item moving at a constant speed does.
According to the question, the given values are :
Time, t = 3.50 sec
Initial Velocity, u = 0 m/s
Use equation of motion :
v = u+at
v = 0+ g sin 18 × 3.5
v = 10.6 m/s.
So, the final velocity will be 10.6 m/s.
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_____________ friction is the interlocking of surfaces due to irregularities on the surfaces preventing those surfaces from moving/sliding against each other. For surfaces moving/sliding on each other, ___________ friction overwhelms kinetic friction to that movement/sliding. Kinetic friction is alway larger than ____________ friction. Kinetic friction is alway equal to _________ friction.
Answer:
STATIC, STATIC
KINETIC friction is less than static friction
Explanation:
In this exercise you are asked to complete the sentences with the correct words.
STATIC friction prevents the relative movement of two surfaces in contact.
For moving surfaces the friction is STATIC is greater than the kinetic friction.
For the last two sentences I think they are misspelled, the correct thing is
KINETIC friction is less than static friction
Consider a sound wave modeled with the equation s(x, t) = 3.00 nm cos(3.50 m−1x − 1,800 s−1t). What is the maximum displacement (in nm), the wavelength (in m), the frequency (in Hz), and the speed (in m/s) of the sound wave?
Answer:
- maximum displacement = 3.00nm
- λ = 1.79m
- f = 286.47 s^-1
Explanation:
You have the following equation for a sound wave:
[tex]s(x,t)=3.00nm\ cos(3.50m^{-1}x- 1,800s^{-1} t)[/tex] (1)
The general form of the equation of a sound wave can be expressed as the following formula:
[tex]s(x,t)=Acos(kx-\omega t)[/tex] (2)
A: amplitude of the wave = 3.00nm
k: wave number = 3.50m^-1
w: angular frequency = 1,800s^-1
- The maximum displacement of the wave is given by the amplitude of the wave, then you have:
maximum displacement = A = 3.00nm
- The wavelength is given by :
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{3.50m^{-1}}=1.79m[/tex]
The values for the wavelength is 1.79m
- The frequency is:
[tex]f=\frac{\omega}{2\pi}=\frac{1,800s^{-1}}{2\pi}=286.47s^{-1}[/tex]
The frequency is 286.47s-1
What will be the volume and density of stone if mass of stone is 10 gram .please tell the answer fast it's very urgent I will mark as a brain me answer if you will answer it correct.
Answer:
[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]
Explanation:
Assume the stone consists of basalt, which has a density of 3.0 g/cm³.
[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]
A medieval city has the shape of a square and is protected by walls with length 500 m and height 15 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 m/s). At what range of angles should you tell your men to set the catapult? (Assume the path of the rocks is perpendicular to the wall. Round your answers to one decimal place. Use g ≈ 9.8 m/s2. Enter your answer using interval notation. Enter your answer in terms of degrees without using a degree symbol.)
Answer:
θ₁ = 85.5º θ₂ = 12.98º
Explanation:
Let's analyze this projectile launch problem, the catapults are 100 m from the wall 15 m high, the objective is for the walls, let's look for the angles for which the rock stops touching the wall.
Let's write the equations for motion for this point
X axis
x = v₀ₓ t
x = v₀ cos θ t
Y axis
y = [tex]v_{oy}[/tex] t - ½ g t2
y = v_{o} sin θ t - ½ g t²
let's substitute the values
100 = 80 cos θ t
15 = 80 sin θ t - ½ 9.8 t²
we have two equations with two unknowns, so the system can be solved
let's clear the time in the first equation
t = 100/80 cos θ
15 = 80 sin θ (10/8 cos θ) - 4.9 (10/8 cos θ)²
15 = 100 tan θ - 7.656 sec² θ
we can use the trigonometric relationship
sec² θ = 1- tan² θ
we substitute
15 = 100 tan θ - 7,656 (1- tan² θ)
15 = 100 tan θ - 7,656 + 7,656 tan² θ
7,656 tan² θ + 100 tan θ -22,656=0
let's change variables
tan θ = u
u² + 13.06 u + 2,959 = 0
let's solve the quadratic equation
u = [-13.06 ±√(13.06² - 4 2,959)] / 2
u = [13.06 ± 12.599] / 2
u₁ = 12.8295
u₂ = 0.2305
now we can find the angles
u = tan θ
θ = tan⁻¹ u
θ₁ = 85.5º
θ₂ = 12.98º
A guitar string 0.65 m long has a tension of 61 N and a mass per unit length of 3.0 g/m. (i) What is the speed of waves on the string when it is plucked? (ii) What is the string's fundamental frequency of vibration when plucked? (iii) At what other frequencies will this string vibrate?
Answer:
i
[tex]v = 142.595 \ m/s[/tex]
ii
[tex]f = 109.69 \ Hz[/tex]
iii1 )
[tex]f_2 =219.4 Hz[/tex]
iii2)
[tex]f_3 =329.1 Hz[/tex]
iii3)
[tex]f_4 =438.8 Hz[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 0.65 \ m[/tex]
The tension on the string is [tex]T = 61 \ N[/tex]
The mass per unit length is [tex]m = 3.0 \ g/m = 3.0 * \frac{1}{1000} = 3 *10^{-3 } \ kg /m[/tex]
The speed of wave on the string is mathematically represented as
[tex]v = \sqrt{\frac{T}{m} }[/tex]
substituting values
[tex]v = \sqrt{\frac{61}{3*10^{-3}} }[/tex]
[tex]v = 142.595 \ m/s[/tex]
generally the string's frequency is mathematically represented as
[tex]f = \frac{nv}{2l}[/tex]
n = 1 given that the frequency we are to find is the fundamental frequency
So
substituting values
[tex]f = \frac{142.595 * 1 }{2 * 0.65}[/tex]
[tex]f = 109.69 \ Hz[/tex]
The frequencies at which the string would vibrate include
1 [tex]f_2 = 2 * f[/tex]
Here [tex]f_2[/tex] is know as the second harmonic and the value is
[tex]f_2 = 2 * 109.69[/tex]
[tex]f_2 =219.4 Hz[/tex]
2
[tex]f_3 = 3 * f[/tex]
Here [tex]f_3[/tex] is know as the third harmonic and the value is
[tex]f_3 = 3 * 109.69[/tex]
[tex]f_3 =329.1 Hz[/tex]
3
[tex]f_3 = 4 * f[/tex]
Here [tex]f_4[/tex] is know as the fourth harmonic and the value is
[tex]f_3 = 4 * 109.69[/tex]
[tex]f_4 =438.8 Hz[/tex]
Calculate the ideal banking angle in degrees for a gentle turn of 1.88 km radius on a highway with a 136.3 km/hr speed limit, assuming everyone travels at the speed limit.
Answer:
Ф = 4.4°Explanation:
given:
radius (r) = 1.88 km
velocity (v) = 136.3 km/hr
required:
banking angle ∡ ?
first:
convert 1.88 km to m = 1.88km * 1000m / 1km
r = 1880 m
convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)
v = 37.86 m/s
now.. calculate the angle
Ф = inv tan (v² / r * g) we know that gravity = 9.8 m/s²
Ф = inv tan (37.86² / (1880 * 9.8))
Ф = 4.4°
2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the
Answer:
for the first interference m = 1 y = 2,839 10-3 m
for the second interference m = 2 y = 5,678 10-3 m
Explanation:
The double slit interference phenomenon, for constructive interference is described by the expression
d sin θ = m λ
where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.
In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles
tan θ = y / L
with the angle it is small,
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
the distance between the central maximum and an interference line is
y = m λ L / d
let's reduce the magnitudes to the SI system
λ = 546 nm = 546 10⁻⁹ m
d = 0.25 mm = 0.25 10⁻³ m
let's substitute the values
y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³
y = m 2,839 10⁻³
the explicit value for a line depends on the value of the integer m, for example
for the first interference m = 1
the distance from the central maximum to the first line is y = 2,839 10-3 m
for the second interference m = 2
the distance from the central maximum to the second line is y = 5,678 10-3 m
Which observation have scientists used to support Einstein's general theory of relativity?
The orbital path of Mercury around the Sun has changed.
O GPS clocks function at the same rate on both Earth and in space.
O The Sun has gotten more massive over time.
Objects act differently in a gravity field than in an accelerating reference frame.
Answer:
Objects act differently in a gravity field than in an accelerating reference frame.
Explanation:
The main thrust of the theory general relativity as proposed by Albert Einstein boarders on space and time as the two fundamental aspects of spacetime. Spacetime is curved in the presence of gravity, matter, energy, and momentum. The theory of general relativity explains gravity based on the way space can 'curve', that is, it seeks to relate gravitational force to the changing geometry of space-time.
The Einstein general theory of relativity has replaced Newton's ideas proposed in earlier centuries as a means of predicting gravitational interactions. This concept is quite helpful but cannot be fitted into the context of quantum mechanics due to obvious incompatibilities.
Answer:
A - The orbital path of mercury around the sun has changed.
Explanation:
got right on edg.
a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric field at the 30 cm mark
Answer:
-1748*10^N/C
Explanation:
See attached file
What is the change in internal energy of an engine if you put 15 gallon of gasoline into its tank? The energy content of gasoline is 1.5 x 106 J/gallon. All other factors, such as the engine’s temperature, are constant. How many hours the engine can work if the power of the engine’s motor is 600 W? (8 marks)
Answer:
ΔU = 2.25 x 10⁸ J
t = 104.17 s
Explanation:
The change in internal energy of the engine can be given by the following formula:
ΔU = (Mass of Gasoline)(Energy Content of Gasoline)
ΔU = (1.5 x 10⁶ J/gallon)(15 gallon)
ΔU = 2.25 x 10⁸ J
Now, for the time of operation, we use the following formula of power.
P = W/t = ΔU/t
t = ΔU/P
where,
t = time of operation = ?
ΔU = Change in internal energy = 2.25 x 10⁸ J
P = Power of motor = 600 W
Therefore,
t = (2.25 x 10⁸ J)/(600 W)
t = (375000 s)(1 h/3600 s)
t = 104.17 s
what is the preferred method of using percentage data by using a circle divided into sections
Answer:
A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole
Explanation:
pie charts are a useful way to organize data in order to see the size of components relative to the whole.
Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the function V 230 sin (100t), where tis measured in seconds and Vin volts.What is the frequency of the voltage
Answer:
[tex]\frac{50}{\pi }[/tex]Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
Now,
From the question,
V(t) = 230 sin (100t) ---------------(ii)
By comparing equations (i) and (ii) the following holds;
V = 230
ω = 100
Ф = 0
But;
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = [tex]\frac{50}{\pi }[/tex]Hz
Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz
The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period
Answer:
Q = 23.49 C
Explanation:
We have,
Electric current drawn by the air conditioner is 16.2 A
Time, t = 1.45 s
It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,
[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]
So, the charge of 23.49 C is passing through the circuit during this period.
An electron, moving west, enters a magnetic field of a certain strength. Because of this field the electron curves upward. What is the direction of the magnetic field?
Answer:
Towards the west.
Explanation:
The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.
Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.
According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward
An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s = 9 cos(t) + 9 sin(t), t ≥ 0, where s is measured in centimeters and t in seconds. (Take the positive direction to be downward.) (a) Find the velocity and acceleration at time t.
Answer:
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
Explanation:
Given that
s = 9 cos(t) + 9 sin(t), t ≥ 0
Then acceleration and velocity is
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man’s shoes and the merry-go-round is µs = 0.5, how far from the axis of rotation can he stand without sliding?
Answer:
0.8 m
Explanation:
Draw a free body diagram. There are three forces:
Weight force mg pulling down,
Normal force N pushing up,
and friction force Nμ pushing towards the center.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Sum of forces in the centripetal direction:
∑F = ma
Nμ = m v²/r
Substitute and simplify:
mgμ = m v²/r
gμ = v²/r
Write v in terms of ω and solve for r:
gμ = ω²r
r = gμ/ω²
Plug in values:
r = (10 m/s²) (0.5) / (2.5 rad/s)²
r = 0.8 m
The distance (radius) from the axis of rotation which the man can stand without sliding is 0.784 meters.
Given the following data:
Angular speed = 2.5 rad/s.Coefficient of static friction = 0.5To determine how far (radius) from the axis of rotation can the man stand without sliding:
We would apply Newton's Second Law of Motion, to express the centripetal and force of static friction acting on the man.
[tex]\sum F = \frac{mv^2}{r} - uF_n\\\\\frac{mv^2}{r} = uF_n[/tex]....equation 1.
But, Normal force, [tex]F_n = mg[/tex]
Substituting the normal force into eqn. 1, we have:
[tex]\frac{mv^2}{r} = umg\\\\\frac{v^2}{r} = ug[/tex]....equation 2.
Also, Linear speed, [tex]v = r\omega[/tex]
Substituting Linear speed into eqn. 2, we have:
[tex]\frac{(r\omega )^2}{r} = ug\\\\r\omega ^2 = ug\\\\r = \frac{ug}{\omega ^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]r = \frac{0.5 \times 9.8}{2.5^2} \\\\r = \frac{4.9}{6.25}[/tex]
Radius, r = 0.784 meters
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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly:___________.
A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c
Answer:
0.85c
Explanation:
Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²
Rest mass of proton [tex]M_{0P}[/tex] = 938 MeV/c²
The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²
for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV
for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV
Recall that the rest energy, and the total energy are related by..
[tex]E[/tex] = γ[tex]E_{0}[/tex]
which can be written in this case as
[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1
where [tex]E[/tex] = total energy of the kaon, and
[tex]E_{0}[/tex] = rest energy of the kaon
γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]
where [tex]\beta = \frac{v}{c}[/tex]
But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...
[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2
where [tex]E_{K}[/tex] is the total energy of the kaon, and
[tex]E_{0P}[/tex] is the rest energy of the proton.
From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²
equ 1 becomes
938c² = γ494c²
γ = 938c²/494c² = 1.89
γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89
1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1
squaring both sides, we get
3.57( 1 - [tex]\beta^{2}[/tex]) = 1
3.57 - 3.57[tex]\beta^{2}[/tex] = 1
2.57 = 3.57[tex]\beta^{2}[/tex]
[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72
[tex]\beta = \sqrt{0.72}[/tex] = 0.85
but, [tex]\beta = \frac{v}{c}[/tex]
v/c = 0.85
v = 0.85c
6. Two forces of 50 N and 30 N, respectively, are acting on an object. Find the net force (in
N) on the object if
the forces are acting in the same direction
b. the forces are acting in opposite directions.
Answer:
same direction = 80 (n)
opposite direction = 20 (n) going one direction
Explanation:
same direction means they are added to each other
and opposite means acting on eachother
what is drift speed ? {electricity}
Answer: In physics a drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
Explanation:
A crane lifts a 425 kg steel beam vertically a distance of 64 m. How much work does the crane do on the beam if the beam accelerates upward at 1.8 m/s2
Answer:
work done= 48.96 kJExplanation:
Given data
mass of load m= 425 kg
height/distance h=64 m
acceleration a= 1.8 m/s^2
The work done can be calculated using the expression
work done= force* distance
but force= mass *acceleration
hence work done= 425*1.8*64= 48,960 J
work done= 48.96 kJ
A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t squared plus 2 t, where v (t )is in meters per second. a) How far does the particle travel during the first 2 sec? b) How far does it travel during the second 2 sec?
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
Consider a conducting rod of length 31 cm moving along a pair of rails, and a magnetic field pointing perpendicular to the plane of the rails. At what speed (in m /s) must the sliding rod move to produce an emf of 0.75 V in a 1.75 T field?
Answer:
The speed of the rod is 1.383 m/s
Explanation:
Given;
length of the conducting rod, L = 31 cm = 0.31 m
induced emf on the rod, emf = 0.75V
magnetic field around the rod, B = 1.75 T
Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.
emef = BLv
where;
B is the magnetic field
L is length of the rod
v is the speed of the rod
v = emf / BL
v = (0.75) / (1.75 x 0.31)
v = 1.383 m/s
Therefore, the speed of the rod is 1.383 m/s