Answer:
12.5 rad/s²
Explanation:
Angular Acceleration: This can be defined as the ratio of linear acceleration and radius. The S.I unit is rad/s²
From the question,
a = αr................... Equation 1
Where a = linear acceleration, α = angular acceleration, r = radius.
But,
a = (v-u)/t.............. Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
(v-u)/t = αr
make α the subject of the equation
α = (v-u)/tr................. Equation 3
Given: v = 30 m/s, u = 0 m/s, t = 6 s, r = 0.4 m
Substitute into equation 3
α = (30-0)/(0.4×6)
α = 30/2.4
α = 12.5 rad/s²
You have a circuit of three resistors in series connected to a battery. You add a fourth resistor, also in series, to the combination. As a result:_______.
A. Power delivered from the battery to combination increases.
B. It is impossible to give the answer without knowing the actual resistances and voltage of the battery.
C. Power delivered from the battery to combination is unchanged.
D. Power delivered from the battery to combination decreases.
Answer:
D
Explanation:
The power equation is P= V^2/R
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As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.
What is a resistors?A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.
Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.
A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.
The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.
Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.
Thus, the correct option is D.
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Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)?
Answer:
94.248 g/sec
Explanation:
For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:
[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]
And, the velocity of blood pumping is 30 cm^2
Now apply the following formula to solve the total current
[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]
Q = 94.248 g/sec
Basically we applied the above formula So, that the total current could come
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 ball is dropped from the top of an eleven-story building to a balcony on the ninth floor. In which case is the change in the potential energy associated with the motion of the ball the greatest
Answer:
at the top of the 9 story building i think
Explanation:
When the ball starts to move, its kinetic energy increases and potential energy decreases. Thus the ball will experience its maximum potential energy at the top height before falling.
What is potential energy?Potential energy of a massive body is the energy formed by virtue of its position and displacement. Potential energy is related to the mass, height and gravity as P = Mgh.
Where, g is gravity m is mass of the body and h is the height from the surface. Potential energy is directly proportional to mass, gravity and height.
Thus, as the height from the surface increases, the body acquires its maximum potential energy. When the body starts moving its kinetic energy progresses and reaches to zero potential energy.
Therefore, at the sate where the ball is at the top of the building it have maximum potential energy and then changes to zero.
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How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Question:
A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Answer:
S = 5.508 × 10¹¹m
V = 2.62 × 10⁸ m/s
Explanation:
The radius of the orbit of Jupiter, Rj is 43.2 light-minutes
radius of the orbit of Mars, Rm is 12.6 light-minutes
Distance travelled S = (Rj - Rm)
= 43.2 - 12.6 = 30.6 light- minutes
= 30.6 × (3 ×10⁸m/s) × 60 s
= 5.508 × 10¹¹m
time = 35mins = (35 × 60 secs)
= 2100 secs
speed = distance/time
V = 5.508 × 10¹¹m / 2100 s
V = 2.62 × 10⁸ m/s
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]
According to Newton, when the distance between two interacting objects doubles, the gravitational force is
Answer:
1/4 of its original value
Explanation:
Newton's law of universal gravitation states that when two bodies of masses M₁ and M₂ interact, the force of attraction (F) between these bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between these bodies. i.e
F ∝ [tex]\frac{M_1 M_2}{r^2}[/tex] ------------(i)
From the equation above, it can be deduced that;
F ∝ [tex]\frac{1}{r^2}[/tex]
=> F = G [tex]\frac{1}{r^2}[/tex] -----------(ii)
Where;
G = constant of proportionality called the gravitational constant
Equation (ii) can be re-written as
Fr² = G
=> F₁r₁² = F₂r₂² -----------(iii)
Where;
F₁ and r₁ are the initial values of the force and distance respectively
F₂ and r₂ are the final values of the force and distance respectively
From the question, if the distance doubles i.e;
r₂ = 2r₁,
Then the final value of the gravitational force F₂ is calculated as follows;
Substitute the value of r₂ = 2r₁ into equation (iii) as follows;
F₁r₁² = F₂(2r₁)²
F₁r₁² = 4F₂r₁² [Divide through by r₁²]
F₁ = 4F₂ [Make F₂ subject of the formula]
F₂ = F₁ / 4 [Re-write this]
F₂ = [tex]\frac{1}{4} F_1[/tex]
Therefore the gravitational force will be 1/4 of its original value when the distance between the bodies doubles.
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
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
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 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
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
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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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
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
The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?
Answer:
The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.
Explanation:
The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.
The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.
Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.
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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
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).
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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supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistance is 207.4 What is the temerature (in degree Celsius of this solution
Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
T₂ = 95.56°C
a point charge q is located at the center of a cube with edge length d. whatis the value of the flux over one face of the cube
Answer:
q/6Eo
Explanation:
See attached file pls
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
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
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Answer:
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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
Read more: https://brainly.com/question/13754413
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 uniform disk of 10 kg and radius 4.0 m can rotate in a horizontal plane about a vertical axis through its center. The disk is rotating at an angular velocity of 15 rad/s when a 5-kg package is dropped vertically on a point that is 2.0 m from the center of the disk. What is the angular velocity of the disk/package system
Answer:
18.75 rad/s
Explanation:
Moment of inertia of the disk;
I_d = ½ × m_disk × r²
I_d = ½ × 10 × 4²
I_d = 80 kg.m²
I_package = m_pack × r²
Now,it's at 2m from the centre, thus;
I_package = 5 × 2²
I_package = 20 Kg.m²
From conservation of momentum;
(I_disk + I_package)ω1 = I_disk × ω2
Where ω1 = 15 rad/s and ω2 is the unknown angular velocity of the disk/package system.
Thus;
Plugging in the relevant values, we obtain;
(80 + 20)15 = 80 × ω2
1500 = 80ω2
ω2 = 1500/80
ω2 = 18.75 rad/s
A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back toward your friend. What should you do in order to maximize your speed right after your interaction with the ball?
A. You should catch the ball.
B. You should let the ball go past you without touching it.
C. You should deflect the ball back toward your friend.
D. More information is required to determine how to maximize your speed.
E. It doesn't matter. Your speed is the same regardless of what you do.
Answer:
C You should deflect the ball back toward your friend.
Explanation:
This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,
with m= mass, V=velocity, i=initial, f=final:
mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)
So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision
Answer:
A. You should catch the ball.
Explanation:
Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.
Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the initial current (at t = 0), and τ is a constant having dimensions of time. Consider a fixed observation point within the conductor.
Required:
a. How much charge passes this point between t = 0 and t = τ?
b. How much charge passes this point between t = 0 and t = 10 τ?
c. What If ? How much charge passes this point between t = 0 and t = [infinity]?
Answer:
Pls see attached file
Explanation: