Answer:
2.92 x 10¹² electrons
Explanation:
given,
distance between two plastic ball, r = 17 cm
r = 0.17 m
Force of attraction = F = 68 mN
F = 0.068 N
number of electron transferred from one ball to another.
using Coulomb Force equation
[tex]F = \dfrac{kq^2}{r^2}[/tex]
[tex]0.068 = \dfrac{9\times 10^9\times q^2}{0.17^2}[/tex]
q² = 2.1835 x 10⁻¹³
q = 4.67 x 10⁻⁷ C
now, number of electron
[tex]N = \dfrac{q}{e}[/tex]
e is the charge of electron which is equal to 1.6 x 10⁻¹⁹ C
[tex]N = \dfrac{4.67\times 10^{-7}}{1.6\times 10^{-19}}[/tex]
N = 2.92 x 10¹² electrons
electrons were transferred from one ball to the other is 2.92 x 10¹²
Field strength is directly proportional to the square of charge. The charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex]
From Coulomb law:[tex]F = \dfrac {kQ^2}{r^2}[/tex]
Where,
[tex]F[/tex]- field strength = 68 mN = 0.06 N
[tex]Q[/tex]- charge
[tex]r[/tex]- distance = 17 cm =0.17 m
[tex]k[/tex]- constant = 9x10⁹
Put the values in the formula,
[tex]0.068 = \dfrac{9\times 10^9 \rimes Q^2}{0.017^2}\\\\q^2 = 2.1835 \times 10^{-13}\\\\q = 4.67 \times 10^{-7} \rm \ C[/tex]
Therefore, the charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex].
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A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this process?
Answer:
[tex]0.005734 s^{-1}[/tex] and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]a_o=a\times e^{-kt}[/tex]
where,
k = rate constant
t = age of sample
[tex]a_o[/tex] = let initial amount of the reactant
a = amount left after decay process
We have :
[tex]a_o=x[/tex]
[tex]a=58\%\times x=0.58x[/tex]
t = 95 s
[tex]0.58x=x\times e^{-k\times 95 s}[/tex]
[tex]\k= 0.005734 s^{-1}[/tex]
Half life is given by for first order kinetics::
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
[tex]=\frac{0.693}{0.005734 s^{-1}}=120.86 s[/tex]
[tex]0.005734 s^{-1}[/tex] and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
Which of the following changes would increase the seperation between the bright fringes in the diffraction pattern formed by diffraction grating?
A. INCREASE THE WAVELENGTH OF THE LIGHT
B. IMMERSE THE APPARATUSE IN WATER
C. INCREASE THE SEPERATION BETWEEN THE SLITS
D. NONE OF THESE
E. MORE THAN ONE OF THESE
The increase in the separation between the bright fringes in the diffraction pattern formed by diffraction grating occurs when there is increase in the wavelength of the light.
The equation for diffraction grating that relates the wavelength and separation of bright fringes is given as;
[tex]d sin\theta = m \lambda \\\\sin \ \theta = \frac{m \lambda}{d}[/tex]
where;
λ is the wavelength d is the spacingThus, we can conclude that the increase in the separation between the bright fringes in the diffraction pattern formed by diffraction grating occurs when there is increase in the wavelength of the light.
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Increasing the wavelength of the light will increase the separation between the bright fringes in the diffraction pattern. The correct option is Option A. Immersion in water or changing slit separation do not contribute similarly.
To increase the separation between the bright fringes in a diffraction pattern, one effective approach is to increase the wavelength of the light. When the wavelength increases, the angle of diffraction becomes larger for each order of maximum, resulting in greater fringe separation. Therefore, the correct answer is Option A: INCREASE THE WAVELENGTH OF THE LIGHT.
A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general there are an infinite number of such modes, each one with a distinctive frequency fi and associated pattern of oscillation.
Consider an example of a system with normal modes: a string of length L held fixed at both ends, located at x=0 and x=L. Assume that waves on this string propagate with speed v. The string extends in the x direction, and the waves are transverse with displacement along the y direction.
In this problem, you will investigate the shape of the normal modes and then their frequency.
The normal modes of this system are products of trigonometric functions. (For linear systems, the time dependance of a normal mode is always sinusoidal, but the spatial dependence need not be.) Specifically, for this system a normal mode is described by
yi(x,t)=Ai sin(2π*x/λi)sin(2πfi*t)
A)The string described in the problem introduction is oscillating in one of its normal modes. Which of the following statements about the wave in the string is correct?
The wave is traveling in the +x direction.
a) The wave is traveling in the -x direction.
b) The wave will satisfy the given boundary conditions for any arbitrary wavelength lambda_i.
c) The wavelength lambda_i can have only certain specific values if the boundary conditions are to be satisfied.
d) The wave does not satisfy the boundary condition y_i(0;t)=0.
B)Which of the following statements are true?
a)The system can resonate at only certain resonance frequencies f_i and the wavelength lambda_i must be such that y_i(0;t) = y_i(L;t) = 0.
b) A_i must be chosen so that the wave fits exactly on the string.
c) Any one of A_i or lambda_i or f_i can be chosen to make the solution a normal mode.
C) Find the three longest wavelengths (call them lambda_1, lambda_2, and lambda_3) that "fit" on the string, that is, those that satisfy the boundary conditions at x=0 and x=L. These longest wavelengths have the lowest frequencies.
D) The frequency of each normal mode depends on the spatial part of the wave function, which is characterized by its wavelength lambda_i.
Find the frequency f_i of the ith normal mode.
Answer:
Explanation:
(A)
The string has set of normal modes and the string is oscillating in one of its modes.
The resonant frequencies of a physical object depend on its material, structure and boundary conditions.
The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.
Given below are the incorrect options about the wave in the string.
• The wave is travelling in the +x direction
• The wave is travelling in the -x direction
• The wave will satisfy the given boundary conditions for any arbitrary wavelength [tex]\lambda_i[/tex]
• The wave does not satisfy the boundary conditions [tex]y_i(0;t)=0 [/tex]
Here, the string of length L held fixed at both ends, located at x=0 and x=L
The key constraint with normal modes is that there are two spatial boundary conditions,[tex]y(0,1)=0 [/tex]
and [tex]y(L,t)=0[/tex]
.The spring is fixed at its two ends.
The correct options about the wave in the string is
• The wavelength [tex]\lambda_i[/tex] can have only certain specific values if the boundary conditions are to be satisfied.
(B)
The key factors producing the normal mode is that there are two spatial boundary conditions, [tex]y_i(0;t)=0[/tex] and [tex]y_i(L;t)=0[/tex], that are satisfied only for particular value of [tex]\lambda_i[/tex] .
Given below are the incorrect options about the wave in the string.
• [tex]A_i[/tex] must be chosen so that the wave fits exactly o the string.
• Any one of [tex]A_i[/tex] or [tex]\lambda_i[/tex] or [tex]f_i[/tex] can be chosen to make the solution a normal mode.
Hence, the correct option is that the system can resonate at only certain resonance frequencies [tex]f_i[/tex] and the wavelength [tex]\lambda_i[/tex] must be such that [tex]y_i(0;t) = y_i(L;t)=0 [/tex]
(C)
Expression for the wavelength of the various normal modes for a string is,
[tex]\lambda_n=\frac{2L}{n}[/tex] (1)
When [tex]n=1[/tex] , this is the longest wavelength mode.
Substitute 1 for n in equation (1).
[tex]\lambda_n=\frac{2L}{1}\\\\2L[/tex]
When [tex]n=2[/tex] , this is the second longest wavelength mode.
Substitute 2 for n in equation (1).
[tex]\lambda_n=\frac{2L}{2}\\\\L[/tex]
When [tex]n=3[/tex], this is the third longest wavelength mode.
Substitute 3 for n in equation (1).
[tex]\lambda_n=\frac{2L}{3}[/tex]
Therefore, the three longest wavelengths are [tex]2L[/tex],[tex]L[/tex] and [tex]\frac{2L}{3}[/tex].
(D)
Expression for the frequency of the various normal modes for a string is,
[tex]f_n=\frac{v}{\lambda_n}[/tex]
For the case of frequency of the [tex]i^{th}[/tex] normal mode the above equation becomes.
[tex]f_i=\frac{v}{\lambda_i}[/tex]
Here, [tex]f_i[/tex] is the frequency of the [tex]i^{th}[/tex] normal mode, v is wave speed, and [tex]\lambda_i[/tex] is the wavelength of [tex]i^{th}[/tex] normal mode.
Therefore, the frequency of [tex]i^{th}[/tex] normal mode is [tex]f_i=\frac{v}{\lambda_i}[/tex]
.
The wave in the string travels in the +x direction and the wavelength lambda_i can have only certain specific values to satisfy the boundary conditions. The system can resonate at certain resonance frequencies and the wavelength must satisfy the boundary conditions. The frequency of each normal mode depends on the wavelength.
Explanation:For the given system of a string held fixed at both ends, the wave in the string travels in the +x direction. The boundary conditions for the wave to satisfy are that yi(0, t) = 0 and yi(L, t) = 0. Therefore, the wavelength lambda_i can have only certain specific values if the boundary conditions are to be satisfied.
It is not necessary for Ai to fit exactly on the string, so statement (b) is incorrect. However, statement (a) is true - the system can resonate at only certain resonance frequencies fi, and the wavelength lambda_i must be such that yi(0, t) = 0 and yi(L, t) = 0. Any one of Ai or lambda_i or fi can be chosen to make the solution a normal mode, so statement (c) is also true.
The three longest wavelengths that fit on the string and satisfy the boundary conditions are lambda_1 = 2L, lambda_2 = L, and lambda_3 = L/2. These wavelengths have the lowest frequencies.
The frequency fi of the ith normal mode depends on the wavelength lambda_i. The frequency can be calculated using the equation fi = (v/lambda_i) * (1/2), where v is the speed of propagation of waves on the string.
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The winning time for the 2005 annual race up 86 floors of the Empire State Building was 10 min and 49 s . The winner's mass was 60 kg . If each floor was 3.7 m high, what was the winner's change in gravitational potential energy? If the efficiency in climbing stairs is 25 %, what total energy did the winner expend during the race? How many food Calories did the winner "burn" during the race? What was the winner's metabolic power in watts during the race up the stairs? Express your answer using two significant figures.
Answer:
[tex]1.9\times 10^5\ J[/tex]
[tex]1.8\times 10^{2}\ kcal[/tex]
[tex]1.3\times 10^{2}\ kcal[/tex]
[tex]1.2\times 10^{3}\ W[/tex]
Explanation:
m = Mass of person = 60 kg
g = Acceleration due to gravity = 9.81 m/s²
t = Time taken = 10 min and 49 s
Total height
[tex]h=3.7\times 86\\\Rightarrow h=318.2\ m[/tex]
Potential energy is given by
[tex]U=mgh\\\Rightarrow U=60\times 9.81\times 318.2\\\Rightarrow U=187292.52\ J[/tex]
The gravitational potential energy is [tex]1.9\times 10^5\ J[/tex]
The energy in the climb
[tex]\dfrac{187292.52}{0.25}=749170.08\ J[/tex]
Converting to kcal or Cal
[tex]\dfrac{749170.08}{4184}=179.05594\ kcal[/tex]
The amount of energy used to climb [tex]1.8\times 10^{2}\ kcal[/tex]
Amount gone to heat
[tex]179.05594\times 0.75=134.291955\ kcal[/tex]
The amount burned [tex]1.3\times 10^{2}\ kcal[/tex]
Power is given by
[tex]P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{749170.08}{10\times 60+49}\\\Rightarrow P=1154.34526\ W[/tex]
The power is [tex]1.2\times 10^{3}\ W[/tex]
Calculate the linear momentum of photons of wavelength 350 nm. What speed does a hydrogen molecule need to travel to have the same linear momentum?
Answer:
a)[tex]p=1.89x10^{-27}kg.m.s^{-1}[/tex]
b)[tex]v=0.565\frac{m}{s}[/tex]
Explanation:
First, we need to obtain the linear momentum of the photons of wavelength 350nm.
We are going to use the following formula:
[tex]\lambda=\frac{h}{p}\\Where:\\\lambda=wavelength\\h=placnk's\_constant\\p=Linear\_momentum[/tex]
So the linear momentum is given by:
[tex]p=\frac{h}{\lambda}\\\\p=\frac{6.626x10^{-34}J.s}{350x10^{-9}m}\\\\p=1.89x10^{-27}kg.m.s^{-1}[/tex]
Having the linear momentum of the photon, we can calculate the speed of the hydrogen molecule to have the same momentum, we can use the classic formula for that:
[tex]p=m.v[/tex]
[tex]where:\\m=mass\\v=speed\\p=linear\_momentum[/tex]
The mass of the hydrogen molecule is given by:
[tex]m=2*(1.0078x10^{-3}\frac{kg}{mol}})x\frac{1}{6.022x10^{23}mol}[/tex]
[tex]3.35x10^{-27}kg[/tex]
What we've done here is to use the molecular weight of the hydrogen, and covert it kilograms, we had to multiply by two because the hydrogen molecule is found in pairs.
so:
[tex]v=\frac{p}{m}\\\\v=\frac{1.89x10^{-27}kg.m.s^{-1}}{3.35x10^{-27}kg}\\\\v=0.565\frac{m}{s}[/tex]
(a) The linear momentum of the photon is 1.89 x 10⁻²⁷ kgm/s.
(b) The speed of hydrogen atom with same momentum is 2,079.35 m/s.
Linear Momentum of the photon
The linear momentum of the photon is calculated as follows;
P = h/λ
P = (6.63 x 10⁻³⁴) / (350 x 10⁻⁹)
P = 1.89 x 10⁻²⁷ kgm/s
Speed of hydrogen atom with same momentump = mv
v = p/m
v = (1.89 x 10⁻²⁷ ) / (9.11 x 10⁻³¹)
v = 2,079.35 m/s
Thus, the speed of hydrogen atom with same momentum is 2,079.35 m/s.
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What best describes the angle between a changing electric field and the electromagnetic wave produced by it?
always equal to a right angle
always less than a right angle
varies from parallel to perpendicular
varies from perpendicular to parallel
The correct answer here is Option A, always equal to a right angle.
Keep in mind that the phase difference that exists between the electric field and the magnetic wave is 90 °, this means that the changing electric field and the electromagnetic wave are always perpendicular and will be forming a right angle.
The parking brake on a 1000 kg Cadillac has failed, and it is rolling slowly, at 5 mph , toward a group of small children. Seeing the situation, you realize you have just enough time to drive your 2000 kg Volkswagen head-on into the Cadillac and save the children. Part A With what speed should you impact the Cadillac to bring it to a halt?
To solve this problem we will apply the concepts related to the conservation of the Momentum. The initial momentum must be equal to the final momentum. Momentum is described as the product between body mass and its respective velocity. Since there is no movement at the end of the collision, the final momentum will be zero.
Our values are given as
Mass of Cadillac
[tex]m_1 = 1000kg[/tex]
mass of VW
[tex]m_2 = 2000kg[/tex]
Let VW impact with speed v, then for conservation of momentum
[tex]m_1v_1+m_2v_2 = 0 \rightarrow[/tex] the Final speed is zero
Replacing,
[tex]1000kg*5mph+2000kg(-v) = 0[/tex]
[tex]v = \frac{1000kg*5mph}{2000kg}[/tex]
[tex]v = 2.5mph[/tex]
Therefore the speed to bring the cadillac to a halt is 2.5mph
The speed that can impact the Cadillac to bring it to a halt is 2.5 mph.
Velocity of the car can be calculated by the conservation of momentum formula,
[tex]\bold {m_1 v_1 =m_2v_2 = 0}[/tex] Since the final speed is zero,
where,
[tex]\bold {m_1 - mass\ of\ the\ cadillac = 1000 kg}\\\bold {m_2 - mass\ of\ the\ Volks\ Vagon = 2000 kg}[/tex]
[tex]\bold {v_1}[/tex] - velocity of the Cadillac - 5 mph.
Put the values in the formula
[tex]\bold {1000\ kg \times 5\ mph + 2000 kg (-v) = 0 }\\\\\bold {v = \dfrac {1000\ kg \times 5\ mph}{2000\ kg}}\\\\\bold {v = 2.5\ mph}[/tex]
Therefore, the speed that can impact the Cadillac to bring it to a halt is 2.5 mph.
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A 75 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 620 m/s . Part A What is the recoil speed of the hunter?
Express your answer to two significant figures and include the appropriate units.
To develop this problem we will apply the concepts related to the conservation of momentum. For this purpose we will define that the initial moment must be preserved and be equal to the final moment. Since the system was at a stationary moment, the moment before the bullet is fired will be 0, while the moment after the bullet is fired will be equivalent to the sum of the products of the mass and the velocity of each object. , this is,
[tex]m_1u_1+m_2u_2 = m_1v_1+m_2v_2[/tex]
Here,
m = mass of each object
u = Initial velocity of each object
v = Final velocity of each object
Our values are,
[tex]\text{Mass of hunter} ) m_1 = 75 kg = 75000 g[/tex]
[tex]\text{Mass of bullet} = m_2 = 42 g[/tex]
[tex]\text{Speed of bullet} = v_2 = 6200 m/s[/tex]
[tex]\text{Recoil speed of hunter} =v_1 = ?[/tex]
Applying our considerations we have that the above formula would become,
[tex]m_1v_1 + m_2v_2 = 0[/tex]
Solving for speed 1,
[tex]v_1 = - \frac{m_2v_2}{m_1 }[/tex]
[tex]v_1 = - \frac{42 * 620}{75000 }[/tex]
[tex]v_1 = - 0.3472 m/s[/tex]
Therefore the recoil speed of the hunter is -0.3472m/s, said speed being opposite to the movement of the bullet.
The recoil speed of the hunter standing on the frictionless ice is 0.35m/s².
Given the data in the question;
Mass of hunter; [tex]m_1 = 75kg[/tex]Mass of bullet; [tex]m_2 = 42g = 0.042kg[/tex]Velocity of bullet; [tex]v_2 = 620m/s[/tex]Recoil speed of the hunter; [tex]v_1 =\ ?[/tex]
From the law of conservation of momentum:
[tex]MV = mv[/tex]
Where m is mass and v is velocity
So
[tex]m_1v_1 = m_2v_2\\\\v_1 = \frac{m_2v_2}{m_1}[/tex]
We substitute our given values into the equation
[tex]v_1 = \frac{ 0.042kg \ *\ 620m/s^2}{75kg} \\\\v_1 = \frac{26.04kg.m/s^2}{75kg} \\\\v_1 = 0.35m/s^2[/tex]
Therefore, the recoil speed of the hunter standing on the frictionless ice is 0.35m/s².
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An electron is moving horizontally in a laboratory when a uniform electric field is suddenly turned on. This field points vertically upward. Which of the paths shown will the electron follow, assuming that gravity can be neglected?
Explanation:
We know that electron always tends to travel in the direction opposite the direction of electric field ( force in directed opposite to the field on electron).
So, since the field points vertically upward , it force the electron move downward. So, the electron will be accelerated in downward in the right direction.
Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you directed a laser beam of wavelength 650 nm onto it at a rate of 1 mol of photons per second from a base on the moon. The spacecraft has a mass of 1.0 kg. Given that, after a period of acceleration from standstill, speed = (force/mass) x time, how many minutes would it take for the craft to accelerate to a speed of 1.0 m/s (about 2.2 mph)?
Answer:
(a) F = 6.14 *10⁻⁴ N
(b) P = 6.14* 10⁻¹⁰ Pa
(c) t = 27.2 min
Explanation:
Area of sail A = 1.0 km² = 1.0 * 10⁶m²
Wavelength of light λ = 650 nm = 650 * 10⁻⁹ m
Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s
(a)
Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹
= 1.0192 * 10⁻²⁷ kg.m/s
Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.
Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.
dp/dt = R * h/ λ
This is the force acting on the sail.
F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N
F = 6.14 *10⁻⁴ N
b)
Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail
P = F/A = 6.14 * 10⁻⁴ / 1.0 * 10⁶ = 6.14* 10⁻¹⁰ Pa
P = 6.14* 10⁻¹⁰ Pa
c)
Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²
As the spacecraft starts from rest, initial speed u=0,m/s ,
final speed is u = 1.0 m/s after time t
v = u+at
t = 1.0 - 0/ 6.14 * 10⁻⁴ = 1629s = 27.2 min
t = 27.2 min
Two identical 2.0-kg objects are involved in a collision. The objects are on a frictionless track. The initial velocity of object A is 12 m/s and the initial velocity of object B is -6.0 m/s. What is the final velocity of the two objects
Answer:
final velocity of the objects will be 3 m/sec
Explanation:
We have given there are two identical objects of mass 2 kg
So [tex]m_1=m_2=2kg[/tex]
Initial velocity of object A [tex]v_1=12m/sec[/tex]
Initial velocity of object B [tex]v_2=-6m/sec[/tex]
According to conservation of momentum
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]2\times 12+2\times -6=(2+2)v[/tex]
[tex]4v=24-12[/tex]
[tex]4v=12[/tex]
v = 3 m/sec
So final velocity of the objects will be 3 m/sec
To find the final velocity of mass B after a collision, you can use the principle of conservation of momentum. The formula for conservation of momentum is (mass of A * initial velocity of A) + (mass of B * initial velocity of B) = (mass of A * final velocity of A) + (mass of B * final velocity of B). By substituting the given values and solving for the final velocity of mass B, we can determine its value.
Explanation:To find the final velocity of mass B after the collision, we can use the principle of conservation of momentum. Since the collision is elastic, the total momentum before the collision is equal to the total momentum after the collision.
Mass A has an initial velocity of 5.0 m/s in the +x-direction and after the collision, moves with a velocity of 3.0 m/s in the -x-direction. Mass B has an initial velocity of 3.0 m/s in the -x-direction.
Let's assume the final velocity of mass B is vB. Using the principle of conservation of momentum, we can set up the equation:
Initial momentum = Final momentum
(mass of A * initial velocity of A) + (mass of B * initial velocity of B) = (mass of A * final velocity of A) + (mass of B * final velocity of B)
Substituting the given values:
(5.0 kg * 5.0 m/s) + (5.0 kg * 3.0 m/s) = (5.0 kg * 3.0 m/s) + (mass of B * vB)
Simplifying the equation:
25.0 m/s + 15.0 m/s = 15.0 m/s + (mass of B * vB)
40.0 m/s = 15.0 m/s + (mass of B * vB)
Subtracting 15.0 m/s from both sides:
25.0 m/s = (mass of B * vB)
Dividing both sides by the mass of B:
vB = 25.0 m/s / mass of B
Therefore, the final velocity of mass B after the collision is 25.0 m/s divided by the mass of B.
Suppose a thin conducting wire connects two conducting spheres. A negatively charged rod is brought near one of the spheres, the wire between them is cut, and the charged rod is taken away. Which one of the following is true? a. The spheres will attract each other. b. The spheres will repel each other. c. There will be no electrostatic force between the spheres
Answer:
a. The spheres will attract each other.
Explanation:
When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.
When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.Hence the two spheres will experience some amount of electrostatic force between them.A 2 kg rock is suspended by a massless
string from one end of a 7 m measuring stick.
What is the weight of the measuring stick
if it is balanced by a support force at the
1 m mark? The acceleration of gravity is
9.81 m/s^2
Answer in units of N.
Answer:
Weight = 7.848 N
Explanation:
Given:
Mass of the rock (m) = 2 kg
Length of the stick (L) = 7 m
Distance of support from the rock (x) = 1 m
The weight of the stick acts at the center, which is at a distance of 3.5 m from one end.
So, let 'M' be mass of the measuring stick.
Distance of 'W' from the supporting force (d) = 3.5 - 1 = 2.5 m
Now, for equilibrium, the sum of clockwise moments must be equal to the sum of anticlockwise moments.
Sum of clockwise moments = Sum of anticlockwise moments
[tex]m\times g\times x=M\times g\timesd\\\\M=\frac{mx}{d}[/tex]
Plug in the given values and solve for 'M'. This gives,
[tex]M=\frac{2\times 1}{2.5}=0.8\ kg[/tex]
Now, weight of the stick is given as:
[tex]W=Mg=0.8\times 9.81=7.848\ N[/tex]
Answer:
The weight at [tex]1[/tex]m mark is 7.848 N.
Explanation:
We have been given,
Mass of the rock= [tex]2[/tex] kg
Acceleration (gravity) = [tex]9.8[/tex] m/s^2
Length of the string = [tex]7[/tex] m
To find the weight at [tex]1[/tex] m mark let it be W.
So,
We know that torque will be balanced here.
Moment arm for the force (W) is [tex]3.5 - 1=2.5[/tex] m
As it is of [tex]7[/tex] m length so [tex]3.5[/tex] m is the COM (center of mass) of the string.
Let the moment be [tex]L_1\ and\ L_2[/tex] on clockwise and anti-clockwise direction.
[tex]L_1=L_2[/tex]
[tex]2.5*W = 2* 9.81*1[/tex]
Dividing both sides with [tex]2.5[/tex]
[tex]W=\frac{2*9.81*1}{2.5} =7.848\N[/tex]
So the weight is 7.848 N.
You crouch from a standing position, lowering your center of mass 18 cm in the process. Then you jump vertically. The average force exerted on you by the floor while you jump is three times your weight. What is your upward speed as you pass through your standing position in leaving the floor?
To solve this problem we will apply the concepts related to balance. Since the force applied is 3 times the weight, and the weight is defined as the multiplication between mass and gravity, we will have that the dynamic equilibrium ratio would be given by the relation,
[tex]\sum F = ma[/tex]
[tex]3mg-mg = ma[/tex]
Rearranging to find a,
[tex]a = 2g[/tex]
Using the linear motion kinematic equations, which express that the final velocity of the body, and in the absence of initial velocity, is equivalent to the product between 2 times the acceleration by the distance traveled, that is
[tex]v^2 = 2as[/tex]
[tex]v^2 = 2(2g)(0.18)[/tex]
[tex]v^2 = 2(2*9.8)(0.18)[/tex]
[tex]v = 2.66m/s[/tex]
Therefore the upward speed is 2.66m/s
Gauss's law is usualy written as :
∫Φ_E = E * dA= qencl / ε0
what does qencl refers in Guass's law?
O the net charge inside the closed surface
O the charge residing on insulators inside the closed surface
O all the charge in the physical system
O any charge inside the closed surface that is arranged symmetrically
Answer:
(a) the net charge inside the closed surface.
Explanation:
In Gauss' Law, Qencl refers to the net charge inside the Gaussian surface. This surface is usually taken as a symmetric geometric surface, but this is merely for simplicity. Gauss' Law holds for any closed surface. Inside this surface there can be insulators as well as conductors. Regardless of the geometry or the materials inside, Qencl refers to the net charge inside the closed surface. The charge outside the surface is irrelevant for Gauss' Law, therefore all the charge in the physical system is not included in Gauss' Law.
A steam power plant operates on an ideal reheat Rankine cycle between the pressure limits of 15 MPa and 10 kPa. The mass flow rate of steam through the cycle is 12 kg/s. Steam enters both stages of the turbine at 500C.
If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10%, determine
(a) The pressure, in kPa ,at which reheating takes place,
(b) The total rate of heat input, in kW, in the boiler,
(c) The thermal efficiency of the cycle. Also, show the cycle on a T-s diagram with respect to the saturation lines.
This is a complex thermodynamics problem about a steam power plant's ideal reheat Rankine cycle. It requires calculation of pressure at reheat stage, total heat input rate and thermal efficiency by using correct thermodynamic practices. Due to the complexity, expert understanding and use of steam tables is necessary.
Explanation:This is a thermodynamics problem associated with the ideal reheat Rankine cycle in a steam power plant. Due to the complex nature of the problem, the aid of thermodynamics and steam table reference books is required to solve it.
Under these ideal cycle conditions, calculating assumed pressure drops and using the steam tables, we can estimate the reheat pressure during the cycle. The exact figures would be subject to critical examination and calculation based on precise cycle parameters following real-life engineering practices.
For part (b), the total rate of heat input can be calculated using the mass flow rate and specific enthalpies at different stages of the cycle. The thermal efficiency for part (c) can be obtained by dividing the net work output by the total heat input.
An actual T-s diagram for the Rankine cycle would aid in visualizing the process but cannot be drawn here. Basically, the cycle is plotted with entropy(s) on the x-axis and temperature(T) on the y-axis, showing the expansion, reheat, and condensation stages of the steam.
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The Ideal Reheat Rankine Cycle is a model for determining efficiency in steam power plants. The pressure at which reheating takes place is typically assumed to be at the midpoint of the high and low-pressure limits. The total heat rate input in the boiler and thermal efficiency cannot be determined without more detailed specification of the conditions.
Explanation:This question is related to the concepts of thermodynamics in the field of Mechanical Engineering. The Ideal Reheat Rankine Cycle is a model used to predict the performance of steam turbine systems, and we will use principles from this model to answer your questions.
(a) The pressure at which reheating takes place cannot be definitively stated without more specific information like the temperature at the exit of the high-pressure turbine or the entropy at key points in the cycle. However, it’s commonly assumed in practice to occur at the mid-point of the high and low-pressure limits, which would be 7.5 MPa or 7500 kPa.
(b) The total rate of heat input in the boiler cannot be definitively stated without more specific information such as the specific enthalpies at key points in the cycle.
(c) Thermal efficiency of the cycle also requires further specific information such as the enthalpy or temperature at key points in the cycle to compute.
Moreover, drawing a T-s (Temperature vs Entropy) diagram with respect to the saturation lines would need a graphical interface, so it's not possible to illustrate it here.
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You throw a glob of putty straight up toward the ceiling, which is 3.00m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 8.40m/s . What is the speed of the putty just before it strikes the ceiling? How much time from when it leaves your hand does it take the putty to reach the ceiling?
The answer provides the speed of the putty just before striking the ceiling and the time it takes for the putty to reach the ceiling. The speed is found to be 5.79 m/s and the time taken is calculated to be 1.5 seconds.
Explanation:The speed of the putty just before it strikes the ceiling:
Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is acceleration, and s is displacement, the speed is found to be 5.79 m/s.
The time it takes for the putty to reach the ceiling:
By using the equation v = u + at, where t is time, a is acceleration, u is initial velocity, and v is final velocity, the time taken is calculated to be 1.5 seconds.
a raft of dimensions 2m wide by 3m long by .5m thick is floating in a pond if one fourth of the raf is submerged find the density of the raft material
Answer:
250 kg/m3
Explanation:
The total volume of the raft is length times width times height
V = lwh = 2 * 3 * 0.5 = 3 m cubed
The volume of the raft that is submerged in water is 1/4 of total volume
3 /4 = 0.75 m cubed
Let water density = 1000 kg/m cubed and g = 10 m/s2
The buoyant force is equal to the weight of water displaced by the raft
F = 0.75 * 1000 * 10 = 7500 N
This force is balanced by the raft weight, so the weight of the raft is also 7500N
Mass of raft is 7500 / g = 7500 / 10 = 750 kg
Raft density is mass divided by volume = 750 / 3 = 250 kg/m3
The temperature of a system rises by 45 °C during a heating process. Express this rise in temperature in Kelvin and Fahrenheit.
Answer:
318K
113°F
Explanation:
45°C = 45 + 273K = 318K
45°C = (45 × 9/5) + 32 = 81 + 32 = 113°F
An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular acceleration of each wheel?
To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.
Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.
The average angular acceleration
[tex]\alpha = \frac{\omega_f - \omega_0}{t}[/tex]
Here
[tex]\alpha[/tex] = Angular acceleration
[tex]\omega_{f,i} =[/tex] Initial and final angular velocity
There is not initial angular velocity,then
[tex]\alpha = \frac{\omega_f}{t}[/tex]
We know that the relation between the tangential velocity with the angular velocity is given by,
[tex]v = r\omega[/tex]
Here,
r = Radius
[tex]\omega[/tex] = Angular velocity,
Rearranging to find the angular velocity
[tex]\omega = \frac{v}{r}}[/tex]
[tex]\omega = \frac{30}{0.20} \rightarrow[/tex] Remember that the radius is half te diameter.
Now replacing this expression at the first equation we have,
[tex]\alpha = \frac{30}{0.20*6}[/tex]
[tex]\alpha = 25 rad /s^2[/tex]
Therefore teh average angular acceleration of each wheel is [tex]25rad/s^2[/tex]
Final answer:
To find the average angular acceleration of each wheel, calculate it using the provided linear acceleration data and the relationship between linear and angular quantities.
Explanation:
Angular acceleration can be calculated using the formula:
α = Δω / Δt
Given the acceleration of the car from 0 to 30 m/s in 6.0 s, we can find the angular acceleration by first calculating the final and initial angular velocities using the relationship between linear and angular quantities.
Substitute the values into the formula and solve for the average angular acceleration of each wheel.
How many photons does a green (532nm) 1mW laser pointer emit? Can we still treat the light as a continuous electromagnetic wave in this case? (2 pts)
Answer:
[tex]n=2.68\times 10^{15}[/tex]photons/s
Yes,we can treat the light as a continuous electromagnetic wave in this case.
Explanation:
We are given that
Wavelength of photon=[tex]\lambda=532 nm=532\times 10^{-9}m[/tex]
1nm=[tex]10^{-9} m[/tex]
Power of photon=1mW=[tex]0.0010J/s[/tex]
We have to find the number of photons emit from laser.
In 1 s, photon emit energy=0.0010 J
We know that
[tex]E=nh\nu=nh\times \frac{c}{\lambda}[/tex]
Where c[tex]\c=3\times 10^8 m/s[/tex]
[tex]E=[/tex]Energy of photon
[tex]\lambda[/tex]=Wavelength of light
[tex]h=6.626\times 10^{-34}Js[/tex]
Substitute the values then, we get
[tex]0.0010=n\times 6.626\times 10^{-34}\times \frac{3\times 10^8}{532\times 10^{-9}}[/tex]
[tex]n=\frac{0.0010\times 532\times 10^{-9}}{6.626\times 10^{-34}\times 3\times 10^8}[/tex]
[tex]n=2.68\times 10^{15}[/tex]photons/s
Visible light is EM wave.
Range of wavelength of visible light =400nm- 700 nm
Green color lies in visible region because its wavelength lies in 400nm-700 nm.
Therefore, we can treat the light as a continuous electromagnetic wave in this case.
It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit Mars and blasted pieces of rock (perhaps containing primitive life) free of the surface. Astronomers know that many Martian rocks have come to Earth this way. One objection to this idea is that microbes would have to undergo an enormous, lethal acceleration during the impact. Let us investigate how large such an acceleration might be. To escape Mars, rock fragments would have to reach its escape velocity of 5.0 km/s, and this would most likely happen over a distance of about 4.0m during the impact.
What would be the acceleration, in m/s, of such a rock fragment?
Answer:
[tex]a=3125000 m/s^2\\a=3.125*10^6 m/s^2[/tex]
Acceleration, in m/s, of such a rock fragment = [tex]3.125*10^6m/s^2[/tex]
Explanation:
According to Newton's Third Equation of motion
[tex]V_f^2-V_i^2=2as[/tex]
Where:
[tex]V_f[/tex] is the final velocity
[tex]V_i[/tex] is the initial velocity
a is the acceleration
s is the distance
In our case:
[tex]V_f=V_{escape}, V_i=0,s=4 m[/tex]
So Equation will become:
[tex]V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2[/tex]
Acceleration, in m/s, of such a rock fragment = [tex]3.125*10^6m/s^2[/tex]
Final answer:
To find the acceleration of a rock fragment blasted from Mars, use the equation for acceleration, considering final velocity, initial velocity, and distance. Given the escape velocity of Mars, the acceleration would be approximately 6250 m/s².
Explanation:
To calculate the acceleration of a rock fragment blasted from Mars, we use the equation for acceleration: a = (v² - u²) / (2s), where v is the final velocity, u is the initial velocity, and s is the distance. Given that the final velocity is the escape velocity of Mars, 5.0 km/s, the initial velocity is 0 km/s (as it starts from rest), and the distance s is 4.0 m, we can compute the acceleration.
Substitute the values into the equation: a = ((5.0 km/s)² - (0 km/s)²) / (2 × 4.0 m) = 6.25 km/s². Converting the units to meters per second squared, we get an acceleration of 6250 m/s² for the rock fragment.
If the speed of sound in air is 340 m/s, what is approximately the length of the shortest air column closed at one end that will respond to a tuning fork of frequency 198 Hz?
Answer:
42.9 cm
Explanation:
given,
speed of sound, v = 340 m/s
Frequency of tuning fork, f = 198 Hz
The length of the shortest air column closed at one end.
this is the case of the standing wave.
L = N λ
N = the number of complete sine waves in the standing wave.
The standing wave in the problem is the first harmonic of a pipe open at one end.
N = 1/4
we know,
v = f λ
340 = 198 x λ
λ = 1.717 m
now,
L = N λ
[tex]L = \dfrac{1}{4}\times 1.717[/tex]
[tex]L = 0.429\ m[/tex]
L = 42.9 cm
minimum length of the required pipe is 42.9 cm.
did you find the answer?
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is 1522 m/s. When the dolphin is swimming directly away at 8.5 m/s, the marine biologist measures the number of clicks occuring per second to be at a frequency of 2770 Hz.
1. What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?
From Doppler effect we have that the frequency observed for the relation between the velocities is equivalent to the frequency observed. That is mathematically,
[tex]F_r = \frac{v}{v+v_s}F_s[/tex]
Here,
Speed of sound in water [tex]v = 1522m/s[/tex]
The Dolphin swims directly away from the observer with a velocity [tex]v_s = 8.5m/s[/tex]
Observed frequency of the clicks produced by the dolphin [tex]F_r = 2770Hz[/tex]
Replacing we have,
[tex]F_r = \frac{v}{v+v_s}F_s[/tex]
[tex]F_s = \frac{v+v_s}{v}[/tex]
[tex]F_s = 2770 (\frac{1522}{1522+8.5})[/tex]
[tex]F_s = 2754.61Hz[/tex]
Therefore the frequency emitted by the dolphin is 2754.61Hz
At time t in seconds, a particle’s distance s(t), in centimeters, from a point is given by s(t)=13sin t+40. What is the average velocity of the particle from t= π3 to t= 13π3?
Answer:
Explanation:
Given
distance [tex]s(t)=13\sin (t)+40[/tex]
at [tex]t=\frac{\pi }{3}[/tex]
[tex]s(\frac{\pi }{3})=13\sin (\frac{\pi }{3})+40[/tex]
[tex]s(\frac{\pi }{3})=13\times \frac{\sqrt{3}}{2}+40[/tex]
at [tex]t=\frac{13\pi}{3}=4\pi+\frac{\pi}{3}[/tex]
[tex]s(4\pi+\frac{\pi}{3})=13\sin (4\pi+\frac{\pi}{3})+40[/tex]
Average velocity is change in position w.r.t time
[tex]v_{avg}=\frac{s(4\pi+\frac{\pi}{3})-s(\frac{\pi }{3})}{4\pi+\frac{\pi}{3}-\frac{\pi }{3}}[/tex]
[tex]v_{avg}=0[/tex]
A rotating wheel requires 10.0 s to rotate 39.0 revolutions. Its angular velocity at the end of the 10.0-s interval is 92.0 rad/s. What is the constant angular acceleration (in rad/s) of the wheel
Answer:
[tex]13.5 rad/s^2[/tex]
Explanation:
39 revolutions = 39 * 2π = 245 rad
Let [tex]\omega_0[/tex] be the initial angular velocity of the wheel before the 10 s interval. We have the following equation of motion for angular velocity
[tex]\omega = \omega_0 + \alpha t[/tex]
where [tex]\omega = 92 rad/s[/tex] is the final angular velocity of the rotating wheel. [tex]\alpha[/tex] is the constant angular acceleration of the wheel, which we are looking for, and t = 10s is the duration time that the wheel rotates
[tex]92 = \omega_0 + 10\alpha[/tex]
[tex]\omega_0 = 92 - 10\alpha[/tex]
We also have the following equation of motion for angles
[tex]\theta = \omega_0t + \alpha t^2/2[/tex]
where [tex]\theta = 245 rad[/tex] is the total angles rotated
we can also substitute [tex]\omega_0 = 92 - 10\alpha[/tex]
[tex]245 = 10(92 - 10\alpha) + \alpha 10^2/2[/tex]
[tex]245 = 920 - 100\alpha + 50\alpha[/tex]
[tex]50\alpha = 675 [/tex]
[tex]\alpha = 675 / 50 = 13.5 rad/s^2[/tex]
A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
(a) Find the electric force on charge q2.
F12 = ? mN
(b) Find the electric force on q1.
F21 = ? mN
(c) What would your answers for Parts (a) and (b) differ if q2 were -6.0 µC?
To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as
[tex]F = \frac{kq_1q_2}{d^2}[/tex]
Here
k = Coulomb's Constant
[tex]q_{1,2} =[/tex] Charge of each object
d = Distance
Our values are given as,
[tex]q_1 = 1 \mu C[/tex]
[tex]q_2 = 6 \mu C[/tex]
d = 1 m
[tex]k = 9*10^9 Nm^2/C^2[/tex]
a) The electric force on charge [tex]q_2[/tex] is
[tex]F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}[/tex]
[tex]F_{12} = 54 mN[/tex]
Force is positive i.e. repulsive
b) As the force exerted on [tex]q_2[/tex] will be equal to that act on [tex]q_1[/tex],
[tex]F_{21} = F_{12}[/tex]
[tex]F_{21} = 54 mN[/tex]
Force is positive i.e. repulsive
c) If [tex]q_2 = -6 \mu C[/tex], a negative sign will be introduced into the expression above i.e.
[tex]F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}[/tex]
[tex]F_{12} = F_{21} = -54 mN[/tex]
Force is negative i.e. attractive
A 7500 kg open train car is rolling on frictionless rails at 25.0 m/s when it starts pouring rain. A few minutes later, the car's speed is 20.0 m/s. What mass of water has collected in the car?
Answer:
Mass of rain collected will be 1875 kg
Explanation:
We have given mass of the train [tex]m_i[/tex] = 7500 kg
Velocity of the car [tex]v_i[/tex] = 25 m/sec
After few minutes velocity of the car = 20 m/sec
From conservation of momentum we know that [tex]m_iv_i=m_fv_f[/tex]
So [tex]7500\times 25=m_f\times 20[/tex]
[tex]m_f=9375kg[/tex]
[tex]m_f[/tex] will be sum of mass of car and mass of rain collected
As mass of car is 7500 kg
So mass of rain collected = 9375 - 7500 = 1875 kg
An electrically neutral balloon is rubbed on cloth and becomes positively charged. What can be said about its mass:_______
Answer:
The charged balloon’s mass will be almost equal to the original mass
Explanation:
When an electrically neutral balloon is rubbed on cloth, it losses some electron and becomes positively charged.
The balloon becomes positively charged by the removal of electrons (electrons from the cloth repels electron in the balloon).
The mass of positively charged balloon, will be equal to the original mass minus the mass of electrons removed.
Since Electrons have negligible mass, the removal of electrons from the neutral balloon will have little effect on the original mass of the balloon.
Therefore, the charged balloon’s mass will be almost equal to the original mass
The volume of a fixed mass of an ideal gas is doubled while the temperature is increased from 100 K to 400 K.
What is the final pressure in terms of its initial pressure?
a. 2 Pi
b. 3 Pi
c. 4 Pi
d. 1/2 Pi
Answer:
c. 4 Pi
Explanation:
The pressure law states that the pressure of a gas is directly proportional to its temperature provided volume and other physical quantities remain constant.
If t is the temperature and p is the pressure of a gas,
p ∝ t
p = kt where k s the constant of proportionality
k = p/t
If t1 and p1 are the initial temperature and pressure respectively and
t2, p2 are the final temperature and pressure
p1/t1 = p2/t2
t1 = 100k, t2 = 400k
p2 = p1 × 400/100
p2 = 4t1
The final pressure of the gas is 4 times the initial pressure.