Answer:
U_f = (U_o)/2)
Explanation:
The capacitance of a given capacitor is given by the formula;
C = ε_o•A/d
While energy stored in plates capacitor is given as; U_o = Q²/2C
Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;
C' = ε_o•4A/2d
Hence, C' = 2C
so capacitance is now doubled
Thus, the final energy stored between the plates of capacitor is given as;
U_f = Q²/2C'
From earlier, we saw that C' = 2C.
Thus;
U_f = Q²/2(2C)
U_f = Q²/4C
Rearranging, we have;
U_f = (1/2)(Q²/2C)
From earlier, U_o = Q²/2C
Hence,
U_f = (1/2)(U_o)
Or
U_f = (U_o/2)
Uses of pressure and the uses of density
Answer:
Pressure is a scalar quantity defined as per unit area.
Density is the objects ,times its the acceleration due to gravity.
Explanation:
Pressure is the alternative object increases the area of contact decrease .
Pressure is the force component to the surface used to calculate pressure.
pressure is that collisions of the gas to container as the per unit time .
pressure is an physical important quantity to play the solid and fluid .
Pressure is the expressed in a number of units depend the context use, pressure exerted by the liquid alone.
Density is the objects, times, volume of the object that times acceleration objects.
Density is the used to the system complex objects and materials.
Density force is the weight of a region or objects static fluid.
the efficiency of a carnot cycle is 1/6.If on reducing the temperature of the sink 75 degrees celcius ,the efficiency becomes 1/3,determine he initial and final temperatures between which the cycle is working.
Answer:
450°C
Explanation: Given that the efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .
η = 1 - T2 / T1
η = 1/6 initially
when T2 is reduced by 65°C then η becomes 1/3
Solution
η = 1/6
1 - T2 / T1 = 1/6 [ using the Formula ]........................(1)
When η = 1/3 :
η = 1 - ( T2 - 75 ) / T1
1/3 = 1 - (T2 - 75)/T1.........................(2)
T2 - T1 = -75 [ because T2 is reduced by 75°C ]
T2 = T1 - 75...........................(3)
Put this in (2) :
> 1/3 = 1 - ( T1 - 75 - 75 ) / T1
> 1/3 = 1 - (T1 - 150 ) /T1
> (T1 - 150) / T1 = 1 - 1/3
> ( T1 -150 ) / T1 = 2/3
> 3 ( T1 - 150 ) = 2 T1
> 3 T1 - 450 = 2 T1
Collecting the like terms
3 T1- 2 T1 = 450
T1 = 450
The temperature initially was 450°C
Suppose a 185 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m.
Randomized Variables
m = 185 kg
v = 29 m/s
h = 32 m
A. How high can it coast up the hill. if you neglect friction in m?
B. How much energy is lost to friction if the motorcycle only gains an altitude of 33 m before coming to rest?
Answer:
a) Height reached before coming to rest is 42.86 m
b) Energy lost to friction is 17902.45 J
Explanation:
mass of the motorcycle = 185 kg
speed of the towards the hill = 29 m/s
The wheels weigh 12 kg each
Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m
a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height
the kinetic energy of the motorcycle is given as
[tex]KE[/tex] = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the motorcycle
v is the velocity of the motorcycle
[tex]KE[/tex] = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J
This will be converted to potential energy
The potential energy up the hill will be
[tex]PE[/tex] = mgh
where m is the mass
g is acceleration due to gravity 9.81 m/s^2
h is the height reached before coming to rest
[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h
equating the kinetic energy to the potential energy for energy conservation, we'll have
77792.5 = 1814.85h
height reached before coming to rest = 77792.5/1814.85 = 42.86 m
b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is
[tex]PE[/tex] = mgh
[tex]PE[/tex] = 185 x 9.81 x 33 = 59890.05 J
The energy lost to friction will be the kinetic energy minus this potential energy.
energy lost = 77792.5 - 59890.05 = 17902.45 J
A) The motorcycle can coast up the hill by ; 42.86m
B) The amount of energy lost to friction : 17902.45 J
A) Determine how high the motorcycle can coast up the hill when friction is neglected
apply the formula for kinetic and potential energies
K.E = 1/2 mv² ---- ( 1 )
P.E = mgH ---- ( 2 )
As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.
∴ K.E = P.E
1/2 * mv² = mgH
∴ H = ( 1/2 * mv² ) / mg ---- ( 3 )
where ; m = 185 kg , v = 29 m/s , g = 9.81
Insert values into equation ( 3 )
H ( height travelled by motorcycle neglecting friction ) = 42.86m
B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest
P.E = mgh = 185 * 9.81 * 33 = 59890.05 J
where : h = 33 m , g = 9.81
K.E = 1/2 * mv² = 77792.5 J ( question A )
∴ Energy lost ( ΔE ) = ( 77792.5 - 59890.05 ) = 17902.45 J
Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction : 17902.45 J.
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Search Results Web results A car of mass 650 kg is moving at a speed of 0.7
Answer:
W = 1413.75 J
Explanation:
It is given that,
Mass of car, m = 650 kg
Initial speed of the car, u = 0.7 m/s
Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s
Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.
[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]
So, the work done by the car is 1413.75 J.
A plane progressive
the expression
in time, ys
where you
progressivo ware is no presented by
(At + A
y- 5 sin
in metre, t es in time the doplicensel
Calculate
the amplitude of the wave.
Answer:
Amplitude, A = 5 m
Explanation:
Let a progressive wave is given by equation :
[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)
The general equation of a progressive wave is given by :
[tex]y=A\sin (\omega t-kx)[/tex] ....(2)
Here,
A is the amplitude of the wave
[tex]\omega[/tex] is the angular frequency
k is propagation constant
We need to find the amplitude of the wave.
If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.
The flywheel of an engine has I of 1.60kg.m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rpm in 8.00s, starting from rest?
Answer:
Torque = 8.38Nm
Explanation:
Time= 8.00s
angular speed (w) =400 rpm
Moment of inertia (I)= 1.60kg.m2 about its rotation axis
We need to convert the angular speed from rpm to rad/ sec for consistency
2PI/60*n = 0.1047*409 = 41.8876 rad/sec
What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?
Then we need to use the formula below for our torque calculation
from basic equation T = J*dω/dt ...we get
Where : t= time in seconds
W= angular velocity
T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm
Therefore, constant torque that is required is 8.38 Nm
Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.
Given-
Inertia of the flywheel is 1.60 kg m squared.
Angular speed of the flywheel [tex]n[/tex] is 400 rpm. Convert it into the rad/sec, we get,
[tex]\omega =\dfrac{2\pi }{60} \times n[/tex]
[tex]\omega =\dfrac{2\pi }{60} \times 400[/tex]
[tex]\omega = 41.89[/tex]
Thus, the angular speed of the flywheel [tex]\omega[/tex] is 41.89 rad/sec.
When a torque [tex]\tau[/tex] is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia [tex]I[/tex]. Mathematically,
[tex]\tau=\dfrac{\Delta \omega }{\Delta t} \times I[/tex]
[tex]\tau=\dfrac{ 41.89 }{8} \times 1.6[/tex]
[tex]\tau=8.38[/tex]
Hence, the required constant torque is 8.38 N-m.
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You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you
Answer:
The speed of light will be c=3x10^8m/s
Explanation:
This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with the Earth's moon, with mass Mm = 3.75 ✕ 1019 kg and radius Rm = 1.98 ✕ 105 m, giving it a free-fall acceleration of g = 0.0636 m/s2. One astronaut, being a baseball fan and having a strong arm, decides to see how high she can throw a ball in this reduced gravity. She throws the ball straight up from the surface of Mimas at a speed of 41 m/s (about 91.7 mph, the speed of a good major league fastball).
(a) Predict the maximum height of the ball assuming g is constant and using energy conservation. Mimas has no atmosphere, so there is no air resistance.
(b) Now calculate the maximum height using universal gravitation.
(c) How far off is your estimate of part (a)? Express your answer as a percent difference and indicate if the estimate is too high or too low.
Answer:
a) h = 13,205.4 m
b) r_f = 2.12 106 m
c) e% = 0.68%
Explanation:
a) This is an exercise we are asked to use energy conservation,
Starting point. On the surface of Mimas
Em₀ = K = ½ m v²
Final point. Where the ball stops
[tex]Em_{f}[/tex] = U = m g h
Em₀ = Em_{f}
½ m v² = m g h
h = ½ v² / g
let's calculate
h = ½ 41² / 0.0636
h = 13,205.4 m
b) For this part we are asked to use the law of universal gravitation, write the energy
starting point. Satellite surface
Em₀ = K + U = ½ m v² - GmM / r_o
final point. Where the ball stops
[tex]Em_{f}[/tex]= U = - G mM / r_f
Em₀ = Em_{f}
½ m v² - G m M / r_o = - G mM / r_f
In this case all distances are measured from the center of the satellite
1 / rf = 1 / GM (-½ v² + G M / r_o)
let's calculate
1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)
1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)
1 / r_f = 4,714 10⁻⁷
r_f = 1 / 4,715 10⁻⁷
r_f = 2.12 106 m
to measure this distance from the satellite surface
r_f ’= r_f - r_o
r_f ’= 2.12 106 - 1.98 105
r_f ’= 1,922 106 m
c) the percentage difference is
e% = 13 205.4 / 1,922 106 100
e% = 0.68%
The estimate of part a is a little low
You connect three resistors with resistances R, 2R, and 3R in parallel. The equivalent resistance of the three resistors will have a value that is
Answer:
The equivalent is 6R/11Explanation:
We know that the equivalent resistance of resistors connected in parallel is expressed as
[tex]\frac{1}{Re} =\frac{1}{R1} +\frac{1}{R2}+\frac{1}{R3}\\\\\frac{1}{Re} =\frac{1}{R} +\frac{1}{2R}+\frac{1}{3R}\\[/tex]
the L.C.M is 6R
[tex]\frac{1}{Re} =\frac{6+3+2}{6R} = \frac{11}{6R} \\\\Re= \frac{6R}{11}[/tex]
Someone help find centripetal acceleration plus centripetal force!
Answer:Centripetal force that acts an object keep it along a moving circular path.
Explanation:Centripetal force along a path circular of radius(r) with velocity(V) acceleration the center of the path.
a=v/r
object will along moving continue a straight path unless by the external force.External force is the centripetal force.
Centripetal force is to moving in horizontal circle,Centripetal force is not a fundamental force.Gravitational force satellite and orbit of centripetal force.
Centripetal acceleration and centripetal force are used to calculate the motion of objects in circular motion. The main answer to the question is given below:The centripetal force is given by:F = mv²/rwhere m is the mass of the object, v is the speed of the object and r is the radius of the circle. The unit of centripetal force is Newtons (N).The centripetal acceleration is given by:a = v²/rThe unit of centripetal acceleration is meters per second squared
(m/s²).Explanation:When an object moves in a circular motion, there is a force that acts upon it. This force is called the centripetal force. This force always points towards the center of the circle. It is responsible for keeping the object moving in a circular motion.The centripetal force is related to the centripetal acceleration.
The centripetal acceleration is the acceleration of an object moving in a circle. It is always directed towards the center of the circle.The magnitude of the centripetal force is given by:F = mv²/rwhere F is the force, m is the mass of the object, v is the speed of the object and r is the radius of the circle.The magnitude of the centripetal acceleration is given by:a = v²/rwhere a is the acceleration, v is the speed of the object and r is the radius of the circle.
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5. A nail contains trillions of electrons. Given that electrons repel from each other, why do they not then fly out of the nail?
Answer:
Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.
If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.
How many electrons circulate each second through the cross section of a conductor, which has a current intensity of 4A.
Answer:
2.5×10¹⁹
Explanation:
4 C/s × (1 electron / 1.60×10⁻¹⁹ C) = 2.5×10¹⁹ electrons/second
How much heat is needed to melt 2.5 KG of water at its melting point? Use Q= mass x latent heat of fusion.
Answer:
Q = 832 kJ
Explanation:
It is given that,
Mass of the water, m = 2.5 kg
The latent heat of fusion, L = 333 kJ/kg
We need to find the heat needed to melt water at its melting point. The formula of heat needed to melt is given by :
Q = mL
[tex]Q=2.5\ kg\times 333\ kJ/kg\\\\Q=832.5\ kJ[/tex]
or
Q = 832 kJ
So, the heat needed to melt the water is 832 kJ.
3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?
O A. Ferromagnetic
O B. Non-magnetic
C. Diamagnetic
O D. Paramagnetic
Answer:
Diamagnetic
Explanation:
Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).
For vanadium V ion, there are 18 electrons which will be arranged as follows;
1s2 2s2 2p6 3s2 3p6.
All the electrons present are spin paired hence the ion is expected to be diamagnetic.
Answer:
its paramagnetic
Explanation:
i took this quiz
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done by gravity during the roll, in Joules
Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
[tex]F = W_{x} = mgsin(\theta)[/tex]
Where:
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
[tex]F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N[/tex]
Now, we can find the work:
[tex]W = F*d = 24.53 N*20 m = 490.6 J[/tex]
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
An object has an acceleration of 12.0 m/s/s. If the net force was doubled and the mass were tripled, then the new acceleration would be _____ m/s/s.
✴ Case - I
⟶ Force = F
⟶ Mass = m
⟶ Acceleration = 12m/s²
✴ Case - II
⟶ Force = 2F
⟶ Mass = 3m
To Find :➳ Acceleration in second case.
Concept :⇒ This question is completely based on the concept of newton's second law of motion.
⇒ As per this law, Force is defined as the product of mass and acceleration.
Mathematically, F = ma
Calculation :[tex]\implies\sf\:\dfrac{F_1}{F_2}=\dfrac{m_1\times a_1}{m_2\times a_2}\\ \\ \implies\sf\:\dfrac{F}{2F}=\dfrac{m\times 12}{3m\times a_2}\\ \\ \implies\sf\:\dfrac{1}{2}=\dfrac{4}{a_2}\\ \\ \implies\sf\:a_2=4\times 2\\ \\ \implies\underline{\boxed{\bf{a_2=8\:ms^{-2}}}}[/tex]
New acceleration would be 12 m/s²
Given that;
Acceleration of object = 12 m/s²
New net force = 2f
New mass = 3m
Find:
New acceleration
Computation:
[tex]\frac{F1}{F2} = \frac{m1a1}{m2a2} \\\\\frac{f}{2f} = \frac{m(12)}{(3m)a2} \\\\\frac{1}{2} = \frac{4}{a2} \\\\a2 = 8 m/s^2[/tex]
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Help me with these question and please explainnn
Explanation:
1. Impulse = change in momentum
J = Δp
J = mΔv
In the x direction:
Jₓ = mΔvₓ
Jₓ = (0.40 kg) (30 m/s cos 45° − (-20 m/s))
Jₓ = 16.5 kg m/s
In the y direction:
Jᵧ = mΔvᵧ
Jᵧ = (0.40 kg) (30 m/s sin 45° − 0 m/s)
Jᵧ = 8.49 kg m/s
The magnitude of the impulse is:
J = √(Jₓ² + Jᵧ²)
J = 18.5 kg m/s
The average force is:
FΔt = J
F = J/Δt
F = 1850 N
2. Momentum is conserved.
m₁u₁ + m₂u₂ = (m₁ + m₂) v
In the x direction:
(1000 kg) (0 m/s) + (1500 kg) (-12 m/s) = (1000 kg + 1500 kg) vₓ
vₓ = -7.2 m/s
In the y direction:
(1000 kg) (20 m/s) + (1500 kg) (0 m/s) = (1000 kg + 1500 kg) vᵧ
vᵧ = 8 m/s
The magnitude of the final speed is:
v = √(vₓ² + vᵧ²)
v = 10.8 m/s
3. Momentum is conserved.
m₁u₁ + m₂u₂ = (m₁ + m₂) v
In the x direction:
(0.8 kg) (18 m/s cos 45°) + (0.36 kg) (9.0 m/s) = (0.8 kg + 0.36 kg) vₓ
vₓ = 11.6 m/s
In the y direction:
(0.8 kg) (-18 m/s sin 45°) + (0.36 kg) (0 m/s) = (0.8 kg + 0.36 kg) vᵧ
vᵧ = -8.78 m/s
The magnitude of the final speed is:
v = √(vₓ² + vᵧ²)
v = 14.5 m/s
an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
Answer:
4°C
Explanation:
Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.
The temperature gradient between the core of Mars and its surface is approximately 0.0003 K/m. Compare this temperature gradient to that of Earth. What can you determine about the rate at which heat moves out of Mars’s core compared to Earth?
Answer:
The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.
Explanation:
Answer:
The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.
Explanation:
Edmentum sample answer
You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By monitoring the capacitor voltage on an oscilloscope, you see that the voltage decays to half its initial value in 3.40 msms . You then use the resistor and capacitor to make a low-pass filter. What is the crossover frequency fcfc
Answer:
The frequency is [tex]f = 0.221 \ Hz[/tex]
Explanation:
From the question we are told that
The time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]
Let the voltage of the capacitor when it is fully charged be [tex]V_o[/tex]
Then the voltage of the capacitor at time t is said to be [tex]V = \frac{V_o}{2}[/tex]
Now this voltage can be mathematical represented as
[tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]
Where RC is the time constant
substituting values
[tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]- \frac{0.5}{RC} = ln (0.5)[/tex]
[tex]-\frac{0.5}{RC} = -0.6931[/tex]
[tex]RC = 0.721[/tex]
Generally the cross-over frequency for a low pass filter is mathematically represented as
[tex]f = \frac{1}{2 \pi * RC }[/tex]
substituting values
[tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]
[tex]f = 0.221 \ Hz[/tex]
One hundred turns of insulated copper wire are wrapped around an iron core of cross-sectional area 0.100m2. As the magnetic field along the coil axis changes from 0.5 T to 1.00T in 4s, the voltage induced is:
Answer:
The voltage induced in the coil is 1.25 V.
Explanation:
Given;
number of turns, N = 100 turns
cross sectional area of the copper coil, A = 0.1 m²
initial magnetic field, B₁ = 0.5 T
final magnetic field, B₂ = 1.00 T
duration of change in magnetic field, dt = 4 s
The induced emf in the coil is calculated as;
[tex]emf = -N\frac{\delta \phi}{\delta t} \\\\emf = - N (\frac{\delta B}{\delta t}) A\\\\emf = -N (\frac{B_1 -B_2}{\delta t} )A\\\\emf = N(\frac{B_2-B_1}{\delta t} )A\\\\emf = 100(\frac{1-0.5}{4} )0.1\\\\emf = 1.25 \ Volts[/tex]
Therefore, the voltage induced in the coil is 1.25 V.
When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-half its original value, and the charge of B to one-tenth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F = F₀ 0.2
Explanation:
For this exercise we apply Coulomb's law with the initial data
F₀ = k q_A q_B / d²
indicate several changes
q_A ’= ½ q_A
q_B ’= 1/10 q_B
d ’= ½ d
let's substitute these new values in the Coulomb equation
F = k q_A ’q_B’ / d’²
F = k ½ q_A 1/10 q_B / (1/2 d)²
F = (k q_A q_B / d2) ½ 1/10 2²
F = F₀ 0.2
An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the electron is 1.48 107 m/s, determine the following.
(a) the radius of the circular path ............ cm
(b) the time interval required to complete one revolution ............ s
Answer:
(a) 3.9cm
(b) 1.66 x 10⁻⁸s
Explanation:
Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,
F = m x a --------------(i)
Where;
m = mass of the particle
a = acceleration of the mass
The centripetal acceleration is given by;
a = v² / r [v = linear velocity of particle, r = radius of circular path]
Therefore, equation (i) becomes;
F = m v²/ r --------------------(ii)
The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;
F = qvBsinθ -------------(iii)
Where;
q = charge of the particle
v = velocity of the particle
B = magnetic field
θ = angle between the velocity and the magnetic field
Combine equations (ii) and (iii) as follows;
m (v² / r) = qvBsinθ [divide both side by v]
m v / r = qBsinθ [make r subject of the formula]
r = (m v) / (qBsinθ) ---------(iv)
(a) From the question;
v = 1.48 x 10⁷m/s
B = 2.14mT = 2.14 x 10⁻³T
θ = 90° [since the direction of velocity is perpendicular to magnetic field]
m = mass of electron = 9.11 x 10⁻³¹kg
q = charge of electron = 1.6 x 10⁻¹⁹C
Substitute these values into equation (iv) as follows;
r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)
r = 3.9 x 10⁻²m
r = 3.9cm
Therefore, the radius of the circular path is 3.9cm
(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by
T = d / v --------------(*)
Where;
d = distance traveled in the circular path in one complete turn = 2πr
v = velocity of the motion = 1.48 x 10⁷m/s
d = 2 π (3.9 x 10⁻²) [Take π = 22/7 = 3.142]
d = 2(3.142)(3.9 x 10⁻²) = 0.245m
Substitute the values of d and v into equation (*) as follows;
T = 0.245 / 1.48 x 10⁷
T = 0.166 x 10⁻⁷s
T = 1.66 x 10⁻⁸s
Therefore, the time interval is 1.66 x 10⁻⁸s
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar.
Answer:
a) 6738.27 J
b) 61.908 J
c) [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
Explanation:
The complete question is
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.
moment of inertia is given as
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]
where m is the mass of the flywheel,
and r is the radius of the flywheel
for the flywheel with radius 1.1 m
and mass 11 kg
moment of inertia will be
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2
The maximum speed of the flywheel = 35 m/s
we know that v = ωr
where v is the linear speed = 35 m/s
ω = angular speed
r = radius
therefore,
ω = v/r = 35/1.1 = 31.82 rad/s
maximum rotational energy of the flywheel will be
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J
b) second flywheel has
radius = 2.8 m
mass = 16 kg
moment of inertia is
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] = [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2
According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels
for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s
for their combination, the rotational momentum is
[tex](I_{1} +I_{2} )w[/tex]
where the subscripts 1 and 2 indicates the values first and second flywheels
[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω
where ω here is their final angular momentum together
==> 69.375ω
Equating the two rotational momenta, we have
211.76 = 69.375ω
ω = 211.76/69.375 = 3.05 rad/s
Therefore, the energy stored in the first flywheel in this situation is
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J
c) one third of the initial energy of the flywheel is
6738.27/3 = 2246.09 J
For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
where m is the mass of the car
[tex]v_{car}[/tex] is the velocity of the car
Equating the energy
2246.09 = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
making m the subject of the formula
mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]f= -75 \ cm = - 0.75 \ m[/tex]
b
[tex]P = -1.33 \ diopters[/tex]
Explanation:
From the question we are told that
The image distance is [tex]d_i = -75 cm[/tex]
The value of the image is negative because it is on the same side with the corrective glasses
The object distance is [tex]d_o = \infty[/tex]
The reason object distance is because the object father than it being picture by the eye
General focal length is mathematically represented as
[tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]
substituting values
[tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]
=> [tex]f= -75 \ cm = - 0.75 \ m[/tex]
Generally the power of the corrective lens is mathematically represented as
[tex]P = \frac{1}{f}[/tex]
substituting values
[tex]P = \frac{1}{-0.75}[/tex]
[tex]P = -1.33 \ diopters[/tex]
An object has an acceleration of 6.0 m/s/s. If the net force was tripled and the mass were doubled, then the new acceleration would be _____ m/s/s.
Answer:
The new acceleration would be 9 m/s².
Explanation:
Acceleration of an object is 6 m/s²
Net force is equal to the product of mass and acceleration i.e.
F = ma
[tex]a=\dfrac{F}{m}\\\\\dfrac{F}{m}=6\ m/s^2[/tex]
If the net force was tripled and the mass were doubled, it means,
F' = 3F
m' = 2m
Let a' is new acceleration. So,
[tex]a'=\dfrac{F'}{m'}\\\\a'=\dfrac{(3F)}{(2m)}\\\\a'=\dfrac{3}{2}\times \dfrac{F}{m}\\\\a'=\dfrac{3}{2}\times 6\\\\a'=9\ m/s^2[/tex]
So, the new acceleration would be 9 m/s².
Suppose you have two point charges of opposite sign. As you move them farther and farther apart, the potential energy of this system relative to infinity:_____________.
(a) stays the same.
(b) Increases.
(c) Decreases.
(d) The answer would depend on the path of motion
Answer:
(b) Increases
Explanation:
The potential energy between two point charges is given as;
[tex]U = F*r = \frac{kq_1q_2}{r}[/tex]
Where;
k is the coulomb's constant
q₁ ans q₂ are the two point charges
r is the distance between the two point charges
Since the two charges have opposite sign;
let q₁ be negative and q₂ be positive
Substitute in these charges we will have
[tex]U = \frac{k(-q_1)(q_2)}{r} \\\\U = - \frac{kq_1q_2}{r}[/tex]
The negative sign in the above equation shows that as the distance between the two charges increases, the potential energy increases as well.
Therefore, as you move the point charges farther and farther apart, the potential energy of this system relative to infinity Increases.
You are at the carnival with you your little brother and you decide to ride the bumper cars for fun. You each get in a different car and before you even get to drive your car, the little brat crashes into you at a speed of 3 m/s.
A. Knowing that the bumper cars each weigh 80 kg, while you and your brother weigh 60 and 30 kg,respectively, write down the equations you need to use to figure out how fast you and your brother are moving after the collision.
B. After the collision, your little brother reverses direction and moves at 0.36 m/s. How fast are you moving after the collision?
C. Assuming the collision lasted 0.05 seconds, what is the average force exerted on you during the collision?
D. Who undergoes the larger acceleration, you or your brother? Explain.
Answer:
a) The equation is [tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) Your velocity after collision is 2.64 m/s
c) The force you felt is 7392 N
d) you and your brother undergo an equal amount of acceleration
Explanation:
Your mass [tex]m_{y}[/tex] = 60 kg
your brother's mass [tex]m_{b}[/tex] = 30 kg
mass of the car [tex]m_{c}[/tex] = 80 kg
your initial speed [tex]u_{y}[/tex] = 0 m/s (since you've not started moving yet)
your brother's initial velocity [tex]u_{b}[/tex] = 3 m/s
your final speed [tex]v_{y}[/tex] after collision = ?
your brother's final speed [tex]v_{b}[/tex] after collision = ?
a) equations you need to use to figure out how fast you and your brother are moving after the collision is
[tex](m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
but [tex]u_{y}[/tex] = 0 m/s
the equation reduces to
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) if your little brother reverses with velocity of 0.36 m/s it means
[tex]v_{b}[/tex] = -0.36 m/s (the reverse means it travels in the opposite direction)
then, imputing values into the equation, we'll have
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
(30 + 80)3 = (60 + 80)[tex]v_{y}[/tex] + (30 + 80)(-0.36)
330 = 140[tex]v_{y}[/tex] - 39.6
369.6 = 140[tex]v_{y}[/tex]
[tex]v_{y}[/tex] = 369.6/140 = 2.64 m/s
This means you will also reverse with a velocity of 2.64 m/s
c) your initial momentum = 0 since you started from rest
your final momentum = (total mass) x (final velocity)
==> (60 + 80) x 2.64 = 369.6 kg-m/s
If the collision lasted for 0.05 s,
then force exerted on you = (change in momentum) ÷ (time collision lasted)
force on you = ( 369.6 - 0) ÷ 0.05 = 7392 N
d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s
your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2
your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s
his deceleration is (3 - 0.36)/0.05 = 52.8 m/s
you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost
A vertically polarized light wave of intensity 1000 mW/m2 is coming toward you, out of the screen. After passing through this polarizing filter, the wave's intensity is
Answer:
The intensity is [tex]I = 500 mW/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]
Generally the intensity of the light emerging from the polarizer is mathematically represented as
[tex]I = \frac{I_o}{2}[/tex]
substituting values
[tex]I = \frac{1000 *10^{-3}}{2}[/tex]
[tex]I = 500 *10^{-3} W/m^2[/tex]
[tex]I = 500 mW/m^2[/tex]
The inner and outer surface temperature of a glass window 10 mm thick are 25 and 5 degree-C, respectively. What is the heat loss through a 1 m x 3 m window
Answer:
The heat loss is [tex]H = 8400\ W[/tex]
Explanation:
From the question we are told that
The thickness is [tex]t = 10 \ mm = 0.01 \ m[/tex]
The inner temperature is [tex]T_i = 25 ^oC[/tex]
The outer temperature is [tex]T_o = 5 ^oC[/tex]
The length of the window is L = 1 m
The width of the window is w = 3 m
Generally the heat loss is mathematically represented as
[tex]H = \frac{k * A * \Delta T}{t}[/tex]
Where k is the thermal conductivity of glass with value [tex]k = 1.4\ W/m \cdot K[/tex]
and A is the area of the window with value
[tex]A = 1 * 3[/tex]
[tex]A = 3 \ m^2[/tex]
substituting values
[tex]H = \frac{1.4 * 3 * (23-5)}{0.01}[/tex]
[tex]H = 8400\ W[/tex]