Answer:
Explanation:
For spring
[tex]n=\sqrt{\frac{k}{m} }[/tex]
where n is frequency of oscillation and k is force constant and m is mass
Putting the values
[tex]1.2=\sqrt{\frac{k}{.28} }[/tex]
k = .4032 N/m
F= k x
where F is force , k is force constant and x is extension
Putting the given values
1 = .4032 x
x = 2.48 m
A block of mass m is suspended by a vertically oriented spring. If the mass of a block is increased to 4m, how does the frequency of oscillation change, if at all
Answer:
The frequency will be reduced by a factor of √2/2
Explanation:
Pls see attached file
The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
Let the initial mass of block be m.
And new mass is, 4m.
The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,
[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
Here,
k is the spring constant.
If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,
[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]
Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
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An air conditioner connected to a 103 V rms AC line is equivalent to a 20 resistance and a 1.68 inductive reactance in series. a) What is the impedance of the air conditioner
Answer:
20.07ohms
Explanation:
Impedance is defined as the opposition to the flow of current through the elements of the circuit.
Impedance for R-L AC circuit is expressed as Z = √R²+XL²
R is the resistance
XL is the inductive reactance.
Given resistance of the air condition = 20 ohms
Inductive reactance XL = 1.68 ohms
Z = √20²+1.68²
Z = √400+2.8224
Z = √402.8224
Z = 20.07 ohms
Hence the impedance of the air conditioner is 20.07ohms
Two particles, of charges q1 and q2, are separated by a distance d on the x-axis with q1 at the origin and q2 in the positive direction. The net electric field due to the particles is zero at x = d/4. With V = 0 at infinity, locate (in terms of d) any point on the x-axis (other than infinity) at which the electric potential due to the two particles is zero.
Answer:
No point on the x-axis
Pls see attached file
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.
Answer:
206.67NExplanation:
The sum of force along both components x and y is expressed as;
[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]
The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
[tex]a_x = \frac{d^2 x }{dt^2}[/tex]
[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]
Similarly,
[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]
[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]
[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
n oscillator is driven by a sinusoidal force. The frequency of the applied force A : must be less than the natural frequency of the oscillator. B : is independent of the natural frequency of the oscillator. C : becomes the natural frequency of the oscillator. D : must be equal to the natural frequency of the oscillator. E : must be greater than the natural frequency of the oscillator
Answer:
B : is independent of the natural frequency of the oscillator
Explanation:
You can apply any force you like to a natural oscillator. It is independent of the natural frequency of the oscillator.
The result you get will depend on how the frequency of the applied force and the natural frequency relate to each other. It will also depend on the robustness of the oscillator with respect to the applied force.
Clearly, if the force is small enough, it will have no effect on the oscillator. If it is large enough, it will overpower any motion the oscillator may attempt. For forces in the intermediate range, there will be some mix of natural oscillation and forced behavior. One may modulate the other, for example.
Rick spends four hours researching on the internet and does 1090 J of work. In the process, his internal energy decreases by 2190 J. Determine the value of Q, including the algebraic sign.
Answer:
Q = -3280J
Explanation:
From the First Law of Thermodynamics, energy cannot be created nor destroyed but it can be converted from one form to another with the interaction of heat. Mathematically, this can be expressed as:
ΔU = Q + W ----------(i)
Where;
ΔU = total change in internal energy of a system.
Q = heat exchanged between the system and the surrounding
W = work done by or on the system.
If heat is lost into the surrounding, then Q = -ve, else Q = +ve
If work is done on the system, then W = -ve, else W = -ve
=> From the question, Rick is the system and does a work of
W = +1090J [since Rick does the work, W = +ve]
=>Also, the internal energy decreases by 2190J, therefore,
ΔU = -2190J [since there is a decrease in internal energy]
Substitute the values of W and ΔU into equation (i) as follows;
-2190 = Q + 1090
=> Q = -2190 - 1090
=> Q = -3280J
Therefore, the value of Q = -3280J
A 900 kg roller coaster car starts from rest at point A. rolls down the track, goes
around a loop (points B and C) and then flies off the inclined part of the track (point D),
Figure 2.
The dimensions are: H =80 m.
r= 15m, h=10m and theta =9.30°
Calculate the
(a) gravitational potential energy at point A.
(b) velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J.
c) distance of the car land (in the horizontal direction) from point D if given the
velocity at point D is 37.06 m/s
I
Answer:
gravitational potential energy at point A.
A) The gravitational potential energy at point A is; 705600 J
B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s
A) Formula for gravitational potential energy is;
PE = mgh
At point A;
mass; m = 900 kg
height; h = 80 m
Thus;
PE = 900 × 9.8 × 80
PE = 705600 J
B) Kinetic energy of the roller coaster at point C is given as;
KE = PE - W
We are given Workdone; W = 264870 J
Thus;
KE = 705600 - 264870
KE = 440730 J
Thus, velocity at point C is gotten from the formula of kinetic energy;
KE = ½mv²
v = √(2KE/m)
v = √(2 × 440730/900)
v = 31.295 m/s
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If the car decelerates uniformly along the curved road from 27 m/s m/s at A to 13 m/s m/s at C, determine the acceleration of the car at B
Answer:
0.9m/s²
Explanation:
See attached files
A solenoid 26.0 cm long and with a cross-sectional area of 0.580 cm^2 contains 490 turns of wire and carries a current of 90.0 A.
Calculate:
(a) the magnetic field in the solenoid;
(b) the energy density in the magnetic field if the solenoid is filled with air;
(c) the total energy contained in the coil’s magnetic field (assume the field is uniform);
(d) the inductance of the solenoid.
Answer:
A.21.3T
B.1.8x 10^6J/m^3
C.0.27x10^2J
D.6.6x10^-3H
Explanation:
Pls see attached file
A flashlight is held at the edge of a swimming pool at a height h = 1.6 m such that its beam makes an angle of θ = 38 degrees with respect to the water's surface. The pool is d = 1.75 m deep and the index of refraction for air and water are n1 = 1 and n2 = 1.33, respectively.
Required:
What is the horizontal distance from the edge of the pool to the bottom of the pool where the light strikes? Write your answer in meters.
one of the answers that i found was 5.83 m i did some more research and it showed the same answer again. good luck with it. hope i was able to help you.
Suppose a space vehicle with a rest mass of 150 000 kg travels past the International Space Station at a constant speed of 2.6 x 108 m/s with respect to the I.S.S. When an observer on the I.S.S. measures the moving vehicle, her measurement of the space vehicle length is 25.0 m. Determine the relativistic mass of the space vehicle. Determine the length of the space vehicle as measured by an astronaut on the space vehicle.
Answer:
m = 300668.9 kg
L₀ = 12.47 m
Explanation:
The relativistic mass of the space vehicle is given by the following formula:
[tex]m = \frac{m_{0}}{\sqrt{1-\frac{v^{2} }{c^{2}} } }[/tex]
where,
m = relativistic mass = ?
m₀ = rest mass = 150000 kg
v = relative speed = 2.6 x 10⁸ m/s
c = speed of light = 3 x 10⁸ m/s
Therefore
[tex]m = \frac{150000kg}{\sqrt{1-\frac{(2.6 x 10^{8}m/s)^{2} }{(3 x 10^{8}m/s)^{2}} } }[/tex]
m = 300668.9 kg
Now, for rest length of vehicle:
L = L₀√(1 - v²/c²)
where,
L = Relative Length of Vehicle = 25 m
L₀ = Rest Length of Vehicle = ?
Therefore,
25 m = L₀√[1 - (2.6 x 10⁸ m/s)²/(3 x 10⁸ m/s)²]
L₀ = (25 m)(0.499)
L₀ = 12.47 m
At t=0 a 2150kg rocketship in outer space fires the engine which exerts a force=At2, and F(1.25s)=781.25N in the x direction. Find the impulse J during the interval t=2.00s and t=3.5s
Answer:
5.81 X 10^3 Ns
Explanation:
Given that
F = At² and F at t = 1.25 s is 781.25 N ?
A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²
d(Impulse) = Fdt
Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s
Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s
Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns
5.81 X 10^3 N.s
A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10
Answer:
It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water
Explanation:
A string passing over a pulley has a 3.85-kg mass hanging from one end and a 2.60-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.5 cm and mass 0.79 kg .
A. If the bearings of the pulley were frictionless, what would be the acceleration of the two masses?
B. In fact, it is found that if the heavier mass is given a downward speed of 0.20 m/s , it comes to rest in 6.4 s . What is the average frictional torque acting on the pulley?
Answer:
Explanation:
Let the acceleration be a of the system
T₁ and T₂ be the tension in the string attached with 3.85 and 2.6 kg of mass
for motion of 3.85 kg , applying newton's law
3.85g - T₁ = 3.85 a
for motion of 2.6 kg
T₂ - 2.6g = 2.6 a
T₂ - T₁ + 1.25 g = 6.45 a
T₁ - T₂ = 1.25 g - 6.45 a
for motion of pulley
(T₁ - T₂ ) x R = I x α where R is radius of pulley , I is its moment of inertia and α is angular acceleration
(T₁ - T₂ ) x R = 1 /2 m R² x a / R
(T₁ - T₂ ) = m x a / 2 = .79 x a / 2 = . 395 a
1.25 g - 6.45 a = .395 a
1.25 g = 6.845 a
a = 1.79 m /s²
B )
When heavier mass is given speed of .2 m /s , it comes to rest in 6.4 s
Average deceleration = .2 / 6.4 = .03125 m /s²
Total deceleration created by frictional torque = 1.79 + .03125
= 1.82125 m /s²
If R be the average frictional torque acting on the pulley
angular deceleration of pulley = a / R
= 1.82125 / .045
= 40.47 rad /s²
Now R = I x 40.47 , I is moment of inertia of pulley
= 1 /2 x .79 x .045² x 40.47
= .0323 N.m
Torque created = .0323 Nm
The acceleration of the two masses hanging from ends of the pulley is 31 m/s².
The average frictional torque acting on the pulley is 0.55 Nm.
The given parameters;
mass of the pulley, = M = 0.79 kgfirst mass, m₁ = 3.85 kgsecond mass, m₂ = 2.6 kgradius, R = 4.5 cm = 0.045 mThe acceleration of the two masses is determined by taking net torque acting on the pulley;
[tex]\tau _{net} = I \alpha[/tex]
[tex]T_1R - T_2R = I \alpha\\\\[/tex]
where;
T is the tension on both stings suspending the masses = mgI is the moment of inertia of the pulley [tex]= \frac{MR^2}{2}[/tex]α is the angular acceleration[tex]R(T_1 - T_2) = (\frac{MR^2}{2} )(\frac{a}{R} )\\\\T_1 - T_2 = (\frac{MR^2}{2} )(\frac{a}{R} ) \times \frac{1}{R} \\\\T_1 - T_2 = \frac{M}{2} \times a\\\\a = \frac{2}{M} (T_1 - T_2)[/tex]
Substitute the given parameters, to solve for the acceleration of the masses;
[tex]a = \frac{2}{M} (m_1g - m_2 g)\\\\a = \frac{2g}{M} (m_1 - m_2)\\\\a = \frac{2 \times 9.8}{0.79} (3.85 - 2.6)\\\\a = 31 \ m/s^2[/tex]
The average frictional torque acting on the pulley when the heavier mass speeds down by 0.2 m/s and stop by 6.4 s.
[tex]a = \frac{v}{t} = \frac{0.2}{6.4} = 0.031 \ m/s^2 \\\\ a_t = 31 m/s^2+ 0.031 m/s^2 = 31.031 m/s^2 \\\\\tau = I \alpha\\\\\tau = (\frac{MR^2}{2} )(\frac{a_t}{R} )\\\\\tau = (\frac{0.79 \times 0.045^2 }{2} ) (\frac{31.031}{0.045} )\\\\\tau = 0.55 \ Nm[/tex]
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Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the puck begins to move to the moment it hits the spring. Use 0.253 m for the displacement of the puck along the ramp and 9.80 m/s2 for the acceleration due to gravity. Assume that the mass of the puck is 0.180 kg. Express your answer using SI units to three significant figures.
Answer:
0.16joules
Explanation:
Using the relation for The gravitational potential energy
E= Mgh
Where,
E= Potential energy
h = Vertical Height
M = mass
g = Gravitational Field Strength
To find the vertical component of angle of launch Where the angle is 22°
h= sin theta
So E = mghsintheta
= 0.18 x 0.98 x 0.253 sin22
=0.16joules
Explanation:
The force required to compress a spring with elastic constant 1500N / m, with a distance of 30 cm is
Explanation:
F = kx
F = (1500 N/m) (0.30 m)
F = 450 N
A 18.0 kg electric motor is mounted on four vertical springs, each having a spring constant of 24.0 N/cm. Find the period with which the motor vibrates vertically.
Answer:
Explanation:
Total mass m = 18 kg .
Spring are parallel to each other so total spring constant
= 4 x 24 = 96 N/cm = 9600 N/m
Time period of vibration
[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]
Putting the given values
[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]
= .27 s .
Gravitational Force: Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one
Answer:
F' = F
Hence, the magnitude of the attractive force remains same.
Explanation:
The force of attraction between two bodies is given by Newton's Gravitational Law:
F = Gm₁m₂/r² --------------- equation 1
where,
F = Force of attraction between balls
G = Universal Gravitational Constant
m₁ = mass of first ball
m₂ = mass of 2nd ball
r = distance between balls
Now, we double the masses of both balls and the separation between them. So, the force of attraction becomes:
F' = Gm₁'m₂'/r'²
here,
m₁' = 2 m₁
m₂' = 2 m₂
r' = 2 r
Therefore,
F' = G(2 m₁)(2 m₂)/(2 r)²
F' = Gm₁m₂/r²
using equation 1:
F' = F
Hence, the magnitude of the attractive force remains same.
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin(kx−ωt)j^,where j^ is the unit vector in the y direction.
What is the Poynting vector S⃗ (x,t), that is, the power per unit area associated with the electromagnetic wave described in the problem introduction?
Given that,
The electric field is given by,
[tex]\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}[/tex]
Suppose, B is the amplitude of magnetic field vector.
We need to find the complete expression for the magnetic field vector of the wave
Using formula of magnetic field
Direction of [tex](\vec{E}\times\vec{B})[/tex] vector is the direction of propagation of the wave .
Direction of magnetic field = [tex]\hat{j}[/tex]
[tex]B=B_{0}\sin(kx-\omega t)\hat{k}[/tex]
We need to calculate the poynting vector
Using formula of poynting
[tex]\vec{S}=\dfrac{E\times B}{\mu_{0}}[/tex]
Put the value into the formula
[tex]\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}[/tex]
[tex]\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]
Hence, The poynting vector is [tex]\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]
Receiver maxima problem. When the receiver moves through one cycle, how many maxima of the standing wave pattern does the receiver pass through
The number of maxima of the standing wave pattern is two.
Maxima problem:At the time when the receiver moves via one cycle so here two maximas should be considered. At the time when the two waves interfere by traveling in the opposite direction through the same medium so the standing wave pattern is formed.
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A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire of length 2L. This long wire is attached to a battery, and a current is flowing through it. If the electric field in the narrow wire is E, the electric field in the wide wire is
Answer:
electric field in the wide wire is
E₂ =[tex]\frac{E}{4}[/tex]
Explanation:
given
length of the copper wire = L
radius of the copper wire r₁ = b
length of the second copper wire = L
radius of the second copper wire r₂ = 2b
electric field in the narrow wire = E₁=E
recall
resistance R = ρL/A
where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.
Resistance of narrow wire, R₁
R₁ = ρL/A
where A = πb²
R₁ = ρL/πb²---------- eqn 1
Resistance of wide wire, R₂
R₂ = ρL/A
where A = π(2b)²
R₂ = ρL/π(2b)²
R₂ = ρL/4πb²-------------- eqn 2
R₂ = ¹/₄(ρL/πb²)
comparing eqn 1 and 2
R₁ = 4R₂
calculating the current in the wire,
I = E/(R₁ + R₂)
recall
R₁ = 4R₂
∴ I = E/(4R₂ + R₂)
I = E/5R₂
calculating the potential difference across R₁ & R₂
V₁ = IR₁
I = E/5R₂
∴ V₁ = ER₁/5R₂
R₁ = 4R₂
V₁ = 4ER₂/5R₂
∴V₁ = ⁴/₅E
potential difference for R₂
V₂= IR₂
I = E/5R₂
∴ V₂ = ER₂/5R₂
V₂ = ER₂/5R₂
∴V₂ = ¹/₅E
so, electric field E = V/L
for narrow wire E₁ = V₁/L ----------- eqn 3
for wide wire, E₂ = V₂/L------------ eqn 4
compare eqn 3 and 4
E₂/E₁ = V₂/V₁( L is constant)
E₂/E₁ = ¹/₅E/⁴/₅E
E₂ = E₁/4
note E₁ = E
∴E₂ =[tex]\frac{E}{4}[/tex]
select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.
Answer:
B.solar panels generate electricity that keeps the satellites running.
Explanation:
Solar panels are a renewable resource because they take energy from the sun.
A circular loop of wire of radius 10 cm carries a current of 6.0 A. What is the magnitude of the magnetic field at the center of the loop
Answer:
3.77x10^-5T
Explanation:
Magnetic field at center of the loop is given as
B=uo*I/2r =(4pi*10-7)*6/2*0.1
B=3.77*10-5Tor 37.7 uTi
Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies. Express your answer in terms of Ki and Kf.
The net work done (W) on a particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies is equal to [tex]W = K_f - K_i[/tex]
The net work done (W) can be defined as the work done in moving an object by a net force, which is the vector sum of all the forces acting on the object.
According to Newton's Second Law of Motion, the net work done (W) on an object or physical body is equal to the change in the kinetic energy possessed by the object or physical body.
Mathematically, the net work done (W) on an object or physical body is given by the formula:
[tex]W =\Delta K_E\\\\W = K_f - K_i[/tex]
Where:
W is the net work done.[tex]K_f[/tex] is the initial kinetic energy.[tex]K_i[/tex] is the final kinetic energy.Read more: https://brainly.com/question/22599382
A 5.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.60 s. Find the force constant of the spring.
Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.
N/m
Explanation:
A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the switch is open and the capacitor is uncharged. What is the voltage across the resistor and the capacitor at the moment the switch is closed
Answer:
The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.
Explanation:
This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery
In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.
Answer:
E = 1.24MeV
Explanation:
The photon travels at the speed of light, 3.0 × [tex]10^{8}[/tex] m/s, and given that its frequency = 1 picometer = 1.0 × [tex]10^{-12}[/tex] m.
Its energy can be determined by;
E = hf
= (hc) ÷ λ
where E is the energy, h is the Planck's constant, 6.626 × [tex]10^{-34}[/tex] Js, c is the speed of the light and f is its frequency.
Therefore,
E = (6.626 × [tex]10^{-34}[/tex]× 3.0 × [tex]10^{8}[/tex]) ÷ 1.0 × [tex]10^{-12}[/tex]
= 1.9878 × [tex]10^{-25}[/tex] ÷ 1.0 × [tex]10^{-12}[/tex]
E = 1.9878 × [tex]10^{-13}[/tex] J
But, 1 eV = 1.6 × [tex]10^{-19}[/tex] J. So that;
E = [tex]\frac{1.9878*10^{-13} }{1.6*10^{-19} }[/tex]
= 1242375 eV
∴ E = 1.24MeV
The energy of the photon is 1.24MeV.
which example describes a nonrenewable resource?
A. everyone in our neighborhood uses solar panels to generate electricity to run their pool pumps.
B. once up and running, the power plant will convert the energy from tides and waves into electricity.
C. there is a long stretch of land in the desert with many windmills that are able to generate enough electricity to run the town.
D. there are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.
Answer:
D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity
Explanation:
Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.
1. The frequency of a wave defines
O A. the minimum height of a wave.
O B. the maximum height of a wave.
O C. how fast the wave is moving in cycles per second.
D. the height of the wave at a given point.
Answer:
The answer is CExplanation:
Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.