Given data
*The speed of the ranger who is driving at 11.4 m/s
*The given acceleration is a = -3.80 m/s^2
The formula for the distance covered by the ranger's vehicle is given by the kinematic equation of motion as
[tex]\Delta x=\frac{v^2-v^2_0}{2a}[/tex]*Here v = 0 m/s is the initial speed of the ranger's vehicle
Substitute the values in the above expression as
[tex]\begin{gathered} \Delta x=\frac{0^2-(11.4)^2}{2\times(3.80)} \\ =17.1\text{ m} \end{gathered}[/tex]The distance is required for the ranger's vehicle to come to rest is calculated as
[tex]\begin{gathered} D=20-17.1 \\ =2.9\text{ m} \end{gathered}[/tex]The stopping time is calculated as
[tex]v=v_0+at[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} 0=11.4+(-3.80)(t) \\ t=3.00\text{ s} \end{gathered}[/tex]Which one would repel a negatively charged object?+АBС
Configuration A will repel a negative charged object.
This comes from the fact that opposite charges attract each other while equal charges repel each other. Now, we notice that configuration B is neutral (it has the same amount of positive and negative charges) then it won't have an effect on a negative charged object; meanwhile configuration C has more positive charges, then it would attract a negative charged object. Configuration A has more negative charges then it will repel a negative charged object.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
When a pole vaulter reaches the top of her vault, how does her potential energy compare to her kinetic energy?
A. The kinetic energy is twice the amount of potential energy.
B. The kinetic energy is greater than the potential energy.
C. The kinetic energy and potential energy are the same.
D. The kinetic energy is less than the potential energy.
Answer: b
Explanation:
A 6 kg sign is suspended by two strings making angles with the ceiling as shown in the diagram. Determine the magnitudes of the tensions in each string. Remember that 1kg exerts a force of 9.8 N. please use sine/cosine law for this question
The triangle representing the given scenario is shown below
T1 and dT2 represents the tension in each spring
We would apply the sine rule which is expressed as
a/SinA = b/SinB = c/SinC
where
A, B and C are the angles of the triangle
a, b and c are the sides opposite the respective angles
Thus,
A = 42
a = T1
B = 23
b = T2
C = 115
c = 58.8
T1/sin42 = 58.8/sin115
By cross multiplying, it becomes
T1sin115 = 58.8sin42
T1 = 58.8sin42/sin115
T1 = 43.41
T2/sin23 = 58.8/sin115
By cross multiplying, it becomes
T2sin115 = 58.8sin23
T2 = 58.8sin23/sin 115
T2 = 25.35
The magnitude of the tension in each string are 43.41 N and 25.35 N
A yo‑yo with a mass of 0.0600 kg and a rolling radius of =1.60 cm rolls down a string with a linear acceleration of 5.20 m/s2.
Calculate the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
What is the moment of inertia of this yo‑yo?
The tension magnitude in the string is 0.312 N.
The angular acceleration magnitude of the yo‑yo is 325 rad/s².
The moment of inertia of the yo-yo is 7.68 x 10⁻⁶ kgm².
What is the moment of inertia of the yo-yo?
The moment of inertia of the yo-yo is calculated by applying the following equation as shown below;
I = ¹/₂MR²
where;
M is the mass of the yo-yoR is the radius of the yo-yoI = ¹/₂(0.06 kg)(0.016)²
I = 7.68 x 10⁻⁶ kgm²
The angular acceleration of the yo-yo is calculated as follows;
α = a/R
where;
a is the linear accelerationα = 5.2/0.016
α = 325 rad/s²
The tension in the string is calculated as follows;
T = ma
where;
m is the mass of the yo-yoa is the linear acceleration of the yo-yoT = 0.06 kg x 5.2 m/s²
T = 0.312 N
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An air-filled pipe is found to have successive harmonics at 800 Hz , 1120 Hz , and 1440 Hz . It is unknown whether harmonics below 800 Hz and above 1440 Hz exist in the pipe. The length is 53.5 cm. Identify the correct pressure variation graph for the 1120 Hz standing wave in the pipe. Note that the closed end of the pipe is on the right.
Using the concept of Harmonic-wave, we got the desired graph which is shown in the image.
It must be an open-closed pipe.
Open-closed pipe
Let the fundamental frequency of f be
The second harmonic is 3f.
3rd harmonic = 5f
Therefore, the difference between harmonics is 3f-f = 2f
So, 2f = (1120-800) = 1440-1120 = 320
Also, f = 160Hz
800Hz = 5th Harmonic
1120 Hz = 7th
1440 = 9th
For an open- or closed-pipe,
f = v/4L
where v is the speed of sound in the air = 343 M/s
So, 160 = 343/4L
Also, L = 0.5336 m = 53.6cm
Hence for the frequency of 1120Hz, we got the desired graph of a wave which is shown in the below graph.
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A scenario where two people are sitting on a see saw is modeled here. Assuming both people are at equal distances from the pivot point, which statements below would result in the unbalanced forces shown above? Select ALL that apply.A)Person A has a force of 70 N and person B has a force of 80 N.B)Person A has a force of 80 N and person B has a force of 70 N.C)Person A has a force of 75 N and person B has a force of 75 N.D)Person A has a force of 50 N and person B has a force of 40 N.E)Person A has a force of 50 N and person B has a force of 70 N.
Find the net force on the 4th part. Given, the value of velocity remain same at every point.
In the fourth case, when the weight and the tension are acting vertically downward.
The net force acting on the bucket is,
[tex]F_{\text{net}}=\frac{mv^2}{r}[/tex]where v is the velocity, r is the radius, and m is the mass,
The net force acting on the bucket will remain the same if the magnitude of the velocity remains the same.
As the direction of the bucket changes thus, the value of net force is not equal to zero.
As the velocity of the bucket is 3 m/s.
Thus, the value of net force acting on the bucket is,
[tex]\begin{gathered} F_{\text{net}}=\frac{0.5\times3^2}{0.6} \\ F_{\text{net}}=7.5\text{ N} \end{gathered}[/tex]Thus, (at the constant velocity at every point case) the net force acting is 7.5 N.
A pitcher throws a 0.140 kg baseball with a speed of 42.3 m/s the batter strikes it with an average force of 5120 N which result in the ball traveling with an initial speed of 31.0 m/s toward the pitcher for how long were the bat and ball in contact
The impulse exerted over an object is equal to the change in its linear momentum:
[tex]I=\Delta p[/tex]On the other hand, the impulse is equal to the force exerted over the object multiplied by the time during which the force was exerted:
[tex]I=F\cdot\Delta t[/tex]First, find the impulse by finding the change in linear momentum of the baseball. The linear momentum is given by the product of the speed of the baseball times its mass:
[tex]p=mv[/tex]Assume that the negative direction is towards the batter and the positive direction is towards the pitcher. Then, the initial velocity of the ball is -42.3 m/s and the final velocity of the ball is 31.0 m/s. Then:
[tex]\begin{gathered} p_i=mv_i=(0.140\operatorname{kg})(-42.3\frac{m}{s})=-5.922\operatorname{kg}\cdot\frac{m}{s} \\ \\ p_f=mv_f=(0.140\operatorname{kg})(31.0\cdot\frac{m}{s})=4.34\operatorname{kg}\cdot\frac{m}{s} \end{gathered}[/tex]Use the initial and final linear momentum to find the change in linar momentum, which is equal to the impulse:
[tex]\begin{gathered} I=\Delta p \\ =p_f-p_i \\ =(4.34\operatorname{kg}\cdot\frac{m}{s})-(-5.922\operatorname{kg}\cdot\frac{m}{s}) \\ =10.262\operatorname{kg}\cdot\frac{m}{s} \end{gathered}[/tex]Isolate Δt from the equation that relates force and impulse and substitute the corresponding values for I and F to find the time during which the bat and the ball were in contact:
[tex]\begin{gathered} \Delta t=\frac{I}{F} \\ =\frac{10.262\operatorname{kg}\cdot\frac{m}{s}}{5120N} \\ =0.00200429687\ldots s \\ \approx0.00200s=2.00ms \end{gathered}[/tex]Therefore, the bat and the ball were in contact during a time interval of 2 miliseconds.
Equations of linear motion
The Newton's equations of motion are stated below
v = u + at
s = ut 1/2at^2
v^2 = u^2 + 2as
s = 1/2(u + v)t
where
u represents initial velocity
v represents final velocity
t represents time
a represents acceleration
Thursday, February 11, 202110:08 AMCSA bullet travelling 420 m/s strikesa 1.5 kg target. How fast is thetarget moving if the bullet has amass of 0.1kg and is travelling300 m/s after the collision.
To solve this problem, we have to use conservation of momentum, which states that the initial momentum is equal to the final momentum. We know that momentum is defined as
[tex]p=mv[/tex]Let's apply it to the problem
[tex]\begin{gathered} p_{i1}+p_{i2}=p_{f1}+p_{f2} \\ m_1\cdot v_{i1}+m_2\cdot v_{i2}=m_1\cdot v_{f1}+m_2\cdot v_{f2} \end{gathered}[/tex]Using all the given information, we have
[tex]0.1\cdot420+1.5\cdot0=0.1\cdot300+1.5\cdot v_{f2}[/tex]Now, we solve the equation for v_f2:
[tex]\begin{gathered} 42=30+1.5\cdot v_{f2} \\ 42-30=1.5\cdot v_{f2} \\ 12=1.5\cdot v_{f2} \\ v_{f2}=\frac{12}{1.5} \\ v_{f2}=8 \end{gathered}[/tex]Hence, the target is moving 8 m/s after the collision.The cores of the terrestrial worlds are made mostly of metal because ______.
A. metals sunk to the centers a long time ago when the interiors were molten throughout
B. the terrestrial worlds as a whole are made mostly of metal
C. the core contained lots of radioactive elements that decayed into metals
D. over billions of years, convection gradually brought dense metals downward to the core
You have a light spring which obeys Hooke's law. This spring stretches 3.02 cm vertically when a 2.50 kg object is suspended from it. Determine the following.(a) the force constant of the spring (in N/m)N/m(b) the distance (in cm) the spring stretches if you replace the 2.50 kg object with a 1.25 kg objectcm(c) the amount of work (in J) an external agent must do to stretch the spring 8.90 cm from its from unstretched positionJ
We are given that a spring stretches 3.02 cm vertically when a 2.5 kg object is suspended.
Part (a) To determine the constant of the spring we need first to determine the weight of the object, to do that we will use the following formula:
[tex]W=mg[/tex]Where:
[tex]\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now we plug in the values:
[tex]W=(2.5\operatorname{kg})(9.8\frac{m}{s^2})[/tex]Solving the operations we get:
[tex]W=24.5N[/tex]Now we use Hooke's law:
[tex]F=kx[/tex]Where:
[tex]\begin{gathered} F=\text{ force} \\ k=\text{ spring contant} \\ x=\text{ distance stretched} \end{gathered}[/tex]Now we solve for "k" by dividing both sides by "x":
[tex]\frac{F}{x}=k[/tex]Now, since the object is placed vertically this means that the only force acting on the spring in the weight of the object, therefore:
[tex]F=W[/tex]Now we plug in the known values:
[tex]\frac{24.5N}{3.02\operatorname{cm}}=k[/tex]Solving the operations we get:
[tex]8.11\frac{N}{\operatorname{cm}}=k[/tex]Since we are required to express the constant in N/m, we need to convert the centimeters into meters. To do that we will use the following conversion factor:
[tex]100\operatorname{cm}=1m[/tex]Now we multiply by the conversion factor in decimal form, placing the centimeters as numerator:
[tex]8.11\frac{N}{\operatorname{cm}}\times\frac{100\operatorname{cm}}{1m}=811.26\frac{N}{m}[/tex]Therefore, the constant of the spring is 811.26 N/m.
part (b) Now we are asked to determine the distance is an object of 1.25 kg is place. First, we determine the weight of the new object:
[tex]W=mg[/tex]Now we plug in the values:
[tex]W=(1.25\operatorname{kg})(9.8\frac{m}{s^2})=12.25N[/tex]Now we use Hooke's law, but we solve for the distance "x" by dividing both sides by the constant "k", we get:
[tex]\frac{F}{k}=x[/tex]Just as before, the only force acting is the weight, therefore, we plug in the values we got:
[tex]\frac{12.25N}{811.26\frac{N}{m}}=x[/tex]Solving the operations:
[tex]0.015m=x[/tex]Now we convert the meters into centimeters using the same conversion factor:
[tex]0.015m\times\frac{100\operatorname{cm}}{1m}=1.5\operatorname{cm}[/tex]Therefore, the new mass stretches the spring 1.5 centimeters.
Part (c) Now we are asked to determine the work that has to be done on the spring to stretch it 8.9 centimeters. To determine that we will use the following formula for the work done on a spring:
[tex]w=\frac{1}{2}kx^2[/tex]Now we replace the values:
[tex]w=\frac{1}{2}(811.26\frac{N}{m})(8.9\operatorname{cm})^2[/tex]We will first convert the 8.9 centimeters into meters:
[tex]8.9\operatorname{cm}\times\frac{1m}{100\operatorname{cm}}=0.089m[/tex]Now we replace this in the formula for the work:
[tex]w=\frac{1}{2}(811.26\frac{N}{m})(0.089m)^2[/tex]Solving the operations we get:
[tex]w=3.21J[/tex]Therefore, the work is 3.21 Joules.
How much time is required for a car engine to do 278 kJ of work, if its maximum power is 95 kW?
W= work = 278 kj
P = Power = 95 Kw
t= time
P = W / t
Isolate t
t = W /P
Replacing:
t= 278kj / 95Kw = 2.93 seconds
It is required 2.93 seconds
Three facts about voltage sources
The voltage source produces potential difference across two ends of a conductor.
The voltage source has two terminals - positive and negative.
Voltage sources can be alternating or direct.
relationship between force of failure and diameter if ultimate tensile strength is same for all material tested i need equation
The relationship between the force and area of the material is,
[tex]F_f=\frac{P}{A}[/tex]where P is the load and A is the area.
As the tensile strength of the material is same for all the material tested.
As the area of the material is directly proportional to the diameter.
Thus, the force of failure is inversely proportional to the diameter.
A certain violet light has a wavelength of 413 nm. Calculate the frequency of the light. Calculate the energy content of one quantum of the light.
We will have the following:
First, we transform from nm to m,t that is:
[tex]\lambda=413nm\cdot\frac{m}{1\cdot10^9nm}\Rightarrow\lambda=4.13\cdot10^{-7}m[/tex]Then:
[tex]4.13\cdot10^{-7}m=\frac{3.0\cdot10^8m/s}{f}\Rightarrow f=\frac{3.0\cdot10^8m/s}{4.13\cdot10^{-7}m}[/tex][tex]\Rightarrow f\approx7.26\cdot10^{14}Hz[/tex]So, the frequency of the ligth is approximately 7.26*10^14 Hz.
Now, the energy will be:
[tex]E=(6.63\cdot10^{-34}m^2Kg/s)(7.26\cdot10^{14}/s)\Rightarrow E\approx4.81\cdot10^{-19}m^2Kg/s^2[/tex][tex]\Rightarrow E\approx4.81\cdot10^{-19}m^2Kg/s^2\cdot(J/m^2Kg/s^2))\Rightarrow E\approx4.81\cdot10^{-19}J[/tex]And the energy is approximately 4.81*10^-19 J.
Calculate the total capacitance of the three capacitors 30µF, 20µF & 12µF connected in series across a d.c. supply
Consider that three capacitors connected in series have the following total capacitance:
[tex]\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}[/tex]where,
C1 = 30µF
C2 = 20µF
C3 = 12µF
Consider that the LCM of the three previous numbers is 60 (to sum the fractions).
Replace the previous values of the parameters into the formula for C and simplify:
[tex]\begin{gathered} \frac{1}{C}=\frac{1}{30}+\frac{1}{20}+\frac{1}{12} \\ \frac{1}{C}=\frac{2+3+5}{60}=\frac{10}{60}=\frac{1}{6} \\ C=6 \end{gathered}[/tex]Hence, the total capacitance is 6µF
3. 1.819 m4. 5.291 m5. 6.321 m39. Answer: B40. Work done by the non conservative forces actingon an object is equal1.to the change in the mechanical energy of the object2. to the change in the kinetic energy of the object3. to the work done by the conservative forces4. to the change in the potential energy of the object5. to the net work done on the object41. Answer: A
We are asked to determine the distance that spring will stretch when a given mass is attached to it.
To do that we will use Hook's law:
[tex]F=kx[/tex]where:
[tex]\begin{gathered} F=\text{ force } \\ k=\text{ spring constant} \\ x=\text{ distance that the spring is stretched} \end{gathered}[/tex]We will determine the constant of the spring "k" first using the fact that the spring is stretched 0.8 meters when a mass of 3kg hangs from it.
Since the only force acting on the spring is the weight of the object we have:
[tex]F=mg[/tex]Where:
[tex]\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now, we substitute and we get:
[tex]mg=kx[/tex]Now, we divide both sides by "x":
[tex]\frac{mg}{x}=k[/tex]Now, we plug in the values:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{0.8m}=k[/tex]Solving the operations:
[tex]36.75\text{ N/m}=k[/tex]Now, we substitute the value of "k":
[tex]F=(36.75\text{ N/m\rparen}x[/tex]Now, we solve for "x":
[tex]\frac{F}{36.75\text{ N/m}}=x[/tex]Now, we substitute the value of the weight of the second object:
[tex]\frac{(14kg)(9.8\frac{m}{s^2})}{36.75\text{ N/m}}=x[/tex]Solving the operations:
[tex]3.733m=x[/tex]Therefore, the spring will stretch by 3.733 meters.
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it is the symbol of dot (.)
Question one I need an explanation Pls. Only part a and b.
Given,
The velocity of the objects, v₁=50.0 m/s and v₂=-25.0 m/s
The initial positions of the objects, x₁_₀=0.00 m and x₂_₀=500 m
The relative velocity of the objects is given by,
[tex]v=v_1-v_2[/tex]On substituting the known values,
[tex]\begin{gathered} v=50.0-(-25.0) \\ =75.0\text{ m/s} \end{gathered}[/tex]The total distance between the objects is,
[tex]\begin{gathered} d=x_{10}+x_{20} \\ =0+500 \\ =500\text{ m} \end{gathered}[/tex](a)
As the two objects are traveling towards each other in a straight line, they will intercept.
The time it takes for the objects to intercept is given by,
[tex]t=\frac{d}{v}[/tex]On substituting the know values,
[tex]\begin{gathered} t=\frac{500}{75} \\ =6.67\text{ s} \end{gathered}[/tex]As t is the time it takes for the objects to intercept, the distance covered by the objects in this time will give us the position x_f where these two objects intercept.
As object 1 starts from the origin, the distance traveled by this object is equal to x_f
The distance traveled by object 1 in that time is,
[tex]x_f=d_1=v_1t[/tex]On substituting the known values,
[tex]\begin{gathered} x_f_{}=50\times6.67 \\ =333.5m\text{ } \end{gathered}[/tex]Thus the objects will intercept at the point x_f=333.5 m from the origin.
(b)
As calculated in part a, the time it takes for the objects to intercept is t=6.67 s
The inductor in the RLC tuning circuit of an AM radio has a value of 10 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 97 kHz
Given:
The inductance is I = 10 mH
The frequency is f = 97 kHz
Required:
Value of capacitor in picofarads.
Explanation:
In order to tune the radio, the condition is
[tex]X_L=X_c[/tex]Here, X_L is the inductive reactance and
X_C is the capacitive reactance.
The capacitance can be calculated as
[tex]\begin{gathered} 2\pi fL=\frac{1}{2\pi fC} \\ C=\frac{1}{(2\pi f)^2L} \\ =\frac{1}{(2\times3.14\times97\times10^3)^2\times10\times10^{-3}} \\ =2.69\text{ }\times10^{-10}\text{ F}\times\frac{10^{12}\text{ pF}}{1\text{ F}} \\ =269\text{ pF} \end{gathered}[/tex]Final Answer: The value of the variable capacitor is 269 picoFarad.
True or false. Just as like magnetic poles attract, unlike poles repel.
Answer: False
Unlike poles attract each other. Think of a south pole of a magnet and the north pole of another. They attract each other, while two of the same poles would repel each other.
I was wondering if I could get help matching these definitions?
We are asked to match the following terms.
1. Accretion: This is the phenomenon when layers of the matter is added to a celestial body due to gravity. Therefore, this is matched with: the "Process of the steady growth of an object by a steady accumulation of material".
2. Interstellar. This is a word of Latin origin that means between the starts.
3. Nebula: These are clouds of hydrogen ions where the stars are formed, therefore, this is matched with a "cloud of gas and dust in space".
4. Nuclear fusion: When the nuclei of atoms combine they emit energy. This is a reaction that occurs inside the stars and is responsible for its emission of radiation. Therefore, this is matched with "a nuclear reaction where nuclei combine and release intense energy".
5. Solar nebula: This is the nebula from which the solar system is formed, therefore, this is matched with: the "cloud of gas and dust from which our solar system is formed"
6. Stellar evolution: These are the changes that a star undergoes during its life cycle, therefore, it is matched with the "life cycle of a star".
7. Supernova. This is a stage in the stellar evolution when a star emits light and energy through an explosion, therefore, this is matched with "an explosion of a star that emits large amounts of matter and energy".
What is the net force on a 4.12kg dog that accelerates at 3.02 m/s^2
Answer: Net force = 12.4 N
Explanation:
We would apply the formula which is expressed as
Net force = mass x acceleration
From the information given,
mass = 4.12
acceleration = 3.02
Net force = 4.12 x 3.02
Net force = 12.4 N
If you drop a 2.3kg ball from the top of a 44 m high building, how much potential energy will it have just before it hits the ground? Round your answer to the nearest whole number and include an appropriate unit
Answer: Potential energy = 0
Explanation:
The formula for calculating potential energy is expressed as
Potential energy = mgh
where
m = mass of object
g = acceleration due to gravity = 9.8m/s^2
h = height of object above the ground
From the information given,
m = 2.3
h = 44
Just before the object hits the ground, all potential energy is converted to kinetic energy. Thus, the answer is
Potential energy = 0
A 5 kg mass, hung onto a spring, causes the spring to stretch 7.0 cm. What is the spring constant? What is the potential energy of the spring?
Given,
The mass is m=5 kg.
The extensionis d=7.0 cm
The force is
F=mg
F=5x0.07=0.35N
Thus the spring constant is:
[tex]k=\frac{F}{d}=\frac{0.35}{0.07}=\frac{5N}{m}[/tex]The potential energy is:
[tex]U=\frac{1}{2}kd^2=\frac{1}{2}\times5\times(0.07)^2=0.012Nm^2[/tex]b. Calculate the electric force that exists between two objects that are 0.500 m apart and carrycharges of 0.00450 C and 0.00240
Given
The two charges,
[tex]\begin{gathered} q_1=0.00450\text{ C} \\ q_2=0.00240\text{ C} \end{gathered}[/tex]Distance between them,
[tex]r=0.5\text{ m}[/tex]To find
The electric force
Explanation
We know the force is given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]Putting the values,
[tex]\begin{gathered} F=9\times10^9\frac{0.00450\times0.00240}{(0.500)^2} \\ \Rightarrow F=388,800\text{ N} \end{gathered}[/tex]Conclusion
The electric force is 388,800 N
You have a coin sitting on
a card on top of a glass. You want to put the
coin into the glass, but you are not allowed
to pick up the card. Think of how you can do
that. Then write a short explanation of why
it works that would make sense to someone
who doesn't remember Newton's laws of
motion.
Answer:
You can flick the car to the side so that the coin can fall through.
Explanation:
The amount of energy in a photon is directly proportional to the ________________ of the electromagnetic wave, according to this equation:a) volumeb) none of these answers are correctc) directiond) motione) frequencyf) wavelength
Given:
[tex]E=h\nu=hf[/tex]Explanation:
The amount of energy in a photon is given as:
[tex]E=h\nu=hf[/tex]Here, E is the energy, h is Planck's constant, and ν (or f) is the frequency of the electromagnetic wave.
As h is a constant, we see that the energy E has a direct dependence on the frequency ν (or f) of the electromagnetic wave. In other words, the energy E of the photon is directly proportional to the frequency ν (or f) of the electromagnetic wave.
Final answer:
The amount of energy in a photon is directly proportional to the frequency of the electromagnetic wave.
Thus, the correct option is (e) frequency.
A standing wave is produced by reflecting a wave off a wall (which acts like a fixed end). If the standing wave consists of 6 anti-nodes and 7 nodes and the wall is 12m away from the source determine the wavelength.
Answer:
4 m
Explanation:
We can represent the standing wave as follows
So, we can divide the total distance into 6 parts as follows
12 m/6 = 2 m
Then, 2 m is the distance from node to node. It means that the wavelength will be twice this distance, so
wavelength = 2 x 2m = 4m
Therefore, the wavelength is 4m