Answer:
No you didn't win the stuffed poodle
Explanation:
From the question we are told that
The distance of the ramp to the hole is d = 3 m \
The angle of inclination is [tex]\theta = 10^o[/tex]
The coefficient of static friction is [tex]\mu_s = 0.1[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.08[/tex]
The mass of the puck is m = 1 kg
The velocity of the first slide is [tex]v_1 = 2.0 m/s[/tex]
Generally the kinetic energy at the bottom of the ramp is equal the energy loss due to friction and this can be mathematically represented as
[tex]\frac{1}{2} m * v^ 2 = \mu_k * [m * g] * cos (theta ) * d[/tex]
=> [tex]\frac{1}{2} * v^ 2 = 0.08 * 9.8 * cos (10 ) * 3[/tex]
=> [tex] v= 2.17 \ m/s [/tex]
Comparing the relative velocity obtained and the velocity of your first throw we can see that you didn't win the stuffed animal
Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by
Answer:
The resultant vector [tex]\vec R = \vec A+\vec B[/tex] is given by [tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex].
Explanation:
Let [tex]\vec A = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})[/tex] and [tex]\vec B = 4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})[/tex], both measured in meters. The resultant vector [tex]\vec R[/tex] is calculated by sum of components. That is:
[tex]\vec R = \vec A+\vec B[/tex] (Eq. 1)
[tex]\vec R = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})+4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})[/tex]
[tex]\vec R = (6\cdot \cos 30^{\circ}-4\cdot \sin 30^{\circ})\,\hat{i}+(6\cdot \sin 30^{\circ}-4\cdot \cos 30^{\circ})\,\hat{j}[/tex]
[tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex]
The resultant vector [tex]\vec R = \vec A+\vec B[/tex] is given by [tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex].
Question 1-1: In each case, lifting or pushing, why must you exert a force to keep the object moving at a constant velocity?
Answer:
We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
Explanation:
LIFTING:
When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.
PUSHING:
When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.
Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
Please help me with this
Answer:
7 Newton's East
Explanation:
when the force is going in the same direction in this case east, you add the forces.
I need help with this answer
decomposition
A decomposition reaction is just the opposite of combination reaction
Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days. If it were to undergo gravitational collapse to a neutron star of radius 13 km , losing three-quarters of its mass in the process, what would its rotation speed be
Answer:
Its rotation will be 3.89x10⁴ rad/s.
Explanation:
We can find the rotation speed by conservation of the angular momentum:
[tex] L_{i} = L_{f} [/tex]
[tex] I_{i}\omega_{i} = I_{f}\omega_{f} [/tex] (1)
The initial angular speed is:
[tex] \omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d} [/tex]
The moment of inertia (I) of a sphere is:
[tex] I = \frac{2}{5}mr^{2} [/tex] (2)
Where m is 9 times the sun's mass and r is the sun's radius
By entering equation (2) into (1) we have:
[tex] \frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f} [/tex]
[tex]9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}[/tex]
[tex]\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s[/tex]
Hence, its rotation will be 3.89x10⁴ rad/s.
I hope it helps you!
7. A 1,500-N force is applied to a 1,000-kg car. What is the car's acceleration?
Answer:
1.5m/s^2
Explanation:
Answer:
1.5 m/s2. accerelation =force ÷mass
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of 1.1 m in a time of 1.8 s. The readout on the display indicates that the average power you are producing is 90 W. What is the magnitude of the force that you exert on the handle?
Answer:
147.27N
Explanation:
Power = workdone/time
Power = Force*distance/time
Given
Power = 90Watts
Distance = 1.1m
Time = 1.8secs
Force = ?
Substitute the given parameters into the formula:
[tex]90 = \frac{1.1d}{1.8}\\cross \ multiply\\ 90 \times 1.8 = 1.1F\\162 = 1.1F\\1.1F = 162\\F = \frac{162}{1.1} \\F = 147.27N[/tex]
Hence the magnitude of the force that you exert on the handle is 147.27N
Complete each statement by dragging the forms of energy into their appropriate boxes.
wind turbine
roller coaster going downhill
toaster
car
A
converts electrical energy into thermal energy.
A
converts rotational energy into electrical energy.
A
converts gravitational energy into mechanical energy.
A
converts rotational energy into mechanical energy.
Statements 1,2,3 and 4 match statements B, C, A, and D respectively.A wind turbine converts rotational energy into electrical energy.
What is the law of conservation of energy?According to the Law of conservation of energy. Energy can not be created nor be destroyed, it can transfer from one to another form.
1.A wind turbine converts rotational energy into electrical energy.
2.A roller coaster going downhill converts gravitational energy into mechanical energy
3. Toaster converts electrical energy into thermal energy
4.A car converts rotational energy into mechanical energy.
Hence,statements 1,2,3 and 4 match statements B, C, A, and D respectively.
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A microwave oven operates at 2.50 GHzGHz . What is the wavelength of the radiation produced by this appliance? Express the wavelength numerically in nanometers.
Answer:
The wavelength is [tex]\lambda = 1.2 * 10^8 nm[/tex]
Explanation:
From the question we are told that
The frequency of operation of the microwave is [tex]f = 2.50 GHz = 2.50 *10^{9} \ Hz[/tex]
Generally the wavelength is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]\lambda = \frac{3.0 *10^{8}}{ 2.50 *10^{9}}[/tex]
=> [tex]\lambda = 0.12 \ m [/tex]
converting to nanometer
[tex]\lambda = 1.2 * 10^8 nm[/tex]
3- For given three vectors a, b and c, c = a x b, then the vector c is:
Answer:
VB
Explanation:
What are two ways that an object can have kinetic energy?
Answer:
The object has to have mass and speed
Explanation:
You can increase both speed and mass to increase the kinetic energy, hope this answers your question.
Happy Halloween!
a tiger leaps with an initial velocity of 55 km/hr at an angle of 13° with respect to the horizontal. what are the components of the tigers velocity?
Answer:
vₓ = 53.6 km/h
vy = 12.4 km/h
Explanation:
if we define two axis perpendicular each other with origin in the point represented by the tiger leaping (assuming we can treat it as a point mass) coincidently with the horizontal (x-axis) and vertical (y-axis) directions, we can obtain the components of the velocity in both independent directions.We can do it simply getting the projections of the velocity vector on both axes, using simple trigonometry, as follows:[tex]v_{x} = v_{o} * cos \theta = 55 km/h * cos 13 = 53.6 km/h[/tex]
[tex]v_{y} = v_{o} * sin\theta = 55 km/h * sin 13 = 12.4 km/h[/tex]
A force of 41 N acts on an object which has a mass of 2.4 kg. What acceleration (in m/s2) is produced by the force
Answer:
The acceleration is [tex] a = 17.083 \ m/s^2 [/tex]
Explanation:
From the question we are told that
The force is [tex]F = 41 \ N[/tex]
The mass of the object is [tex]m = 2.4 \ kg[/tex]
Generally the force is mathematically represented as
[tex]F = m* a[/tex]
=> [tex] 41 = 2.4* a[/tex]
=> [tex] a = 17.083 \ m/s^2 [/tex]
Question C) needs to be answered, please help (physics)
(a) Differentiate the position vector to get the velocity vector:
r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k
v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j
(b) The velocity at t = 2.00 s is
v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j
(c) Compute the electron's position at t = 2.00 s:
r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k
The electron's distance from the origin at t = 2.00 is the magnitude of this vector:
||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that
tan(θ) = (-16.0 m/s) / (3.00 m/s) ==> θ ≈ -79.4º
or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.
what is the force produced on a body of 30kg mass when a body moving with the velocity of 26km/hr is acceleted to gain the velocity of 54 km/hr in 4 sec
Answer:
F = 58.35 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.
We have to convert speeds from kilometers per hour to meters per second
[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]
[tex]v_{f}=v_{o}+(a*t) \\[/tex]
where:
Vf = final velocity = 15 [m/s]
Vi = initial velocity = 7.22 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Note: the positive sign of the above equation is because the car increases its speed
15 = 7.22 + (a*4)
a = 1.945 [m/s^2]
Now we can use the Newton's second law:
F = m*a
F = 30*1.945
F = 58.35 [N]
What are the four basics for knowledge called "innate knowledge"?
Answer: The four basics of innate knowledge are as follows:
Explanation:
1. This knowledge is learned from birth and it is inborn knowledge. It is allows the organism to act naturally. For example, a dog is not taught to pant, but it pants to reduce heat from the body.
2. It is inherent.
3. It is essential for survival.
4. It arises from intellectual knowledge rather than being learned via experiences.
If a projectile hits a stationary target, and the projectile continues to travel in the same direction, the mass of the projectile is less than the mass of the target. the mass of the projectile is equal to the mass of the target. the mass of the projectile is greater than the mass of the target. nothing can be said about the masses of the projectile and target without further information. this is an unphysical situation and will not actually happen.
The correct arrangement of the question is;
If a projectile hits a stationary target, and the projectile continues to travel in the same direction,
A) the mass of the projectile is less than the mass of the target.
B) the mass of the projectile is equal to the mass of the target.
C) the mass of the projectile is greater than the mass of the target.
D) nothing can be said about the masses of the projectile and target without further information.
E) this is an unphysical situation and will not actually happen.
Answer:
Option C: The mass of the projectile is greater than the mass of the target.
Explanation:
We want to find what will happen when a projectile continues in motion after it hits a target.
Now, for the projectile to keep moving in that direction after it hits the target, it means it had a force bigger than the force of the target to overpower it and force it to move with it.
Now, from law of inertia, Force = ma.
But in this case acceleration is 0 because the speed of the projectile is constant.
Thus, the force depends on the mass. So for a higher force, the mass of the projectile has to be more than that of the stationary object.
Thus, option C is correct
A soccer player kicking a ball; the ball soaring through the air and landing on the ground
If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p
Answer:
8.93*10^13 N.
Explanation:
Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:[tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]
where, m1 = mass of the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:[tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]
Fg = 8.93*10^13 N.While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?
Analysing the question:
Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s
We are given:
height of the tower (h) = 66 m
mass of the stone (m) = 0.5 kg
initial velocity of the stone (u) = 0 m/s
time taken by the stone to reach the ground (t) = t seconds
acceleration due to gravity = 10 m/s²
** Neglecting air resistance**
Finding the time taken by the stone to reach the ground:
from the second equation of motion
h = ut + 1/2at²
replacing the variables
66 = (0)(t) + 1/2 (10)(t)²
66 = 5t²
t² = 13.2
t = 3.6 seconds
I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds
but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved
What (rather remarkable!) equation relates the speed of light to other fundamental electromagnetic constants?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equation is [tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
The value of c is [tex]c = 2.998 *10^{8} \ m/s [/tex]
Explanation:
From the question we are told that
Generally the equation that relates the speed of light to other fundamental electromagnetic constants is
[tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
Here c is the speed of light
[tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
and [tex]\epsilon_o[/tex] is the permittivity of free space with value
[tex]\epsilon_o = 8.85*10^{-12} \ C/V \cdot m[/tex]
So
[tex]c = \frac{1}{\sqrt{ 4\pi *10^{-7} * 8.85*10^{-12}} }[/tex]
=> [tex]c = 2.998 *10^{8} \ m/s [/tex]
Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery?
a. All the current is used up in the bulb, so the connecting wires don't matter.
b. Very little energy is dissipated in the thick connecting wires.
c. Electric field in the connecting wires is zero, so emf = E_bulb * L_bulb.
d. Current in the connecting wires is smaller than current in the bulb.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
Answer:
Options B & E are correct
Explanation:
Looking at all the options, B & E are the correct ones.
Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.
Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.
Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.
Option C is not correct because although the Electric field along the wire is not zero, it is very small.
Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.
The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :
b. Very little energy is dissipated in the thick connecting wires.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
"Energy"The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.
The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.
Thus, the correct answer is B and E.
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An FM radio station, 20 miles away, broadcast at a 93.4 MHz frequency(a) What is the wavelength of the radio wave associated with this signal ?(b) How long does it take for the signal to reach your radio from the station ?
Answer:
(a) Wavelength = 3.21 m (b) Time = [tex]1.07\times 10^{-4}\ s[/tex]
Explanation:
Given that,
The frequency of FM radio station, f = 93.4 MHz
(a) We need to find the wavelength of the radio wave associated with this signal. The relation between wavelength and frequency is given by :
[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{93.4\times 10^6}\\\\\lambda=3.21\ m[/tex]
(b) It is given that, an FM radio station, 20 miles away. Let t is time taken for signal to reach your radio from the station. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{20\times 1609.34}{3\times 10^8}\\\\t=1.07\times 10^{-4}\ s[/tex]
Hence, this is the required solution.
If the shoe has less mass, it will experience _______________ (more, less, the same) friction as it would with more mass.
You are studying circular motion by placing pennies on a turntable and then turning it on so that it will spin. You keep increasing the speed until one of the pennies slips off. You repeat this procedure and observe that the penny close to the outer edge always slip off first. What is the best inference?
1.The penny near the edge has a greater tangential velocity than the one in the center, so it experiences more air resistance. It’s the effect of the air “blowing” it off.
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
3.The centrifugal force acting on the pennies is stronger on the one near the edge than the one near the center.
Answer:
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
Explanation:
centripetal force = m ω² R
here m is mass , ω is angular velocity and R is distance of penny from centre
So this force depends upon R
penny on the outer edge will require greater centripetal force to move in circular path .
The centripetal force will be provided by frictional force of table which is same for both the coin . Hence the penny on the outer edge will slip off first the moment , frictional force reach its maximum value for it . But it will be sufficient to keep in balance the penny nearer to the centre .
It takes 525 J of work to compress a spring 25 cm. What is the force constant of the spring (in kN/m)?
Answer:
1.680kN/m
Explanation:
Work done by the spring is expressed as shown:
[tex]W = \frac{1}{2}ke^2[/tex] where:
k is the spring constant
e is the extension
Given
W = 525Joules
extension = 25cm = 0.25m
Substitute into the formula:
[tex]525 = \frac{1}{2}k(0.25)^{2} \\525 = \frac{0.0625k}{2}\\ 525 = 0.03125k\\k = \frac{525}{0.3125}\\k = 1680N/m\\k = 1.680kN/m[/tex]
Hence the force constant of the spring is 1.680kN/m
A spring has natural length 16 cm. A force of 3 N is required to holdthe spring compressed compressed to 11 cm. Find the amount ofwork instretching the spring from 17 cm to 19 cm.
Answer:
W = 0.012 J
Explanation:
For this exercise let's use Hooke's law to find the spring constant
F = K Δx
K = F / Δx
K = 3 / (0.16 - 0.11)
K = 60 N / m
Work is defined by
W = F. x = F x cos θ
in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1
W = ∫ F dx
W = k ∫ x dx
we integrate
W = k x² / 2
W = ½ k x²
let's calculate
W = ½ 60 (0.19 -0.17)²
W = 0.012 J
21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?
Answer:
hello your question is incomplete attached below is the complete question
21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Explanation:
The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Explain why atoms only emit certain wavelengths of light when they are excited. Check all that apply. Check all that apply. Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms are not quantized. When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. The energies of atoms are quantized.
Answer:
Explanation:
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. FALSE. The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.
The energies of atoms are not quantized. FALSE. The energies of the atoms are in specific levels.
When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. FALSE. During absorption, a specific wavelength of light is absorbed, not emmited.
Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. TRUE. Again, you can observe just the transition due the change of energy of an electron in the quantized energy level
When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. TRUE. The electron decreases its energy releasing a specific wavelength of light.
The energies of atoms are quantized. TRUE. In fact, the energy of all subatomic, atomic, and molecular particles is quantized.
The reason why atoms emit only specific wavelengths is because the energy levels in atoms are quantized.
Max Plank introduced the idea of quantization of energy in the early 1900s. He introduced the idea that energy can only take on certain specific values. This idea was later extended to atoms by Neils Bohr.
The following statements explain why atoms only emit certain wavelengths of light when they are excited;
When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms are quantized.Learn more: https://brainly.com/question/24381583
For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same. Describe how they are different.
Answer:
he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
Explanation:
In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.
In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.
From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.