Answer:
The displacement is [tex]A_r = 6.071 \ cm[/tex]
Explanation:
From the question we are told that
The initial displacement is [tex]A_o = 10 \ cm[/tex]
The displacement at the end of first oscillation is [tex]A_d = 9.05 \ cm[/tex]
Generally the damping constant of this damped oscillator is mathematically represented as
[tex]\eta = \frac{A_d}{A_o}[/tex]
substituting values
[tex]\eta = \frac{9.05}{10}[/tex]
[tex]\eta = 0.905[/tex]
The displacement after 4 more oscillation is mathematically represented as
[tex]A_r = \eta^4 * A_d[/tex]
substituting values
[tex]A_r = (0.905)^4 * (9.05)[/tex]
[tex]A_r = 6.071 \ cm[/tex]
Answer:
Displacement reached is 6.0708 cm
Explanation:
Formula for damping Constant "C"
[tex]C^n=\frac{A_2}{A_1}[/tex] where n=1,2,3,........n
Where:
[tex]A_2[/tex] is the displacement after first oscillation
[tex]A_1\\[/tex] is the initial Displacement
[tex]A_1=10\ cm\\A_2=9.05\ cm\\[/tex]
In our case, n=1.
[tex]C=\frac{9.05}{10}\\C=0.905[/tex]
After 4 more oscillation, n=4:
[tex]C^4=\frac{A_6}{A_2}[/tex]
Where:
[tex]A_6[/tex] is the final Displacement after 4 more oscillations.
[tex]A_6=(0.905)^4*(9.05)\\A_6=6.0708\ cm[/tex]
Displacement reached is 6.0708 cm
The maximum gauge pressure in a hydraulic system is 15 atm. What is the largest mass that could be lifted by this system if the diameter of the piston is 65 cm
Answer:
The maximum force that can be lifted by this system is 51,478.4 kg
Explanation:
Given;
maximum gauge pressure of the hydraulic system, Hp = 15 atm = 1.52 x 10⁶ N/m²
diameter of the piston, d = 65 cm = 0.65 m
The maximum gauge pressure of the piston is given as;
[tex]Hp = \frac{F}{A}[/tex]
Where;
F is the maximum force of the piston
A is the area of the piston
[tex]A = \pi (\frac{0.65}{2} )^2\\\\A = 0.3319 \ m^2[/tex]
F = Hp x A
F = 1.52 x 10⁶N/m² x 0.3319m²
F = 504488 N
Force is given as;
F = mg
m = F/g
m = 504488/9.8
m = 51,478.4 kg
Therefore, the maximum force that can be lifted by this system is 51,478.4 kg
1. Why do you see colors when you look at reflected light from a CD or DVD disk, or when you look at a soap bubble or oil film on water?
2. What do you think causes the colors on the artwork panels on the side of HLS2 (Health Sciences building) which change with time of day and the angle from which you view them?
Explanation:
1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.
2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.
Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.
We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.
These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.
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A long bar slides on two contact points and is in motion with velocity ν. A steady, uniform, magnetic field B is present. The induced current through resistor R is:
Answer:
The induced current in the resistor is I = BLv/R
Explanation:
The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by
ε = BLv.
Now, the current I in the resistor is given by
I = ε/R where ε = induced emf in circuit and R = resistance of resistor.
So, the current I = ε/R.
substituting the value of ε the induced emf, we have
I = ε/R
I = BLv/R
So, the induced current through the resistor is given by I = BLv/R
hat a 15 kg body is pulled along a horizontal fictional table by a force of 4N what is the acceleration of the body
Answer:
Acceleration of the body is:
[tex]a=0.27\,\,m/s^2[/tex]
Explanation:
Use Newton's second Law to solve for the acceleration:
[tex]F=m\,\,a\\a=\frac{F}{m} \\a=\frac{4\,N}{15\,\,kg} \\a=0.27\,\,m/s^2[/tex]
A loop of wire in the shape of a rectangle rotates with a frequency of 143 rotation per minute in an applied magnetic field of magnitude 2 T. Assume the magnetic field is uniform. The area of the loop is A = 2 cm2 and the total resistance in the circuit is 7 Ω.
1. Find the maximum induced emf.
e m fmax =
2. Find the maximum current through the bulb.
Imax
Answer:
1. e m fmax = 0.00598 Volt
2. Imax = 0.000854 Amp
Explanation:
1. Find the maximum induced emf.
e m fmax =
Given that e m fmax = N*A*B*w
N = 1
A = 2 cm^2 = 0.0002 m^2
f = 143 rotation per minute = 143/min
f = (143/min) * (1 min/60 sec) = 2.38/sec
w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec
B = 2T
e m fmax = N*A*B*w
e m fmax = 1 * 0.0002 * 2 * 14.95
e m fmax = 0.00598 Volt.
2. Find the maximum current through the bulb.
Imax = e m fmax / R
Where R is the total resistance in the circuit is 7 Ω.
Imax = 0.00598/7 = 0.000854 Amp.
Imax = 0.000854 Amp
1) The maximum induced EMF in the loop of wire is; EMF_max = 9.52 × 10^(-4) V
2) The maximum current through the bulb is;
I_max = 1.36 × 10^(-4) A
We are given;
Number of turns; N = 1
Magnitude of magnetic field; B = 2 T
Area; A = 2 cm² = 0.0002 m²
Angular frequency; ω = 143 /min = 2.38 /s
Resistance; R = 7 Ω.
1) Formula for maximum induced EMF is;
EMF_max = NAωB
Plugging in the relevant values gives;
EMF_max = 1 × 0.0002 × 2.38 × 2
EMF_max = 9.52 × 10^(-4) V
2) Formula for maximum current through the bulb is given as;
I_max = EMF_max/R
Plugging in the relevant values;
I_max = (9.52 × 10^(-4))/7
I_max = 1.36 × 10^(-4) A
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Consider a bus traveling to the west (negative x direction) that begins to slow down as it approaches a traffic light. Which statement concerning its acceleration in the x direction is correct
Complete question is;
Consider a bus traveling to the west (-x direction) begins to slow down as it approaches a traffic light. Which statement concerning its acceleration in the x direction is correct
a) The bus is decelerating and its acceleration is positive.
b) The bus is decelerating, and its acceleration is negative.
c) The acceleration is zero.
d) A statement cannot be made using the information given.
Answer:
Option A - bus is decelerating and acceleration is positive.
Explanation:
We are told that the bus is travelling in (negative x direction) and begins to slow down. Since the bus is slowing down, it means that the bus is undergoing a negative acceleration which is called deceleration.
Thus, the bus is decelerating.
Since it is moving in the negative x-axis, it means acceleration is now; -(-a) which gives +a.
Thus, bus is decelerating and acceleration is positive.
On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance does the bus cover in this time frame
Answer:
Explanation:
initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s .
acceleration = (v - u) / t
= (36 - 20) / 4
a = 4 m / s ²
from the formula
v² - u² = 2 a s , s is distance covered .
putting the values
36² - 20² = 2 x 4 x s
1296 - 400 = 8 x s
s = 112 m .
Answer:112
Explanation:
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45 yards.He’s thinking of running in a 5km race in a few months,but doesn’t know if he can maintain his maximum speed for the entire 5 km.Can you help him determine how far he can?
Answer:
Kindly check explanation
Explanation:
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
(1093.613 * x) = 45
x = 45 / 1093.613
x = 0.0411480 km
Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.
Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
The calculation is as follows;
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
[tex](1093.613 \times x) = 45[/tex]
[tex]x = 45 \div 1093.613[/tex]
x = 0.0411480 km
here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.
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You have three resistors: R1 = 1.00 Ω, R2 = 2.00 Ω, and R3 = 4.00 Ω in parallel. Find the equivalent resistance for the combination
Answer:
4 / 7
Explanation:
1/total resistance = 1/1 + 1/2 + 1/4
= 1¾
total resistance = 1 ÷ 1¾
= 4/7
Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom
Answer:
They both have the same total K.E at the bottom
Explanation:
This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved
So
K1 + U1 = K2 + U2
But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference
thus
K1 = 0 , U2 = 0
ΔK = ΔU = m g. h
But if the two inclines have the same height, and both hoops have the same mass m,
So difference in kinetic energy, must be the same for both hoops.
1.2miles=__________km
Answer:
1.931 kilometres is the answer of 1.2 miles
Answer and Explanation:
1 mile = 1.609 km
Set up a fraction to cancel the miles to get the kilometers.
[tex]\frac{1.2mi}{?km} *\frac{1.609}{1mi} = 1.9308km[/tex] <- This is the answer.
#teamtrees #PAW (Plant And Water)
how does a system naturally change over time
Answer:
The movement of energy and matter in a system differs from one system to another. On the other hand, in open system both the matter and energy move into and out of the system. Therefore, matter and energy in a system naturally change over time will decrease in entropy.
Explanation:
Answer:
Decrease in entropy
Explanation:
Various systems which exist in nature possess energy and matter that move through these system continuously. The movement of energy and matter in a system differs from one system to another.
In a closed system for example, only energy flows in and out of the system while matter does not enter or leave the system.
On the other hand, in open system both the matter and energy move into and out of the system.
A planar electromagnetic wave is propagating in the x direction. At a certain point P and at a given instant, the magnitude of the electric field of the wave is 0.082 V/m . What is the magnetic vector of the wave at point P at that instant?
A) (0.27 nT)k
B) (-0.27 nT)k
C) (0.27 nTİ
D) (6.8 nT)k
E) (-6.8 nT))
Answer:
b
Explanation:
A doctor counts 68 heartbeats in 1.0 minute. What are the corresponding period and frequency of the heart rhythm
Answer:
[tex]f=1.13s^{-1}=1.13Hz[/tex]
Explanation:
Hello,
In this case, a frequency stands for a rate in which some action is done per unit of time. In this case, for the heartbeat, since 68 actions (heartbeats) occur in 1.0, the frequency turns out:
[tex]f=\frac{68}{1.0min}=68min^{-1}[/tex]
Or as most commonly used in Hz ([tex]s^{-1}[/tex]):
[tex]f=68\frac{1}{min} *\frac{1min}{60s}=1.13s^{-1}=1.13Hz[/tex]
Best regards.
A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire?
Answer:
The potential will be Va/b
Explanation:
So Let sphere A charged Q to potential V.
so, V= KQ/a. ....(1
Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.
therefore, potential will be ,
V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]
If the rods with diameters and lengths listed below are made of the same material, which will undergo the largest percentage length change given the same applied force along its length?a. d, 3L b. 3d, L c. 2d, 2L d. 4d, L
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½ )/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Một thang máy chuyển động thẳng đứng hướng xuống dưới chậm dần đều với gia tốc a= -4m/s2. Trần thang máy treo một vật nhỏ bằng một sợi dây mảnh, khoảng cách từ vật tới sàn thang máy h= 2m. Thang máy đang chuyển động thì dây đứt. Tính thời gian từ lúc dây đứt đến khi vật chạm sàn thang máy ? Lấy g= 10m/s2
Answer:
ejkshdkashalsflasfaksg
Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen if the spacing between the first and second dark fringes is to be 4.2 mm?
Answer:
The distance is [tex]D = 2.6 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 476.1 \ nm = 476.1 *10^{-9} \ m[/tex]
The distance between the slit is [tex]d = 0.29 \ mm = 0.29 *10^{-3} \ m[/tex]
The between the first and second dark fringes is [tex]y = 4.2 \ mm = 4.2 *10^{-3} \ m[/tex]
Generally fringe width is mathematically represented as
[tex]y = \frac{\lambda * D }{d}[/tex]
Where D is the distance of the slit to the screen
Hence
[tex]D = \frac{y * d}{\lambda }[/tex]
substituting values
[tex]D = \frac{ 4.2 *10^{-3} * 0.29 *10^{-3}}{ 476.1 *10^{-9} }[/tex]
[tex]D = 2.6 \ m[/tex]
A pair of narrow, parallel slits separated by 0.230 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.50 m away from the plane of the parallel slits.
A) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B) Calculate the distance between the first and second dark bands in the interference pattern.
Answer:
A) y = 3.56 mm
B) y = 3.56 mm
Explanation:
A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:
[tex] y = \frac{m\lambda L}{d} [/tex]
Where:
λ: is the wavelength = 546.1 nm
m: is first bright region = 1
L: is the distance between the screen and the plane of the parallel slits = 1.50 m
d: is the separation between the slits = 0.230 mm
[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
B) The distance between the first and second dark bands is:
[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]
Where:
[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]
[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
I hope it helps you!
a baseball is given an initial velocity with magnitude v at the angle beta above the surface of an incline which in turn inclined at angle teta above horizontal calculate the distance measured along incline from the launch point to where the baseball strike the incline
Explanation:
The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 θ i 2 g .
A 4.00-Ω resistor, an 8.00-Ω resistor, and a 24.0-Ω resistor are connected together. (a) What is the maximum resistance that can be produced using all three resistors? (b) What is the minimum resistance that can be produced using all three resistors? (c) How would you connect these three resistors to obtain a resistance of 10.0 Ω? (d) How would you connect these three resistors to obtain a resistance of 8.00 Ω?
Answer:a) 4+8+24=36
B) 1/4+1/8+1/24=10
C) yu will connect them in parallel connection.
D) you will connect two in parallel then the remaining one in series to the ons connected in parallel.
Explanation:
(a)The maximum resistance that can be produced using all three resistors will be 36 ohms.
(b)The minimum resistance that can be produced using all three resistors will be 10 ohms.
(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.
(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.
What is resistance?Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.
The maximum resistance that can be produced using all three resistors is obtained by adding all the given resistance;
[tex]\rm R_{max}=(4 +8+24 )\ ohms \\\\ R_{max}=36 \ ohms[/tex]
The minimum resistance that can be produced using all three resistors is obtained when connected in the parallel.
[tex]\rm R_{min}=\frac{1}{4} +\frac{1}{8} +\frac{1}{24} \\\\ R_{min}=10 \ ohm[/tex]
(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.
(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.
Hence,the maximum resistance that can be produced using all three resistors will be 36 ohms.
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What is the maximum wavelength of incident light for which photoelectrons will be released from gallium
Answer:
292 nm
Explanation:
The work function of gallium ∅ = 94.25 eV = 6.81 x 10^-19 J
at maximum wavelength, the energy of the photons is equal to its work function
Energy of the electron = hf
but hf = hc/λ
where h is the planck's constant = 6.63 × 10-34 m^2 kg/s
c is the speed of light = 3 x 10^8 m/s
λ is the wavelength that this occurs, which is the maximum wavelength
Equating, we have
hc/λ = ∅
substituting, we have
(6.63 × 10-34 x 3 x 10^8)/λ = 6.81 x 10^-19
(1.989 x 10^-25)/(6.81 x 10^-19) = λ
λ = 292.07 x 10^-9 = 292 nm
Discuss the phase change condition due to reflection of light from a surface. Summarize equations of interference for thin film.
Answer:
if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º
Explanation:
When a ray of light falls on a surface if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º this can be explained by Newton's third law, the light when arriving pushes the atoms of the medium that is more dense, and these atoms respond with a force of equal magnitude, but in the opposite direction.
When the fractional index is lower than that of the medium where the reflacted beam travels, notice a change in phase.
Also, when light penetrates the medium, it modifies its wavelength
λ = λ₀ / n
We take these two aspects into account, the condition for contributory interference is
d sin θ = (m + 1/2) λ
for destructive interference we have
d sin θ = m λ
in general this phenomenon is observed at 90º
2 d = (m +1/2) λ° / n
2nd = (m + ½) λ₀
what is used to measure the amount of sunshine
Answer:
Campbell-Stokes sunshine recorders or modern sunshine sensors.
Explanation:
Answer:
modern sunshine sensorssunshine recordersCampbell-Stokes
A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine the height the ball will rise after the bounce. Show all your work.
Answer:
12.2 m
Explanation:
Given:
v₀ = 15.6 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy
Δy = 12.2 m
[tex] \LARGE{ \boxed{ \rm{ \green{Answer:}}}}[/tex]
Given,
The initial speed is 15.6 m/s The mass of the ball is 42g = 0.042kgFinding the initial kinetic energy,
[tex]\large{ \boxed{ \rm{K.E. = \frac{1}{2}m {v}^{2}}}}[/tex]
⇛ KE = (1/2)mv²
⇛ KE = (1/2)(0.042)(15.6)²
⇛ KE = 5.11 J
|| ⚡By conservation of energy, the potential energy at the highest point will also be 5.11 J, since there is no kinetic energy at the highest point because the ball is not moving (we neglect energy lost due to air resistance, heat, sound, etc.) ⚡||
So, we have:
[tex] \large{ \boxed{ \rm{P.E. = mgh}}}[/tex]
⇛ h = PE/(mg)
⇛ h = 5.11 J /(0.042 × 9.8)
⇛ h = 12.41 m
✏The ball will rise upto a height of 12.41 m
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Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.75
Earth years in its circular orbit around its sun, which is a star with a measured mass of 3.23×1030kg
. Find the radius of the exoplanet's orbit.
Answer:
[tex]r=4.24\times 10^{11}\ m[/tex]
Explanation:
Given that,
Orbital time period, T = 3.75 earth years
Mass of star, [tex]m=3.23\times 10^{30}\ kg[/tex]
We need to find the radius of the exoplanet's orbit. It is a concept of Kepler's third law of motion i.e.
[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]
r is the radius of the exoplanet's orbit.
Solving for r we get :
[tex]r=(\dfrac{T^2GM}{4\pi^2})^{1/3}[/tex]
We know that, [tex]1\ \text{earth year}=3.154\times 10^7\ \text{s}[/tex]
So,
[tex]r=(\dfrac{(3.75\times 3.154\times 10^7)^2\times 6.67\times 10^{-11}\times 3.23\times 10^{30}}{4\pi^2})^{1/3}\\\\r=4.24\times 10^{11}\ m[/tex]
So, the radius of the exoplanet's orbit is [tex]4.24\times 10^{11}\ m[/tex].
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C
Answer:
Option (c) : 20°C
Explanation:
[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]
T(final) = 500* 10 + 100*70/600 = 20°C
Rank these electromagnetic waves on the basisof their speed (in vacuum).
Rank from fastest to slowest. To rankitems as equivalent, overlap them.
yellow light
FM radio wave
green light
X-ray
AM radio wave
infrared wave
Answer:
All electromagnetic waves travel at the same speed in a vacuum
Explanation:
All the wave listed in the question are electromagnetic waves. The speed of electromagnetic waves (collectively called light) in a vacuum is fixed. Its value is 3×10^8 ms^-1. This is a constant for all electromagnetic waves irrespective of their frequency.
Hence for any electromagnetic wave, its speed is 3×10^8 ms^-1, this will be the common velocity of all the electromagnetic waves listed in the question in a vacuum thus we can not rank them according to speed.
A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about
Answer:
-50N
Explanation:
F=ma=m(Vf-Vi)/t
m=10kgVf=0m/sVi=10m/st=2sF=(10)(-10)/(2)=-50N
So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.
What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making