Answer:
Ultraviolet light from the sun.
Explanation:
This is an example of radiation.
Answer:
X-Ray
Explanation:
x-Ray is an example of radiation.
When light is either reflected or refracted, the quantity that does not change in either process is its
Answer:
Frequency
Explanation:
When waves travel from one medium to another, it is only the frequency of the wave that remains constant . when a wave is refracted at the boundary between two media, the wave will slow down and its wavelength decreases. The wave usually bends at the interface between the two media. The wavelength and speed of a wave may change at the boundary between two media but its frequency remains the same.
Hence the frequency of light is its only property that remains constant.
Which scientist proposed a mathematical solution for the wave nature of light?
Answer:
Explanation:
Christian Huygens
Light Is a Wave!
Then, in 1678, Dutch physicist Christian Huygens (1629 to 1695) established the wave theory of light and announced the Huygens' principle.
help... Please help!!!!!!!!!!!
Answer:
a) 6.8--5.10 thats equal 11.9
b) m=ris/run +10 equal 0.06/8 =7.5*10^-3
Suppose a certain laser can provide 82 TW of power in 1.1 ns pulses at a wavelength of 0.24 μm. How much energy is contained in a single pulse?
Answer:
The energy contained in a single pulse is 90,200 J.
Explanation:
Given;
power of the laser, P = 82 TW = 82 x 10¹² W
time taken by the laser to provide the power, t = 1.1 ns = 1.1 x 10⁻⁹ s
the wavelength of the laser, λ = 0.24 μm = 0.24 x 10⁻⁶ m
The energy contained in a single pulse is calculated as;
E = Pt
where;
P is the power of each laser
t is the time to generate the power
E = (82 x 10¹²)(1.1 x 10⁻⁹)
E = 90,200 J
Therefore, the energy contained in a single pulse is 90,200 J
What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round window
Water is draining from an inverted conical tank with base radius 8 m. If the water level goes down at 0.03 m/min, how fast is the water draining when the depth of the water is 6 m
Answer:
0.03/π m/min
Explanation:
See attached file pls
The roller coaster car reaches point A of the loop with speed of 20 m/s, which is increasing at the rate of 5 m/s2. Determine the magnitude of the acceleration at A if pA
Answer and Explanation:
Data provided as per the question is as follows
Speed at point A = 20 m/s
Acceleration at point C = [tex]5 m/s^2[/tex]
[tex]r_A = 25 m[/tex]
The calculation of the magnitude of the acceleration at A is shown below:-
Centripetal acceleration is
[tex]a_c = \frac{v^2}{r}[/tex]
now we will put the values into the above formula
= [tex]\frac{20^2}{25}[/tex]
After solving the above equation we will get
[tex]= 16 m/s^2[/tex]
Tangential acceleration is
[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]
Please help!
Much appreciated!
Answer:
F = 2.7×10¯⁶ N.
Explanation:
From the question given:
F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²
Thus we can obtain the value value of F by carrying the operation as follow:
F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²
F = 2.7648×10¯⁷ / 0.1024
F = 2.7×10¯⁶ N.
Therefore, the value of F is 2.7×10¯⁶ N.
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
a. 1J.
b. 0.50J.
c. 0.25J.
d. 0.
e. dependent upon the resistance of the inductor.
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
A 1.2-m length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x= 5.0m on x-axis.
a. 1.6 nt in the negative z direction
b. 1.6 nt in the positive z direction
c. 2.4 T in the positive z direction
d. 2.4 nt in the negative z direction
e. None of the above
Answer:
None of the above
Explanation:
The formula of the magnetic field of a point next to a wire with current is:
B = 2×10^(-7) × ( I /d)
I is the intensity of the current.
d is the distance between the wire and the point.
● B = 2*10^(-7) × (20/5) = 8 ×10^(-7) T
In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force
Answer:
E = VdB
Explanation:
This is because canceling the electric and magnetic force means
q.vd. B= we
E= Vd. B
A charge of 15 is moving with velocity of 6.2 x17 which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.
a. 06.0T.
b. 08.0T.
c. 07.0T.
d. 05.0 T.
Complete question:
A charge of 15C is moving with velocity of 6.2 x 10³ m/s which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.
a. 0.06 T
b. 0.08 T
c. 0.07 T
d. 0.05 T
Answer:
The magnitude of the magnetic field is 0.07 T.
Explanation:
Given;
magnitude of the charge, q = 15C
velocity of the charge, v = 6.2 x 10³ m/s
angle between the charge and the magnetic field, θ = 48°
the force on the particle, F = 4838 N
The magnitude of the magnetic field can be calculated by applying Lorentz force formula;
F = qvBsinθ
where;
B is the magnitude of the magnetic field
B = F / vqsinθ
B = (4838) / (6.2 x 10³ x 15 x sin48)
B = 0.07 T
Therefore, the magnitude of the magnetic field is 0.07 T.
Calculate the density of the following material.
1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³
Answer:
[tex]density \: = \frac{mass}{volume} [/tex]
1 / 5.587 is equal to 0.179 kg/m³
Hope it helps:)
Answer:
The answer is
0.179 kg/m³Explanation:
Density of a substance is given by
[tex]Density \: = \frac{mass}{volume} [/tex]
From the
mass = 1 kg
volume = 5.583 m³
Substitute the values into the above formula
We have
[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]
We have the final answer as
Density = 0.179 kg/m³Hope this helps you
A rectangular coil having N turns and measuring 15 cm by 25 cm is rotating in a uniform 1.6-T magnetic field with a frequency of 75 Hz. The rotation axis is perpendicular to the direction of the field. If the coil develops a sinusoidal emf of maximum value 56.9 V, what is the value of N?
A) 2
B) 4
C) 6
D) 8
E) 10
Answer:
A) 2
Explanation:
Given;
magnetic field of the coil, B = 1.6 T
frequency of the coil, f = 75 Hz
maximum emf developed in the coil, E = 56.9 V
area of the coil, A = 0.15 m x 0.25 m = 0.0375 m²
The maximum emf in the coil is given by;
E = NBAω
Where;
N is the number of turns
ω is the angular velocity = 2πf = 2 x 3.142 x 75 = 471.3 rad/s
N = E / BAω
N = 56.9 / (1.6 x 0.0375 x 471.3)
N = 2 turns
Therefore, the value of N is 2
A) 2
Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first dark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit
Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×[tex]10^{-9}[/tex] m)
d = (3.30 × 563 ×[tex]10^{-9}[/tex] ) ÷ (0.0047)
= 1.8579 × [tex]10^{-6}[/tex] ÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).
Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz
Answer:
380 kHz
Explanation:
The speed of sound is taken as 1500 m/s
The length of the fetus is 1.6 cm long
The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.
For this 1.6 cm baby, the wavelength must not exceed
λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =
0.4 cm = 0.004 m this is the wavelength of the required ultrasonic sound.
we know that
v = λf
where v is the speed of a wave
λ is the wavelength of the wave
f is the frequency of the wave
f = v/λ
substituting values, we have
f = 1500/0.004 = 375000 Hz
==> 375000/1000 = 375 kHz ≅ 380 kHz
Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?
Answer:
Fluoroscopy
Explanation:
A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.
Answer:
the new length is 17.435cm
Explanation:
the new length is 17.435cm
pls give brainliest
The new length of aluminum rod is 17.435 cm.
The linear expansion coefficient is given as,
[tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]
Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.
and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]
Substitute, [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]
[tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]
Hence, The new length of aluminum rod is 17.435 cm.
Learn more:
https://brainly.com/question/19495810
An electron has an initial velocity to the south but is observed to curve upward as the result of a magnetic field. This magnetic field must have a component:___________
a) north
b) upwards
c) downwards
d) east
e) west
Answer:
e) west
Explanation:
According to Lorentz left hand rule, the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.
In this case, if we point the thumb to the south (towards your body), with the palm facing up, then the fingers will point west.
Suppose a 1300 kg car is traveling around a circular curve in a road at a constant
9.0 m/sec. If the curve in the road has a radius of 25 m, then what is the
magnitude of the unbalanced force that steers the car out of its natural straight-
line path?
Answer:
F = 4212 N
Explanation:
Given that,
Mass of a car, m = 1300 kg
Speed of car on the road is 9 m/s
Radius of curve, r = 25 m
We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N[/tex]
So, the force has a magnitude of 4212 N
Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How far away (in m) is the planet Venus if the echo time is 900 s? m (b) What is the echo time (in µs) for a car 80.0 m from a Highway Patrol radar unit? µs (c) How accurately (in nanoseconds) must you be able to measure the echo time to an airplane 12.0 km away to determine its distance within 11.5 m? ns
Answer:
a) 1.35 x 10^11 m
b) 0.53 µs
c) 8 ns
Explanation:
Radar involves the use of radio wave which has speed c = 3 x 10^8 m/s
a) for 900 s,
The distance for a round trip = v x t
==> (3 x 10^8) x 900 = 2.7 x 10^11 m
The distance of Venus is half this round trip distance = (2.7 x 10^11)/2 = 1.35 x 10^11 m
b) for a 80.0 m distance of the car from the radar source, the radar will travel a total distance of
d = 2 x 80 = 160 m
the time taken = d/c = 160/(3 x 10^8) = 5.3 x 10^-7 s = 0.53 µs
c) accuracy in distance Δd = 11.5 m
Δt = accuracy in time = Δd/c = 11.5/(3 x 10^8) = 3.8 x 10^-8 = 38 ns
You need to repair a broken fence in your yard. The hole in your fence is
around 3 meters in length and for whatever reason, the store you go to
has oddly specific width 20cm wood. Each plank of wood costs $16.20,
how much will it cost to repair your fence? (Hint: 1 meter = 100 cm) *
Answer:
cost = $ 243.00
Explanation:
This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m
#_tables = 3 m (1 / 0.20 m)
#_tables = 15 tables
Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?
Cost = 15 (16.20 / 1)
cost = $ 243.00
Two identical planets orbit a star in concentric circular orbits in the star's equatorial plane. Of the two, the planet that is farther from the star must have
Answer:
The planet that is farther from the star must have a time period greater.
Explanation:
We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} [/tex] (1)
Where:
r: is the radius of the circular orbit of the planet and the star
T: is the period
G: is the gravitational constant
M: is the mass of the planet
From equation (1) we have:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} = k*r^{3/2} [/tex] (2)
Where k is a constant
From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.
I hope it helps you!
If mirror M2 in a Michelson interferometer is moved through 0.233 mm, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?
Answer:
The wavelength is [tex]\lambda = 589 nm[/tex]
Explanation:
From the question we are told that
The distance of the mirror shift is [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]
The number of fringe shift is n = 792
Generally the wavelength producing this fringes is mathematically represented as
[tex]\lambda = \frac{ 2 * k }{ n }[/tex]
substituting values
[tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]
[tex]\lambda = 5.885 *10^{-7} \ m[/tex]
[tex]\lambda = 589 nm[/tex]
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Answer:
the two ice skater have the same momentum but the are in different directions.
Paula will have a greater speed than Ricardo after the push-off.
Explanation:
Given that:
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off
The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.
So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.
Momentum is the product of mass and velocity.
SO, from the information given:
Let represent the mass of Paula with [tex]m_{Pa}[/tex] and its initial velocity with [tex]u_{Pa}[/tex]
Let represent the mass of Ricardo with [tex]m_{Ri}[/tex] and its initial velocity with [tex]u_{Ri}[/tex]
At rest ;
their velocities will be zero, i.e
[tex]u_{Pa}[/tex] = [tex]u_{Ri}[/tex] = 0
The initial momentum for this process can be represented as :
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = 0
after push off from each other then their final velocity will be [tex]v_{Pa}[/tex] and [tex]v_{Ri}[/tex]
The we can say their final momentum is:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
Using the law of conservation of momentum as states earlier.
Initial momentum = final momentum = 0
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Since the initial velocities are stating at rest then ; u = 0
[tex]m_{Pa}[/tex](0) + [tex]m_{Pa}[/tex](0) = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = - [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Given that Ricardo weighs more than Paula
So [tex]m_{Ri} > m_{Pa}[/tex] ;
Then [tex]\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}[/tex]
The magnitude of their momentum which is a product of mass and velocity can now be expressed as:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
The ratio is
[tex]\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1[/tex]
[tex]v_{Pa} >v_{Ri}[/tex]
Therefore, Paula will have a greater speed than Ricardo after the push-off.
(A) Both the skaters have the same magnitude of momentum.
(B) Paula has greater speed after push-off.
Conservation of momentum:Given that two skaters Paula and Ricardo are initially at rest.
Ricardo weighs more than Paula.
Let us assume that the mass of Ricardo is M, and the mass of Paula is m.
Let their final velocities be V and v respectively.
(A) Initially, both are at rest.
So the initial momentum of Paula and Ricardo is zero.
According to the law of conservation of momentum, the final momentum of the system must be equal to the initial momentum of the system.
Initial momentum = final momentum
0 = MV + mv
MV = -mv
So, both of them have the same magnitude of momentum, but in opposite directions.
(B) If we compare the magnitude of the momentum of Paula and Ricardo, then:
MV = mv
M/m = v/V
Now, we know that M>m
so, M/m > 1
therefore:
v/V > 1
v > V
So, Paula has greater speed.
Learn more about conservation of momentum:
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A skull believed to belong to an ancient human being has a carbon-14 decay rate of 5.4 disintegrations per minute per gram of carbon (5.4 dis/min*gC). If living organisms have a decay rate of 15.3 dis/min*gC, how old is this skull
Answer:
9.43*10^3 year
Explanation:
For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation
To start with, we use the formula
t(half) = In 2/k,
if we make k the subject of formula, we have
k = in 2/t(half), now we substitute for the values
k = in 2 / 5730
k = 1.21*10^-4 yr^-1
In(A/A•) = -kt, on rearranging, we find out that
t = -1/k * In(A/A•)
The next step is to substitite the values for each into the equation, giving us
t = -1/1.21*10^-4 * In(5.4/15.3)
t = -1/1.21*10^-4 * -1.1041
t = 0.943*10^4 year
"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"
Answer:
A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if
the dispersion is great
A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 7.4 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?
Answer:
The value is [tex]\mu = 0.76[/tex]
Explanation:
From the question we are told that
The acceleration is [tex]a = 7.4 \ m /s^2[/tex]
Generally the force by which the truck bed (truck) is moving with is mathematically represented as
[tex]F = ma[/tex]
Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet is equal the force by which the the truck bed is moving with that is
[tex]F_f = F[/tex]
Here [tex]F_f[/tex] is the frictional force which is mathematically represented as
[tex]F_f = \mu * m * g[/tex]
substituting into above equation
[tex]\mu * m * g = ma[/tex]
=> [tex]\mu = \frac{a}{g}[/tex]
substituting values
[tex]\mu = \frac{ 7.4 }{ 9.8}[/tex]
[tex]\mu = 0.76[/tex]
Consult Interactive Solution 27.18 to review a model for solving this problem. A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in visible light with wavelength 653 nm in vacuum. Assuming that the visible spectrum extends from 380 to 750 nm, what is the longest visible wavelength (in vacuum) for which the film will appear bright due to constructive interference
Answer:
Explanation:
In the given case for destructive interference , the condition is,
path difference = (2n+1)λ /2 where n is an integer and λ is wavelength
2 μ d = (2n+1)λ /2
Putting λ = 653 nm
for minimum thickness n = 0
2 μ d = 653 / 2 nm
= 326.5 nm
For constructive interference the condition is
2 μ d = n λ₁
326.5 nm = n λ₁
λ₁ = 326.5 / n
For n = 1
λ₁ = 326.5 nm ,
or , 326.5nm .
Longest wavelength possible is 326.5
CAN SOMEONE HELP ME PLEASE ITS INTEGRATED SCIENCE AND I AM STUCK
Answer:
[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]
Explanation:
Two forces are acting on the object.
Subtracting 2 N from both forces.
2 N → Object ← 5 N
- 2 N - 2N
0 N → Object ← 3 N
The force 3 N is pushing the object to the left side.
The mass of the object is 10 kg.
Applying formula for acceleration (Newton’s Second Law of Motion).
a = F/m
a = 3/10
a = 0.3
