Answer:
Δx = distance between central bright fringe and first bright fringe = 19.85 x 10⁻⁴ m = 1.985 mm
distance between central bright fringe and second dark fringe = 2.978 mm
Explanation:
We have the following data:
λ = wavelength of light = 641 nm = 6.41 x 10⁷ m
L = Distance of Screen from slits = 3.19 m
d = slit separation = 1.03 mm = 1.03 x 10⁻³ m
Δx = distance between consecutive bright fringes = fringe spacing = ?
Using formula:
[tex]\Delta x = \frac{\lambda L}{d}\\\\\Delta x = \frac{(6.41\ x\ 10^{-7}\ m)(3.19\ m)}{1.03\ x\ 10^{-3}\ m}[/tex]
Δx = distance between central bright fringe and first bright fringe = 19.85 x 10⁻⁴ m = 1.985 mm
distance between central bright fringe and second dark fringe = 1.5Δx
distance between central bright fringe and second dark fringe = (1.5)(1.985 mm)
distance between central bright fringe and second dark fringe = 2.978 mm
What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)
total lunar eclipse
total solar eclipse
partial lunar eclipse
partial solar eclipse
Answer: I'm not sure, but I think it would be a total lunar eclipse
When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.
What is partial lunar eclipse?A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.
During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.
A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.
The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.
Thus, the correct option is C.
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A child of mass 28 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and the speed is 7 m/s. At this instant the cord is 3.30 m long. (Take the +x direction to be horizontal and to the right, the +y direction upward and +z direction to be out of the page.)
a. At this instant, what is the magnitude of the rate of change of the child's momentum?
b. At this instant, what is the (vector) net force acting on the child?
Answer:
Explanation:
(A)
Force is said to be the rate of change of momentum.
At the instant, the parallel component of the change of momentum is zero. SInce centripetal force always acts towards the center, hence the direction upwards ( along the positive y-axis).
[tex]\dfrac{dp}{dt} = F(+\hat y)[/tex]
[tex]= \dfrac{mv^2}{r}(+\hat y) \ \\ \\ = \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]
(B)
The net force acting on the child is:
[tex]F_{net} = \dfrac{mv^2}{r}( + \hat y) \\ \\[/tex]
[tex]= \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]
what is the distance a train can travel if its speed is 20mph over a time of 5.6 hours (show all 3 steps)
Answer:
distance = 112 miles
Explanation:
its 12 miles every 0.6 in a hour
A physics professor wants to demonstrate the large size of the henry unit. On the outside of a 16-cm-diameter plastic hollow tube, she wants to wind an air-filled solenoid with self-inductance of 1.0 H using copper wire with a 0.79-mm diameter. The solenoid is to be tightly wound with each turn touching its neighbor (the wire has a thin insulating layer on its surface so the neighboring turns are not in electrical contact).
Required:
a. How long will the plastic tube need to be?
b. How many kilometers of copper wire will be required?
c. What will be the resistance of this solenoid?
Answer:
a) the plastic tube need to be 24.7 m long
b) the kilometer of copper wire required is 15.7
c) the resistance of this solenoid is 5538.8 2 ohms
Explanation:
Given the data in the question;
we determine the length of the plastic tube. assuming the solenoid is long.
the self inductance of a long solenoid is;
L = μ₀n²πr²l
μ₀ = 4π × 10⁻⁷ T-m/A
where
n = number of turns per unit length
r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)
l = length of the solenoid or the length of the plastic tube
we find n = number of turns per unit length
given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter
number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter
So from our previous formula, we find l
L = μ₀n²πr²l
we substitute
1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)
1 = 0.04048 × l
l = 1 / 0.04048
l = 24.7 m
Therefore, the plastic tube need to be 24.7 m long
b)
n = number of turns per unit length = 1265.8 turns/metre
so, the length of the plastic tube over which the copper wire is to be wound,
number of turns of copper wire required = n × l
= 1265.8 turns/meter × 24.7 m
= 31,265.26 turns
Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.
so for each turn length of copper wire required = 2π × r
= 2π × 0.08 m
= 0.5026548 m
So copper wire required for 31,265.26 turns will be;
⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km
Therefore, the kilometer of copper wire required is 15.7
c)
p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m
Resistance = pl/a
where l is length of copper wire, a is cross sectional area;
diameter of copper wire is 0.79-mm
radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m
area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²
Resistance = pl/a
we substitute
Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]
Resistance = 5538.8 2 ohms
Therefore, the resistance of this solenoid is 5538.8 2 ohms
What happens in the crushing can experiment?
Explanation:
When hot water is poured on the can in a bucket of cold water, the can crushes off means it gets unshaped
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Which statements explain the relationship between mass, velocity and kinetic energy? * Select the TWO (2) that apply.*
Consider the equation to calculate kinetic energy: KE = ½mv2
Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.
Kinetic energy decreases if either the mass or the velocity of the object increases or if both increase.
Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease.
Kinetic energy increases if either the mass or the velocity of the object decreases or if both decrease.
Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease. and Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.
Explanation:
These statements explain the relationship between mass, velocity, and kinetic energy .i.e KE = ½mv², (3) Kinetic energy decreases if either the mass or the velocity of the object decreases or if b decrease. (4)Kinetic energy increases if either the mass or the velocity of the object decreases or if both decrease.
What is kinetic energy?When a body has a mass of m kg and travels a velocity of v m/s then the kinetic energy is given by,
KE=1/2mv²
In the above relation, it is clearly shown that,
Kinetic energy is directly proportional to the mass of the body,
i.e
KE∝m
and also Kinetic energy directly varies with the square of the velocity of the body.
I.e
KE∝v²
Therefore, From the given expression we can conclude that when mass and velocity increased either one of them or both, the kinetic energy also gets increased.
And when mass and velocity decreased either one of them or both, the kinetic energy also gets decreased. Hence options (3) and (4) are correct.
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A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude is 0.45 m, what is the frequency of the oscillation?
Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s[/tex]
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.145 kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 14.5 m/s , shatters the glass as it passes through, and leaves the window at 10.9 m/s with no change of direction. What is the direction of the impulse that the glass imparts to the baseball
Answer:
J = -0.522 m/s
Explanation:
Given that,
The mass of the baseball, m = 0.145 kg
Initial velocity, u = 14.5 m/s
Final velocity, v = 10.9 m/s
aWe need to find the direction of the impulse that the glass imparts to the baseball. Impulse is equal to the change in momentum such that,
[tex]J=m(v-u)[/tex]
Substitute all the values,
[tex]J=0.145\times (10.9-14.5)\\\\=-0.522\ kg-m/s[/tex]
The direction of impulse is opposite to the direction of velocity.
How much energy would be required to move the earth into a circular orbit with a radius 2.0 kmkm larger than its current radius
Answer:
[tex]3.52\times 10^{25}\ \text{J}[/tex]
Explanation:
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]
m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]
[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]
[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]
Energy required is given by
[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]
The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].
A camper hears her echo off a
hillside 1.52 s after yelling. How
far is she from the hill?
Answer:
261
Explanation:
Acellus
Even in the most advanced circuits, we cannot oscillate electrons back and forth at that rate through wires. But we can oscillate charges back and forth quickly enough to broadcast TV using radio wave signals. At what frequency do the electronics at the TV station need to have the charges oscillate back and forth on a TV broadcast antenna to transmit a typical TV signal (say a radio wave transmission signal with a wavelength of 1 meter)
Answer:
the oscillations of the electrons must be in the 10⁸ Hz = 100 MHz range
Explanation:
The speed of a wave of radio, television, light, heat, all are manifestations of electromagnetic waves that are oscillations of electric and magnetic fields that support each other, the speed of all these waves is the same and the vacuum is equal to c = 3 108 m / s
All waves have a relationship between the speed of the wave, its frequency and wavelength
c = λ f
f = c /λ
for this case lam = 1 m
f = 3 10⁸/1
f = 3 10⁸ Hz
the oscillations of the electrons must be in the MHz range
It should be clarified that the speed of light in air is a little lower
n = c / v
v = c / n
the refractive index of vacuum is n = 1 and the refractive index of air is n = 1.000002
At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?
Answer:
8.45 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 90 Kg
Initial velocity (u) = 13 m/s
Final velocity (v) = 0 m/s
Height (h) =?
NOTE: Acceleration due to gravity (g) = 10 m/s²
The height of the hill can be obtained as follow:
v² = u² – 2gh (since the cart is going against gravity)
0² = 13² – (2 × 10 × h)
0 = 169 – 20h
Rearrange
20h = 169
Divide both side by 20
h = 169/20
h = 8.45 m
Therefore, the height of the hill is 8.45 m
Which is an example of a noncontract force?
(A) elastic force
(B) Normal force
(C)Applied force
(D) Electric Force
Answer:
electric force
Explanation:
its a contact and noncontact force
(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?
Answer:
a = 72 m/s^2
Explanation:
Since the object is moving in circle, the acceleration points TOWARDS the centre of the circle (it is centripetal acceleration)
a = v^2 / r
a = 60^2 / 50
a = 72 m/s^2
Would sound travel faster in an oven or a freezer?
Answer:
An Oven
Explanation:
The heat is higher, so it moves faster. Shile in a freezer the particles are extremely slow!
A baseball with a mass of 142 g is thrown horizontally with a speed of 38.6 m/s (86 mi/h) at a bat. The ball is in contact with the bat for 1.00 ms and then travels straight back at a speed of 46.1 m/s (103 mi/h). Determine the average force (in N) exerted on the ball by the bat. Neglect the weight of the ball (it is much smaller than the force of the bat) and choose the direction of the incoming ball to be positive. (Indicate the direction with the sign of your answer.)
Answer:
-11.600N
Explanation:
lol
Find all of those elements on the periodic table
Answer:
hydrogen, helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, argon, potassium, calcium, scandium, titanium, vanadium, chromium, maganese, iron, cobalt, nickel, copper, zinnc, gallium, germanium, aresnic, selenium, bromine, krypton, rubidium, strongtium, yttrium, zirconium, niobium, molybdenum, technetium, ruthenium, rhodium, palladium, silver, cadmium, indium, tin, atimony, tellurium, iodine, xenon, cesium, barium, lanthanum, cerium, praseodyumium, neodymim, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, luteium, hafnium, tantalum, tungsten, rhenium, osmium, irdium, platinum, gold, mercury, thallium, lead bismuth, polnium, astatine, radon, rancium, radium, actinum, thorium, protactinium
Explanation:
a sharp image is formed when light reflects from a
Answer:
Regular reflection
Explanation:
Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.
i hope this helps a bit.
According to the context, a sharp image is formed when light reflects from a regular reflection.
What is regular reflection?It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.
This reflection of light happens when the angles that the two rays determine with the surface are equal.
Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.
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Water enter the horizontal, circular cross-sectional, sudden-contraction nozzle sketched below at section (1) with a uniformly distributed velocity of 30 ft/s and a pressure of 80 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determinethe axial component of the anchoring force required to hold the contraction in place.
This question is incomplete, the missing image is uploaded along this answer.
Answer:
the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
Explanation:
Given the data in the question and as illustrated in the image below;
first we calculate the area at section 1
A₁ = (πD²)/4
we substitute
A₁ = (π(3 in)²)/4
A₁ = 7.06858 in²
we know that; 1 ft = 12 in
A₁ = ( 7.06858 / (12²) ) ft²
A₁ = ( 7.06858 / 144 ) ft²
A₁ = 0.0491 ft²
now, we write the elation for area at section 2
A₂ = πd²/4
here, d is the diameter at section 2
next, we use the conservation of mass equation between the two section;
m" = pV₁A₁ = pV₂A₂
we calculate the mass flow rate;
m" = pV₁A₁
= (1.94[tex]\frac{slug}{ft^2}[/tex]) × 30[tex]\frac{ft}{s}[/tex] × 0.0491 ft²
= 2.8576 slug/s
Now, Apply the linear momentum along the horizontal direction for the control volume between 1 - 2
-pV₁A₁V₁ = pV₂A₂V₂ = P₁A₁ - F[tex]_A[/tex] - P₂A₂
m"( V₂ - V₁ ) = P₁A₁ - F[tex]_A[/tex] - P₂A₂
F[tex]_A[/tex] = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )
we substitute
F[tex]_A[/tex] = ((80×[tex]\frac{144 in^2}{1 ft^2}[/tex])×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s
F[tex]_A[/tex] = 565.632 - 0 - 200.032
F[tex]_A[/tex] = 565.632 - 200.032
F[tex]_A[/tex] = 365.6 lb
Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
discuss two reasons why people find transition between school and university
Answer:
Is that your answer
How can Newton’s laws of motion and the law of gravitation predict the motion of an object, and how can frames of reference be used to describe that motion?
Cells use nutrients and oxygen to supply the body with the energy it needs. What three-body systems are working together in this situation?
A
nervous, digestive, and circulatory systems
B
digestive, circulatory, and excretory systems
C
circulatory, immune, and respiratory systems
D
digestive, respiratory, and circulatory systems
Answer:
The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.
Explanation:
so the answer is D
Select four of the following that would increase the magnetic field of an electromagnet
Answer:
The correct answers are: A, C, D, E
Explanation:
The magnetic field is a solenoid is given by
B = μ₀ [tex]\frac{N}{L}[/tex] I
where N is the number of turns, I the current and L length of the solenoid.
Using this equation let's examine the different responses to permute increasing the magnetic field
A) True. a thicker wire decreases the resistance and the current can increase the system.
B) False. If there is no voltage source there is no current in the system
C) True. the field is proportional to the number of turns
D) True. the magnetic moments of the core align with the field increasing its value
E) True. When the loops are closer together, more of them can fit per unit length
F) False. If the wire is shorter the number of turns decreases.
The correct answers are: A, C, D, E
३.रात में घूमने वाला write one word substitute
Explanation:
रात में घूमने वाला arthaarat निशाचर
Four children climb on a carousel that is initially at rest. If the carousel accelerates to 0.4 rad/s within 10 seconds, what is the angular acceleration? If the carousel has a radius of 20 meters, what is the linear velocity of the children?
Answer:
If the angular acceleration is a constant A, then this constant times the time will be equal to the angular velocity.
We know that after accelerating for 10 seconds, the angular velocity is 0.4 rad/s
Then:
A*10s = 0.4 rad/s
If we solve this for A we get:
A = (0.4 rad/s)/10s = 0.04 rad/s^2
This is the angular acceleartion.
Now we assume that the angular speed remains constant at 0.4 rad/s, and we know that the radius of the carousel is 20m, we want to find the linear velocity of the children.
We know that for an angular velocity W and for a radius R, the linear velocity is:
V = R*W
Then in this case the linear velocity is:
V = 20m*(0.4 s^(-1)) = 8 m/s
A 40 kg rock is rolling toward a town at 4 m/s after an earthquake. Calculate the KE.
Be sure to show your work and include units!
Use the formula KE = 1/2mv2 will brainlist
Answer:
320 J
Explanation:
From the question,
KE = 1/2mv².................. Equation 1
Where KE = Kinetic Energy, m = mass of the rock, v = velocity of the rock
Given: m = 40 kg, v = 4m/s
Substitute these values into equation 1
KE = 1/2(40)(4²)
KE = 20×16
KE = 320 J
Hence the kinetic energy of the rock is 320 J
What type of fault is shown in the image below?
In this fault, the hanging wall and footwall are side by side and moving in opposite directions.
Public Domain
Lateral fault
Normal fault
Reverse fault
Thrust fault
Answer:
this is a lateral fault
Explanation:
(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?
Answer:
5.0/s
Explanation:
Answer:
b and a it is this that abewsr
A block is sliding along a horizontal surface with 100 J of kinetic energy. The kinetic frictional force stops the block in 5 seconds.
What is the rate (in J/s) at which the force of kinetic friction does work?
Answer:
20 J/s
Explanation:
Given data
Kinetic energy=100 J
Time= 5 seconds
Hence the rate at which the kinetic friction work is
=100/5
=20 J/s
Therefore the answer is 20 J/s
A large wooden block of weight 30n, is observed to push down a ramp. The block is pushed down an inclined plane as shown below and the applied force is parallel to the plane and has a magnitude of 5 n as shown. The force of kinetic friction is 17n the block is pushed 6m down along the inclination of the ramp is 30 degrees with the horizontal
Answer:
OMG IM ON THE SAME QUESTION
Explanation: