Answer:
To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to a value 2D that is twotimes D
A microwave oven operates at 2.4 GHz with an intensity inside the oven of 2300 W/m2 . Part A What is the amplitude of the oscillating electric field
Answer:
The amplitude of the oscillating electric field is 1316.96 N/C
Explanation:
Given;
frequency of the wave, f = 2.4 Hz
intensity of the wave, I = 2300 W/m²
Amplitude of oscillating magnetic field is given by;
[tex]B_o = \sqrt{\frac{2\mu_o I}{c} }[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
I is intensity of wave
c is speed of light = 3 x 10⁸ m/s
[tex]B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T[/tex]
The amplitude of the oscillating electric field is given by;
E₀ = cB₀
E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶
E₀ = 1316.96 N/C
Therefore, the amplitude of the oscillating electric field is 1316.96 N/C
The heat of fusion of water is 79.5 cal/g. This means 79.5 cal of energy are required to:_________
A) raise the temperature of 1 g of water by 1K
B) turn 1 g of water to steam .
C) raise the temperature of 1 g of ice by 1 K .
D) melt 1 g of ice .
E) increase the internal energy of 1 g of water by 1 J .
Answer:
D) melt 1 g of ice
Explanation:
Heat of fusion is the energy required to change the state of a substance from solid state to liquid state at a constant pressure.
Heat of fusion of water occurs when solid ice acquires energy and turn into liquid water through a process known as melting.
Thus, if the heat of fusion of water is 79.5 cal/g, it means 79.5 cal of energy are required to melt 1 g of ice.
D) melt 1 g of ice
During the new moon phase, why is the Moon not visible in the sky?
Answer:
Explanation:
The moon gets the light from the sun. When the moon lies between the sun and the earth, only the back portion of the moon gets the light from the sun. So the side facing the sun does not get any light and appears to be dark or does not appear at all.
Hope this helps
plz mark as brainliest!!!!!!!
Answer:
The moon is between the sun, and Earth and reflects light back towards the sun.
Explanation:
A P E X test answer. Just took the test and this is the correct answer.
The velocity function (in meters per second) is given for a particle moving along a line. Find the total distance traveled by the particle during the given interval
Answer:
s=((vf+vi)/2)t vf is final velocity and vi is initial velocity
A mass weighing 64 pounds stretches a spring 6 inches. The mass moves through a medium offering a damping force that is numerically equal to β times the instantaneous velocity. Determine the values of β > 0 for which the system will exhibit oscillatory motion. (Enter your answer as a single inequality.)
Answer:
chk photo
Explanation:
The values of β for which the system will exhibit oscillatory motion are those greater than 0.
What is oscillatory motion?To determine the values of β for which the system will exhibit oscillatory motion, you need to perform a damped oscillation analysis on the system.
The equation of motion for a damped oscillator is given by:
F = ma = -kx - βv
where F is the force, m is the mass, a is the acceleration, k is the spring constant, x is the displacement of the mass from its equilibrium position, β is the damping coefficient, and v is the velocity of the mass.
To determine the values of β for which the system will exhibit oscillatory motion, you need to solve for the roots of the characteristic equation, which is obtained by setting the acceleration (a) equal to zero:
0 = -kx - βv
βv + kx = 0
x(β + k) = 0
The roots of this equation are x = 0 and x = -(β + k). If x = 0 is a root of the characteristic equation, then the system will exhibit steady-state behavior (i.e., the mass will come to rest at its equilibrium position and not oscillate). If x = -(β + k) is a root of the characteristic equation, then the system will exhibit oscillatory behavior (i.e., the mass will oscillate around its equilibrium position).
Therefore, the values of β for which the system will exhibit oscillatory motion are those for which x = -(β + k) is a root of the characteristic equation. This means that β must be such that:
-(β + k) < 0
β > -k
Since k is a positive value (the spring constant is always positive), the inequality simplifies to:
β > 0
Learn more about oscillation, here:
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A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)sin[(500rad/s)t]. What is the voltage across the resistor at 2.09 x 10-3 s? Group of answer choices 205 V 515 V 636 V 542 V
Answer:
205 V
V[tex]_{R}[/tex] = 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is
[tex]V_{L} = - IwLsin(wt)[/tex]
w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V[tex]_{R}[/tex] = IR
= (0.044 A) (93 Ω)
V[tex]_{R}[/tex] = 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
Putting V[tex]_{R}[/tex] = 4.092 V and w = 500 rad/s
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V[tex]_{R}[/tex] = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V[tex]_{R}[/tex] = 2.05 V
Sammy is 5 feet and 5.3 inches tall.tall.what is sammy's height in metres?
Answer:
65.3
Explanation:
1 foot = 12 inches
Sammy is 5 feet tall.
5 feet = ? inches
Multiply the feet value by 12 to find in inches.
5 × 12
= 60
Add 5.3 inches to 60 inches.
60 + 5.3
= 65.3
Answer:
It will be 》》》》1.664716m
Warm blooded animals are homeothermic; that is, they maintain an approximately constant body temperature. (Forhumans it's about 37 oC.) When they are in an environment that is below their optimum temperature, they use energy derived from chemical reactions within their bodies to warm them up. One of the ways that animals lose energy to their environment is through radiation. Every object emits electromagnetic radiation that depends on its temperature. For very hot objects like the sun, that radiation is visible light. For cooler objects, like a house or a person, that radiation is in the infrared and is invisible. Nonetheless, it still carries energy. Other ways that energy is lost by a warm animal to a cool environment includes conduction (direct touching of a cooler object) and convection (cooler air moving and carrying thermal energy away). See Heat Transfer for a discussion of all three.
For this problem, we'll just consider how much energy an animal needs to burn (obtain from internal chemical reactions) in order to stay warm just from radiation losses. The rate at which an object loses energy through radiation is given by the Stefan-Boltzmann equation:
Rate of energy loss = AεσT4
where T is the absolute (Kelvin) temperature, A is the area of the object, ε is the emissivity (unitless and =1 for a perfect emitter, less for anything else), and σ is the Stefan-Boltzmann constant:
σ = 5.67 x 10-8 J/(s m2 K4)
Consider a patient trying to sleep naked in a cool room (55 oF = 13 oC). Assume that the person being considered is a perfect emitter and absorber of radiation (ε = 1), has a surface area of about 2.5 m2, and a mass of 80 kg.
a. A person emits thermal radiation at a rate corresponding to a temperature of 37 oC and absorbs radiation at a rate (from the air and walls) corresponding to a temperature of 13 oC. Calculate the individual's net rate of energy loss due to radiation (in Watts = Joules/second).
net rate of energy loss = Watts
b. Assume the patient produces no energy to keep warm. If they have a specific heat about equal to that of water (1 Cal/kg-oC) how much would their temperature fall in one hour? (1 Cal = 1kcal = 103 cal)
ΔT = oC
c. Given that the energy density of fat is about 9 Cal/g, how many grams of fat would the person have to utilize to maintain their body temperature in that environment for one hour?
amount of fat needed = g
Answer:
a) 360.7 J/s
b) 16.23 °C
c) 34.48 g
Explanation:
The mass of the person = 80 kg
The person is a perfect emitter, ε = 1
surface area of the person = 2.5 m^2
a) If he emits radiation at 37 °C, [tex]T_{out}[/tex] = 37 + 273 = 310 K
and receives radiation at 13 °C, [tex]T_{in}[/tex] = 13 + 273 = 286 K
Rate of energy loss E = Aεσ([tex]T^{4} _{out}[/tex] - [tex]T^{4} _{in}[/tex] )
where σ = 5.67 x 10^-8 J/(s m^2 K^4)
substituting values, we have
E = 2.5 x 1 x 5.67 x 10^-8 x ([tex]310^{4}[/tex] - [tex]286^{4}[/tex]) = 360.7 J/s
b) If they have specific heat about equal to that of water = 1 Cal/kg-°C
but 1 Cal = 1 kcal = 10^3 cal
specific heat of person is therefore = 10^3 cal/kg-°C
heat loss = 360.7 J/s = 360.7 x 3600 = 1298520 J/hr
heat lost in 1 hour = 1 x 1298520 = 1298520 J
This heat lost = mcΔT
where ΔT is the temperature fall
m is the mass
c is the specific heat equivalent to that of water
the specific heat is then = 10^3 cal/kg-°C
equating, we have
1298520 = 80 x 10^3 x ΔT
1298520 = 80000ΔT
ΔT = 1298520/80000 = 16.23 °C
c) 1298520 J = 1298520/4184 = 310.35 Cal
density of fat = 9 Cal/g
gram of fat = 310.35/9 = 34.48 g
A woman was told in 2020 that she had exactly 15 years to live. If she travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:
Answer:
2035
Explanation:
The doctor does not travel with the woman, and therefore, he won't experience any relativistic effect on his time. The doctor will judge time by the time here on earth. Technically, the last new year's day the doctor, who is here on earth, would expect the woman to celebrate will be in 2020 + 15 years = 2035
An object is placed in a room where the temperature is 20 degrees C. The temperature of the object drops by 5 degrees C in 4 minutes and by 7 degrees C in 8 minutes. What was the temperature of the object when it was initially placed in the room
Answer:
28.3°C
Explanation:
Using
T(t) = (T(0) - 20)*(e^(-k*t)) + 20
for some positive number k, and some initial temperature T(0).
Boundary conditions:
T(4) = T(0) - 5 _______ (i)
T(8) = T(0) - 7 _______ (ii)
==> solving for T(0) and k :
(i):
(T(0) - 20)*(e^(-k*4)) + 20 = T(0) - 5 ==>
(T(0) - 20)*(e^(-k*4)) = T(0) - 20 - 5
(T(0) - 20)*(e^(-k*4)) = (T(0) - 20) - 5
5 = (T(0) - 20) - (T(0) - 20)*(e^(-k*4))
5 = (T(0) - 20) * ( 1 - e^(-k*4) )
(ii):
(T(0) - 20)*(e^(-k*8)) + 20 = T(0) - 7
(T(0) - 20)*(e^(-k*8)) = (T(0) - 20) - 7
7 = (T(0) - 20) - (T(0) - 20)*(e^(-k*8))
7 = (T(0) - 20) * (1 - e^(-k*8))
In both results, subsitute x = e^(-4k) and C = (T(0) - 20)
(i): 5 = C * (1 - x)
(ii): 7 = C * (1 - x^2) = C * (1-x)*(1+x)
Substitute C*(1-x) from (i) into (ii):
(ii): 7 = 5*(1+x) ==> (1+x) = 7/5 ==> x = 2/5
back into (i):
(i): 5 = C * (1 - 2/5) ==> 5 = C * 3/5 ==> C = 25/3
C = T(0) - 20 ==>
T(0) = C + 20 = 25/3 + 20 = 25/3 + 60/3 = 85/3
= 28.3°C
How much time will elapse if a radioisotope with a half-life of 88 seconds decays to one-sixteenth of its original mass?
Answer:
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
Explanation:
The decay of radioisotopes are represented by the following ordinary differential equation:
[tex]\frac{dm}{dt} = -\frac{t}{\tau}[/tex]
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]\tau[/tex] - Time constant, measured in seconds.
[tex]m[/tex] - Mass of the radioisotope, measured in grams.
The solution of this expression is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]m_{o}[/tex] is the initial mass of the radioisotope, measured in kilograms.
The ratio of current mass to initial mass is:
[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
The time constant is now calculated in terms of half-life:
[tex]\tau = \frac{t_{1/2}}{\ln2}[/tex]
Where [tex]t_{1/2}[/tex] is the half-life of the radioisotope, measured in seconds.
Given that [tex]t_{1/2} = 88\,s[/tex], the time constant of the radioisotope is:
[tex]\tau = \frac{88\,s}{\ln 2}[/tex]
[tex]\tau \approx 126.957\,s[/tex]
Now, if [tex]\frac{m(t)}{m_{o}(t)} = \frac{1}{16}[/tex] and [tex]\tau \approx 126.957\,s[/tex], the time is:
[tex]t = -\tau \cdot \ln\frac{m(t)}{m_{o}}[/tex]
[tex]t = -(126.957\,s)\cdot \ln \frac{1}{16}[/tex]
[tex]t \approx 352\,s[/tex]
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
A neutron star has a mass of between 1.4-2.8 solar masses compressed to the size of:
A. Earth
B. The state of Oregon
C. North America
D. An average city
The correct answer is D. An average city
Explanation:
A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.
A certain digital camera having a lens with focal length 7.50 cmcm focuses on an object 1.70 mm tall that is 4.70 mm from the lens.
Part A. How far must the lens be from the photocells?
s = cm
Part B. Is the image on the photocells erect or inverted? Real or virtual?
a. The image is erect and real.
b. The image is inverted and real.
c. The image is erect and virtual.
d. The image is inverted and virtual.
Part C. How tall is the image on the photocells?
|h?| = cm
Part D. A SLR digital camera often has pixels measuring 8.00?m
Answer:
a. 7.62cm
b. Real and inverted
c. 2.76 cm
d. 3450
Explanation:
We proceed as follows;
a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;
1/f = 1/p + 1/q
making q the subject of the formula;
q = pf/p-f
From the question;
p = 4.70m
f = 7.5cm = 0.075m
Substituting these, we have ;
q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm
b. The image is real and inverted since the image distance is positive
c. We want to calculate how tall the image is
Mathematically;
h1 = (q/p)h0
h1 = (7.62/4.70)* 1.7
h1 = 2.76 cm
d. We want to calculate the number of pixels that fit into this image
Mathematically:
n = h1/8 micro meter
n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450
Convierta 164 decimetros a hectometros
Answer:
sinco
Explanation:
Proposed Exercises: Strength and Acceleration in Circular Movement In the situation illustrated below, a 7kg sphere is connected to a rope so that it can rotate in a vertical plane around an O axis perpendicular to the plane of the figure. When the sphere is in position A, it has a speed of 3m/s. Determine for this position the modulus of tension on the string and the rate at which the tangential velocity is increased.
Answer:
81 N
7.1 m/s²
Explanation:
Draw a free body diagram of the sphere. There are two forces:
Weight force mg pulling straight down,
and tension force T pulling up along the rope.
Sum of forces in the centripetal direction:
∑F = ma
T − mg sin 45° = m v² / r
T = m (g sin 45° + v² / r)
T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)
T = 81 N
Sum of forces in the tangential direction:
mg cos 45° = ma
a = g cos 45°
a = (10 m/s²) cos 45°
a = 7.1 m/s²
PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.
Answer:
1.The temperature of each sample will increase by the same amount
Explanation:
This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.
This is shown by the calculation below
Q = mcΔT
ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.
Since Q, m and c are the same for both substances, thus ΔT will be the same.
So, the temperature of each sample will increase by the same amount
Polarized sunglasses:
a. block most sunlight because sunlight is polarized
b. are better but work the same way as non-polarized sunglasses
c. are polarized to filter out certain wavelengths of light
d. block reflected light because reflected light is partially polarized.
Polarized sunglasses creates filter of vertical openings for light. The light rays will reach the eyes of human vertically only.
The sun rays will not reach human eye directly which will create a shield for sun light burden on human eye.
Polarized sunglasses are best used for blocking and eliminating certain wavelengths of light.
Therefore the correct answer is option C. Polarizes Sunglasses are polarized and it filter out certain wavelengths of light.
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A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x axis at 37 m/s in the positive direction. The second, with mass 22 g , moves along the y axis at 34 m/s in the positive direction. Find the velocity of third piece.
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
1. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
2. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling with slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
Answer:
The hoop
Explanation:
Because it has a smaller calculated inertia of 2/3mr² compares to the disc
Two identical trucks have mass 5500 kg when empty, and the maximum permissible load for each is 8000 kg. The first truck, carrying a 3900 kg, is at rest. The second truck plows into it at 64 km/h, and the pair moves away at 44 km/h. As an expert witnes, you're asked to determine whether the second truck was overloaded. What do you report? Yes the truck is overloaded, or no, the truck is not overloaded?
Answer:
no, the truck is not overloaded
Explanation:
The computation is shown below;
Let us assume the mass of the loan in the second truck be M
So, the equation is as follows
{(Mass + M) × second truck × 1000 ÷ 3,600} = {(Mass + M + mass + first truck) × Pair moves away × 1,000 ÷ 3,600}
{(5500 + M) × 64 × 1,000 ÷ 3,600 = {(5,500 + M + 5,500 + 3,900) × 44 × 1,000 ÷ 3,600}
(5500 + M) × 64 = (14,900 + M) × 44
352,000 + 64 M = 655,600 + 44 M
After solving this
M = 15,180 kg
Therefore the second truck is not overloaded
Water flows at 0.00027 m3/s through a 10-m long garden hose lying on the ground, with a radius of 0.01 m. Water has a viscosity of 1 mPa.s What is the magnitude of gauge pressure in Pa of the water entering the hose
Answer:
The gauge pressure is [tex]P = 687.4 \ Pa[/tex]
Explanation:
From the question we are told that
The rate of flow is [tex]Q = 0.00027 m^3 /s[/tex]
The height is h = 10 m
The radius is r = 0.01 m
The viscosity is [tex]\eta = 1mPa \cdot s = 1 *10^{-3} \ Pa\cdot s[/tex]
Generally the gauge pressure according to Poiseuille's equation is mathematically represented as
[tex]P = 8 \pi \eta * \frac{L * v }{ A}[/tex]
Here v is the velocity of the water which is mathematically represented according to continuity equation as
[tex]v = \frac{Q}{A }[/tex]
Where A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 *(0.01)^2[/tex]
[tex]A = 3.142 *10^{-4} \ m^2[/tex]
So
[tex]v = \frac{ 0.00027}{3.142*10^{-4}}[/tex]
[tex]v = 0.8593 \ m/s[/tex]
So
[tex]P = 8 * 3.142 * 1.0*10^{-3}* \frac{10 * 0.8593 }{ 3.142*10^{-4}}[/tex]
[tex]P = 687.4 \ Pa[/tex]
When light travels from one medium to another with a different index of refraction, how is the light's frequency and wavelength affected
Answer:
The frequency does not change, but the wavelength does
Explanation:
Here are the options
A. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, the frequency changes and the wavelength does not.
B. The frequency does change, but the wavelength remains unchanged.
C. Both the frequency and wavelength change.
D. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, neither the wavelength nor the frequency changes.
E. The frequency does not change, but its wavelength does.
When light goes through one medium to the next, the frequency doesn't really change seeing as frequency is dependent on wavelength and light wave velocity. And when the wavelength shifts from one medium to the next.
[tex]n= \frac{C}{V} \ and\ \frac{\lambda_o}{\lambda_m}[/tex]
where [tex]\lambda_o[/tex] indicates wavelength in vacuum
[tex]\lambda_m[/tex] indicates wavelength in medium
n indicates refractive index
v indicates velocity of light wave
c indicates velocity of light
And wavelength is medium-dependent. Frequency Here = v[tex]\lambda[/tex] and shift in wavelength and velocity, not shifts in overall frequency.
Therefore the correct option is E
if an object weighs 550 n and the area is 1 cube
Two parallel slits are illuminated with monochromatic light of wavelength 567 nm. An interference pattern is formed on a screen some distance from the slits, and the fourth dark band is located 1.83 cm from the central bright band on the screen. (a) What is the path length difference corresponding to the fourth dark band? (b) What is the distance on the screen between the central bright band and the first bright band on either side of the central band? (Hint: The angle to the fourth dark band and the angle to the first bright band are small enough that tan θ ≈ sin θ.)
Answer:
a)1984.5nm
b)523mm
Explanation:
A)A destructive interference can be explained as when the phase shifting between the waves is analysed by the path lenght difference
θ=(m+0.5)λ where m= 1,2.3....
Where given from the question the 4th dark Fringe which will take place at m= 3
θ=7/2y
Where y= 567nm
= 7/2(567)=1984.5nm
But
B)tan θ ≈ y/d
And sinθ = mλ/d
y=mλd when m= 1 which is the first bright we have
Then y=(1× 567.D)/d
But the distance from Central to the 4th dark Fringe is 1.83cm then
y= 7λD/2d= 1.83cm
D/d=(2)×(1.83×10^-2)/(7×567×10^-9)
=92221.5
y= (567×10^-9)× (92221.5)
=0.00523m
Therefore, the distance between the first and center is y1-y0= 523mm
A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down.
In the 10.0 s period following the inital spin, the bike wheel undergoes 60.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ????s will it take the bike wheel to come to a complete stop?
The bike wheel has a mass of 0.625 kg0.625 kg and a radius of 0.315 m0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ????fτf that was acting on the spinning wheel.
Answer:
a) Δt = 24.96 s , b) τ = 0.078 N m
Explanation:
This is a rotational kinematics exercise
θ = w₀ t - ½ α t²
Let's reduce the magnitudes the SI system
θ = 60 rev (2π rad / 1 rev) = 376.99 rad
w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s
α = (w₀ t - θ) 2 / t²
let's calculate the annular acceleration
α = (43.98 10 - 376.99) 2/10²
α = 1,258 rad / s²
Let's find the time it takes to reach zero angular velocity (w = 0)
w = w₀ - alf t
t = (w₀ - 0) / α
t = 43.98 / 1.258
t = 34.96 s
this is the total time, the time remaining is
Δt = t-10
Δt = 24.96 s
To find the braking torque, we use Newton's law for angular motion
τ = I α
the moment of inertia of a circular ring is
I = M r²
we substitute
τ = M r² α
we calculate
τ = 0.625 0.315² 1.258
τ = 0.078 N m
The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
Given data:
The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex] (rps means rotation per second).
The time interval is, t' = 10.0 s.
The number of rotations made by wheel is, n = 60.0.
The mass of bike wheel is, m = 0.625 kg.
The radius of wheel is, r = 0.315 m.
The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,
[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]
Here, [tex]\theta[/tex] is the angular displacement, and its value is,
[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]
And, angular speed is,
[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]
Solving as,
[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]
Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.
[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]
Then total time is,
T = t - t'
T = 35.18 - 10
T = 25.18 s
Now, use the standard formula to obtain the value of braking torque as,
[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]
Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
Learn more about the rotational motion here:
https://brainly.com/question/1388042
beam of white light goes from air into water at an incident angle of 58.0°. At what angles are the red (660 nm) and blue (470 nm) parts of the light refracted? (Enter your answer to at least one decimal place.) red ° blue °
Answer:
For red light= 39.7°
Blue light 39.2°
Explanation:
Given that refractive index for red light is 1.33 and that of blue light is 1.342
So angle of refraction for red light will be
Sinစi/ (sinစ2) =( nw)r/ ni
Sin 58° x 1.000293/1.33. =( sinစ2)r
0.64= sinစ2r
Theta2r = 39.7°
For blue light
Sinစi/ (sinစ2) =( nw)b/ ni
Sin 58° x 1.000293/1.342 =( sinစ2)b
0.632= sinစ2r
Theta2b= 39.19°
An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically grounded object. Positively charged paint droplets are repelled away from the paint sprayer's positively charged sphere and towards the grounded object. What charge must a 0.168-mg drop of paint have so that it will arrive at the object with a speed of 18.8 m/s
Answer:
The charge is [tex]Q = 2.177 *10^{-9} \ C[/tex]
Explanation:
From the question we are told that
The electric potential is [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]
The mass of the drop is [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]
The speed is [tex]v = 18.8 \ m/s[/tex]
Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question is mathematically represented as
[tex]Q = \frac{m v^2 }{ 2 * V }[/tex]
Substituting values
[tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]
[tex]Q = 2.177 *10^{-9} \ C[/tex]
At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field to have a maximum value of the induced emf equal to 8.0 V
Answer:
The frequency of the coil is 7.07 Hz
Explanation:
Given;
number of turns of the coil, 200 turn
cross sectional area of the coil, A = 300 cm² = 0.03 m²
magnitude of the magnetic field, B = 30 mT = 0.03 T
Maximum value of the induced emf, E = 8 V
The maximum induced emf in the coil is given by;
E = NBAω
Where;
ω is angular frequency = 2πf
E = NBA(2πf)
f = E / 2πNBA
f = (8) / (2π x 200 x 0.03 x 0.03)
f = 7.07 Hz
Therefore, the frequency of the coil is 7.07 Hz
Two instruments produce a beat frequency of 5 Hz. If one has a frequency of 264 Hz, what could be the frequency of the other instrument
Answer:
259 Hz or 269 Hz
Explanation:
Beat: This is the phenomenon obtained when two notes of nearly equal frequency are sounded together. The S.I unit of beat is Hertz (Hz).
From the question,
Beat = f₂-f₁................ Equation 1
Note: The frequency of the other instrument is either f₁ or f₂.
If the unknown instrument's frequency is f₁,
Then,
f₁ = f₂-beat............ equation 2
Given: f₂ = 264 Hz, Beat = 5 Hz
Substitute into equation 2
f₁ = 264-5
f₁ = 259 Hz.
But if the unknown frequency is f₂,
Then,
f₂ = f₁+Beat................. Equation 3
f₂ = 264+5
f₂ = 269 Hz.
Hence the beat could be 259 Hz or 269 Hz
which objects would have a greater gravitational force between them, Objects A and B, or Objects B and C
Answer:
Objects that are closer together have a stronger force of gravity between them.
Explanation:
For example, the moon is closer to Earth than it is to the more massive sun, so the force of gravity is greater between the moon and Earth than between the moon and the sun.