Answer:
C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays
Explanation:
radio waves have lowest energy , lowest frequency and highest wavelength
gamma rays have highest energy , highest frequency and least wavelength
Answer: C
Explanation:
g One of the harmonics in an open-closed tube has frequency of 500 Hz. The next harmonic has a frequency of 700 Hz. Assume that the speed of sound in this problem is 340 m/s. a. What is the length of the tube
Answer:
The length of the tube is 85 cm
Explanation:
Given;
speed of sound, v = 340 m/s
first harmonic of open-closed tube is given by;
N----->A , L= λ/₄
λ₁ = 4L
v = Fλ
F = v / λ
F₁ = v/4L
Second harmonic of open-closed tube is given by;
L = N-----N + N-----A, L = (³/₄)λ
[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]
Third harmonic of open-closed tube is given by;
L = N------N + N-----N + N-----A, L = (⁵/₄)λ
[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]
The difference between second harmonic and first harmonic;
[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]
The difference between third harmonic and second harmonic;
[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]
Thus, the difference between successive harmonic of open-closed tube is
v / 2L.
[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]
Therefore, the length of the tube is 85 cm
You are holding on to one end of a long string that is fastened to a rigid steel light pole. After producing a wave pulse that was 5 mm high and 4 em wide, you want to produce a pulse that is 4 cm wide but 7 mm high. You must move your hand up and down once,
a. a smaller distance up, but take a shorter time.
b. the same distance up as before, but take a shorter time.
c. a greater distance up, but take a longer time.
d. the same distance up as before, but take a longer time.
e. a greater distance up, but take the same time.
Answer:
It will take. the same distance up as before, but take a longer time
A 384 Hz tuning fork produces standing waves with a wavelength of 0.90 m inside a resonance tube. The speed of sound at experimental conditions is
Answer:
v = 345.6m/s
Explanation:
v = 384 x 0.9 = 345.6
v = 345.6m/s
All household circuits are wired in parallel. A 1140-W toaster, a 270-W blender, and a 80-W lamp are plugged into the same outlet. (The three devices are in parallel when plugged into the same outlet.) Assume that this is the standard household 120-V circuit with a 15-A fuse.
a. What current is drawn by each device?
b. To see if this combination will blow the 15-A fuse, find the total current used when all three appliances are on.
Answer:
total current = 12.417 A
so it will not fuse as current is less than 15 A
Explanation:
given data
toaster = 1140-W
blender = 270-W
lamp = 80-W
voltage = 120 V
solution
we know that current is express as
current = power ÷ voltage ......................1
here voltage is same in all three device
so
current by toaster is
I = [tex]\frac{1140}{120}[/tex]
I = 9.5 A
and
current by blender
I = [tex]\frac{270}{120}[/tex]
I = 2.25 A
and
current by lamp is
I = [tex]\frac{80}{120}[/tex]
I = 0.667 A
so here device in parallel so
total current is = 9.5 A + 2.25 A + 0.667 A
total current = 12.417 A
so it will not fuse as current is less than 15 A
If two identical wires carrying a certain current in the same direction are placed parallel to each other, they will experience a force of repulsion.
a) true
b) false
Answer:
The answer is B. falseExplanation:
Current in the same direction
When current flow through to parallel conductors of a given length, when the current flows in the same direction
1. A force of attraction between the wires occurs and this tends to draw the wires inward
2. A magnetic field in the same direction is produced.
Current in opposite direction
when the current is in opposite direction
1. Force of repulsion between the two wires occurs, draws the wire outward
2. A magnetic field in opposite direction occurs
If 50 mL of each of the liquids in the answer choices were poured into a 250 mL beaker, which layer would be directly above a small rubber ball with a density of 0.960 g/mL? A. sea water – density of 1.024 g/mL B. mineral oil – density of 0.910 g/mL C. distilled water – density of 1.0 g/mL D. petroleum oil – density of 0.820 g/mL
Answer:
B. mineral oil – density of 0.910 g/mL.
Explanation:
Hello,
In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.
In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.
Best regards.
During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 31.7 rad/s. Find the angular displacement Δθ of the tub during a spin of 98.3 s, expressed both in radians and in revolutions.
Answer:
[tex]\Delta \theta = 3116.11\,rad[/tex] and [tex]\Delta \theta = 495.944\,rev[/tex]
Explanation:
The tub rotates at constant speed and the kinematic formula to describe the change in angular displacement ([tex]\Delta \theta[/tex]), measured in radians, is:
[tex]\Delta \theta = \omega \cdot \Delta t[/tex]
Where:
[tex]\omega[/tex] - Steady angular speed, measured in radians per second.
[tex]\Delta t[/tex] - Time, measured in seconds.
If [tex]\omega = 31.7\,\frac{rad}{s}[/tex] and [tex]\Delta t = 98.3\,s[/tex], then:
[tex]\Delta \theta = \left(31.7\,\frac{rad}{s} \right)\cdot (98.3\,s)[/tex]
[tex]\Delta \theta = 3116.11\,rad[/tex]
The change in angular displacement, measured in revolutions, is given by the following expression:
[tex]\Delta \theta = (3116.11\,rad)\cdot \left(\frac{1}{2\pi} \frac{rev}{rad} \right)[/tex]
[tex]\Delta \theta = 495.944\,rev[/tex]
A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.945 rad/s. You, with a mass of 69.7 kg, walk clockwise around the platform along its edge at the speed of 1.01 m/s with respect to the platform. Your 20.7 kg poodle also walks clockwise around the platform, but along a circle at half the platform's radius and at half your linear speed with respect to the platform. Your 17.7 kg mutt, on the other hand, sits still on the platform at a position that is 3/4 of the platform's radius from the center. Model the platform as a uniform disk with mass 93.1 kg and radius 1.93 m. Calculate the total angular momentum of the system.
Answer:
317.22
Explanation:
Given
Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s
You 69.7kg, cw 1.01m/s, at r
Poodle 20.2 kg, cw 1.01/2 m/s, at r/2
Mutt 17.7 kg, 3r/4
You
Relative
ω = v/r
= 1.01/1.93
= 0.522
Actual
ω = 0.945 - 0.522
= 0.42
I = mr^2
= 69.7*1.93^2
= 259.6
L = Iω
= 259.6*0.42
= 109.4
Poodle
Relative
ω = (1.01/2)/(1.93/2)
= 0.5233
Actual
ω = 0.945- 0.5233
= 0.4217
I = m(r/2)^2
= 20.2*(1.93/2)^2
= 18.81
L = Iω
= 18.81*0.4217
= 7.93
Mutt
Actual
ω = 0.945
I = m(3r/4)^2
= 17.7(3*1.93/4)^2
= 37.08
L = Iω
= 37.08*0.945
= 35.04
Disk
I = mr^2/2
= 93.1(1.93)^2/2
= 173.39
L = Iω
= 173.39*0.945
= 163.85
Total
L = 109.4+ 7.93+ 36.04+ 163.85
= 317.22 kg m^2/s
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
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 diffraction grating 19.2 mm wide has 6010 rulings. Light of wavelength 337 nm is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of θ at which maxima appear on a distant viewing screen?
Answer:
(a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
Explanation:
Given that,
Width of diffraction grating [tex]w= 19.2\ mm[/tex]
Number of rulings[tex]N=6010[/tex]
Wavelength = 337 nm
We need to calculate the distance between adjacent rulings
Using formula of distance
[tex]d=\dfrac{w}{N}[/tex]
Put the value into the formula
[tex]d=\dfrac{19.2\times10^{-3}}{6010}[/tex]
[tex]d=3.19\times10^{-6}\ m[/tex]
We need to calculate the value of m
Using formula of constructive interference
[tex]d \sin\theta=m\lambda[/tex]
[tex]\sin\theta=\dfrac{m\lambda}{d}[/tex]
Here, m = 0,1,2,3,4......
[tex]\lambda[/tex]=wavelength
For largest value of θ
[tex]\dfrac{m\lambda}{d}>1[/tex]
[tex]m>\dfrac{d}{\lambda}[/tex]
Put the value into the formula
[tex]m>\dfrac{3.19\times10^{-6}}{337\times10^{-9}}[/tex]
[tex]m>9.46[/tex]
[tex]m = 9[/tex]
(a). We need to calculate the largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{9\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=71.9^{\circ}[/tex]
(b). We need to calculate the second largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{8\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=57.7^{\circ}[/tex]
(c). We need to calculate the third largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{7\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=47.7^{\circ}[/tex]
Hence, (a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
A major strike-slip earthquake on the San Andreas fault in California will cause a catastrophic tsunami affecting residents of San Francisco.
a) true
b) false
Answer:
a) true
Explanation:
The san andres is a transform fault that forms boundary between the Pacific and the North Atlantic plate and this slip strike is characterized by the latex motion the fault runs in the length of the California state. This plate is widely estimated for the high magnitude of earthquakes and varies from 7.7 to 8.3 magnitude. They are capable of producing a deadly tsunami that can devastate the pacific northwest.A bucket filled with water has a mass of 23 Kg and is attached to a rope, which in turn is wound around a 0.050 m radius cylinder at the top of a well. What torque does the weight of water and bucket produce on the cylinder if the cylinder is ont permitted to rotate? (g= 9.8 m/s2)
Answer:
The torque is 11.27 N m
Explanation:
Recall that torque is the vector product of the force times the distance to the pivoting point. So in our case, the distance to the pivoting point is the radius of the cylinder (0.05 m), and the force is given by the weight of the bucket full of water (W = 9.8 * 23 N = 225.4 N)
Then the torque is: 0.05 * 225.4 N m = 11.27 N m
How much energy is required to accelerate a spaceship with a rest mass of 121 metric tons to a speed of 0.509 c?
Answer
1.07E22 Joules
Explanation;
We know that mass expands by a factor
=>>1/√[1-(v/c)²]
But v= 0.509c
So
1/√(1 - 0.509²)
=>>> 1/√(1 - 0.2591)
= >> 1/√(0.7409) = 1.16
But given that 121 tons is rest mass so 121- 1.16= 119.84 tons is kinetic energy
And we know that rest mass-energy equivalence is 9 x 10^19 joules per ton.
So Multiplying by 119.84
Kinetic energy will be 1.07x 10^22 joules
¿Cómo podrías utilizar el sistema de posicionamiento global para proponer recorridos alternativos para llegar a un lugar específico (centro educativo, supermercado, el hogar, el parque, entre otros)?
Answer:
El sistema de posicionamiento global (conocido mundialmente como GPS) podría utilizarse para proponer recorridos alternativos para llegar a un lugar específico, como un parque, a través de la creación de un recorrido guiado por una aplicación móvil con diferentes rutas de acceso al lugar.
Así, por ejemplo, se crearían diferentes rutas de acceso desde un punto A hasta un punto B, teniendo en cuenta factores como: rapidez, congestión vehicular, pago o no de peajes, posibilidad de acceso a pie y determinados factores extra que influyan en la forma de llegar al lugar. Todo ello plasmado en un mapa interactivo en el cual se señalen las rutas disponibles mediante el marcado del mapa en cuestión.
To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.0 m/s. How much time does it take for the glove to return to the pitcher
Answer:
The glove takes 1.02s to return to the pitchers hand.
Explanation:
Given;
initial velocity the pitcher's glove, u = 5 m/s
Apply kinematic equation
s = ut - ¹/₂gt²
where;
g is acceleration due to gravity = 9.8 m/s²
t is the time takes the glove to return to the pitchers hand
s is the displacement of the glove, which will be equal to zero when the glove returns to the pitchers hand. (s = 0)
0 = ut - ¹/₂gt²
ut = ¹/₂gt²
u = ¹/₂gt
gt = 2u
t = (2u) / g
t = (2 x 5) / 9.8
t = 1.02 s
Therefore, the glove takes 1.02s to return to the pitchers hand.
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 70 kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.
Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?
Answer:
Explanation:
A) Vair = 1.3 L
B) Volume is not reasonable
Explanation:
A)
Assume
m to be total mass of the man
mp be the mass of the man that pulled out of the water
m1 be the mass above the water with the empty lung
m2 be the mass above the water with full lung
wp be the weight that the buoyant force opposes as a result of the air.
Va be the volume of air inside man's lungs
Fb be the buoyant force due to the air in the lung
given;
m = 78.5 kg
m1 = 3.2% × 78.5 = 2.5 kg
m2 = 4.85% × 78.5 = 3.8kg
But, mp = m2- m1
mp = 3.8 - 2.5
mp = 1.3kg
So using
Archimedes principle, the relation for formula for buoyant force as;
Fb = (m_displaced water)g = (ρ_water × V_air × g)
Where ρ_water is density of water = 1000 kg/m³
Thus;
Fb = wp = 1.3× 9.81
Fb = 12.7N
But
Fb = (ρ_water × V_air × g)
So
Vair = Fb/(ρ_water × × g)
Vair = 12.7/(1000 × 9.81)
V_air = 1.3 × 10^(-3) m³
convert to litres
1 m³ = 1000 L
Thus;
V_air = 1.3× 10^(-3) × 1000
V_air = 1.3 L
But since the average lung capacity of an adult human being is about 6-7litres of air.
Thus, the calculated lung volume is not reasonable
Explanation:
What will be the nature of the image formed from both a convex lens and a concave
lens of 20 centimeter focus distance, when the object is placed at a distance of
10 centimeters?
Answer:
Explanation:
Using the lens formula
1//f = 1/u+1/v
f is the focal length of the lens
u is the object distance
v is the image distance
For convex lens
The focal length of a convex lens is positive and the image distance can either be negative or positive.
Given f = 20cm and u = 10cm
1/v = 1/f - 1/u
1/v = 1/20-1/10
1/v = (1-2)/20
1/V = -1/20
v = -20/1
v = -20 cm
Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image
For concave lens
The focal length of a concave lens is negative and the image distance is negative.
Given f = -20cm and u = 10cm
1/v = 1/f - 1/u
1/v = -1/20-1/10
1/v = (-1-2)/20
1/V = -3/20
v = -20/3
v = -6.67 cm
Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image
Air flows through a converging-diverging nozzle/diffuser. A normal shock stands in the diverging section of the nozzle. Assuming isentropic flow, air as an ideal gas, and constant specific heats determine the state at several locations in the system. Solve using equations rather than with the tables.
Answer:
HELLO your question has some missing parts below are the missing parts
note: The specific heat ratio and gas constant for air are given as k=1.4 and R=0.287 kJ/kg-K respectively.
--Given Values--
Inlet Temperature: T1 (K) = 325
Inlet pressure: P1 (kPa) = 560
Inlet Velocity: V1 (m/s) = 97
Throat Area: A (cm^2) = 5.3
Pressure upstream of (before) shock: Px (kPa) = 207.2
Mach number at exit: M = 0.1
Answer: A) match number at inlet = 0.2683
B) stagnation temperature at inlet = 329.68 k
C) stagnation pressure = 588.73 kPa
D) ) Throat temperature = 274.73 k
Explanation:
Determining states at several locations in the system
A) match number at inlet
= V1 / C1 = 97/ 261.427 = 0.2683
C1 = sound velocity at inlet = [tex]\sqrt{K*R*T}[/tex] = [tex]\sqrt{1.4 *0.287*10^3}[/tex] = 361.427 m/s
v1 = inlet velocity = 97
B) stagnation temperature at inlet
= T1 + [tex]\frac{V1 ^2}{2Cp}[/tex] = 325 + [tex]\frac{97^2}{2 * 1.005*10^{-3} }[/tex]
stagnation temperature = 329.68 k
C) stagnation pressure
= [tex]p1 ( 1 + 0.2Ma^2 )^{3.5}[/tex]
Ma = match number at inlet = 0.2683
p1 = inlet pressure = 560
hence stagnation pressure = 588.73 kPa
D) Throat temperature
= [tex]\frac{Th}{T} = \frac{2}{k+1}[/tex]
Th = throat temperature
T = stagnation temp at inlet = 329.68 k
k = 1.4
make Th subject of the relation
Th = 329.68 * (2 / 2.4 ) = 274.73 k
Which of
of
these
following material is
used as fuse material?
carbon,
silver
Copper
Aluminium
The provided question is not correct as, there is more than one options are correct, however the explaining every correct option -
Answer:
The correct answer are - silver, copper and aluminium all three used as fuse material.
Explanation:
A safety device in any electric circuit of that prevents the electric system in case of short circuit by breaking the connection of electric system or circuit termed as the Fuse or fuse element. Normally the fuse are made up of wire or element of material that are low in melting point and high in resistance.
Zinc, lead, tin, silver, copper, aluminium, and alloy of tin and alloy are used as fuse element or material for their low melting point and high resistance these are easily breaks the electric path in case of short circuit.
In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen is D=48.0 cm away from the slits. What is the linear distance Δx between the eighth order maximum and the fourth order maximum on the screen?
Answer:
Δx = 4.68 x 10⁻³ m = 4.68 mm
Explanation:
The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:
Δx = λD/d
So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:
Δx = 4λD/d
where,
Δx = distance between eighth order maximum and fourth order maximum=?
λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m
d = slit separation = 0.2 mm = 2 x 10⁻⁴ m
D = Distance between slits and screen = 48 cm = 0.48 m
Therefore,
Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)
Δx = 4.68 x 10⁻³ m = 4.68 mm
At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m
Answer:
a) The value of g at such location is:
[tex]g=9.8005171\,\frac{m}{s^2}[/tex]
b) the period of the pendulum with the length is 0.970 m is:
[tex]T=1.9767 sec[/tex]
Explanation:
Recall the relationship between the period (T) of a pendulum and its length (L) when it swings under an acceleration of gravity g:
[tex]L=\frac{g}{4\,\pi^2} \,T^2[/tex]
a) Then, given that we know the period (2.0 seconds), and the pendulum's length (L=0.993 m), we can determine g at that location:
[tex]g=\frac{4\,\pi^2\,L}{T^2}\\g=\frac{4\,\pi^2\,0.993}{(2)^2}\\g=\pi^2\,(0.993)\,\frac{m}{s^2} \\g=9.8005171\,\frac{m}{s^2}[/tex]
b) for this value of g, when the pendulum is shortened to 0.970 m, the period becomes:
The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps and dips in the road. When a 68 kg (about 150 lb) person sits on the left front fender of a small car, this corner of the car dips by about 1.2 cm (about 1/2 in).
If we treat the spring assembly as a single spring, what is the approximate spring constant?
k= ____________
Answer:
The approximate spring constant is [tex]k = 55533.33 \ N/m[/tex]
Explanation:
From the question we are told that
The mass of the person is [tex]m = 68 \ kg[/tex]
The dip of the car is [tex]x = 1.2 \ cm = 0.012 \ m[/tex]
Generally according to hooks law
[tex]F = k * x[/tex]
here the force F is the weight of the person which is mathematically represented as
[tex]F = m * g[/tex]
=> [tex]m * g = k * x[/tex]
=> [tex]k = \frac{m * g }{x }[/tex]
=> [tex]k = \frac{68 * 9.8}{ 0.012}[/tex]
=> [tex]k = 55533.33 \ N/m[/tex]
Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by 0.490 mm . In the resulting interference pattern on a screen 2.12 m away, adjacent bright fringes are separated by 2.86 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Determining wavelength. Part A What is the wavelength of the light that falls on the slits
Answer:
λ = 6.61 x 10⁻⁷ m = 661 nm
Explanation:
From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:
Δx = λL/d
where,
Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m
L = Distance between slits and screen = 2.12 m
d = slit separation = 0.49 mm = 0.49 x 10⁻³ m
λ = wavelength of light = ?
Therefore,
2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)
(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ
λ = 6.61 x 10⁻⁷ m = 661 nm
A double-convex thin lens is made of glass with an index of refraction of 1.52. The radii of curvature of the faces of the lens are 60 cm and 72 cm. What is the focal length of the lens
Answer:
63 cm
Explanation:
Mathematically;
The focal length of a double convex lens is given as;
1/f = (n-1)[1/R1 + 1/R2]
where n is the refractive index of the medium given as 1.52
R1 and R2 represents radius of curvature which are given as 60cm and 72cm respectively.
Plugging these values into the equation, we have:
1/f = (1.52-1)[1/60 + 1/72)
1/f = 0.0158
f = 1/0.0158
f = 63.29cm which is approximately 63cm
A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.
Answer:
Power factor = 0.87 (Approx)
Explanation:
Given:
Load = 1 Kw = 1000 watt
Current (I) = 5 A
Supply (V) = 230 V
Find:
Power factor.
Computation:
Power factor = watts / (V)(I)
Power factor = 1,000 / (230)(5)
Power factor = 1,000 / (1,150)
Power factor = 0.8695
Power factor = 0.87 (Approx)
Two waves are traveling in the same direction along a stretched string. The waves are 45.0° out of phase. Each wave has an amplitude of 7.00 cm. Find the amplitude of the resultant wave.
Answer:
The amplitude of the resultant wave is 12.93 cm.
Explanation:
The amplitude of resultant of two waves, y₁ and y₂, is given as;
Y = y₁ + y₂
Let y₁ = A sin(kx - ωt)
Since the wave is out phase by φ, y₂ is given as;
y₂ = A sin(kx - ωt + φ)
Y = y₁ + y₂ = 2A Cos (φ / 2)sin(kx - ωt + φ/2 )
Given;
phase difference, φ = 45°
Amplitude, A = 7.00 cm
Y = 2(7) Cos (45 /2) sin(kx - ωt + 22.5° )
Y = 12.93 cm
Therefore, the amplitude of the resultant wave is 12.93 cm.
A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.220 s, during which it produces an average 0.520 W from an average 3.00 V.
A. How much charge moves through the lamp (C)?
B. Find the capacitance (F).
C. What is the resitance of the lamo?
Answer:
A. 0.0374C
B. 0.012F
C. 18 ohms
Explanation:
See attached file
Helium-neon laser light (λ = 6.33 × 10−7 m) is sent through a 0.30 mm-wide single slit. What is the width of the central maximum on a screen 1.0 m from the slit?
Answer:
The width is [tex]w_c = 0.00422 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 6.33*10^{-7} \ m[/tex]
The width of the slit is [tex]d = 0.3\ mm = 0.3 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 1.0 \ m[/tex]
Generally the central maximum is mathematically represented as
[tex]w_c = 2 * y[/tex]
Here y is the width of the first order maxima which is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{6.33*10^{-7} * 1.0}{ 0.30}[/tex]
[tex]y = 0.00211 \ m[/tex]
So
[tex]w_c = 2 *0.00211[/tex]
[tex]w_c = 0.00422 \ m[/tex]
