Answer:
The answer is A
Explanation:
Answer:
implausible or untestable scientific claims
Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?
Answer:
The final temperature is 61.65 °C
Explanation:
mass of copper pot [tex]m_{c}[/tex] = 2 kg
temperature of copper pot [tex]T_{c}[/tex] = 20 °C (the pot will be in thermal equilibrium with the room)
specific heat capacity of copper [tex]C_{c}[/tex]= 385 J/kg-°C
The heat content of the copper pot = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{c}[/tex] = 2 x 385 x 20 = 15400 J
mass of boiling water [tex]m_{w}[/tex] = 200 g = 0.2 kg
temperature of boiling water [tex]T_{w}[/tex] = 100 °C
specific heat capacity of water [tex]C_{w}[/tex] = 4182 J/kg-°C
The heat content of the water = [tex]m_{w}[/tex][tex]C_{w}[/tex][tex]T_{w}[/tex] = 0.2 x 4182 x 100 = 83640 J
The total heat content of the water and copper mix [tex]H_{T}[/tex] = 15400 + 83640 = 99040 J
This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation
[tex]H_{T}[/tex] = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{f}[/tex] + [tex]m_{w}[/tex][tex]C_{w}[/tex]
where [tex]T_{f}[/tex] is the final temperature of the water and the copper
substituting values, we have
99040 = (2 x 385 x [tex]T_{f}[/tex]) + (0.2 x 4182 x
99040 = 770[tex]T_{f}[/tex] + 836.4
99040 = 1606.4[tex]T_{f}[/tex]
[tex]T_{f}[/tex] = 99040/1606.4 = 61.65 °C
A single slit 1.5 mm wide is illuminated by 420- nm light. Part A What is the width of the central maximum (in cm ) in the diffraction pattern on a screen 4.5 m away
Answer:
The width is [tex]w_c = 0.00252 \ m[/tex]
Explanation:
From the question we are told that
The width of the single slit is [tex]a = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 420 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 4.5 \ m[/tex]
Generally the width of the central maximum is
[tex]w_c = 2 * y[/tex]
where y is the width of the first maxima which is mathematically represented as
[tex]y = \frac{\lambda * D}{a}[/tex]
=> [tex]y = \frac{ 420 *10^{-9} * 4.5}{ 1.5*10^{-3}}[/tex]
=> [tex]y = 0.00126 \ m[/tex]
So
[tex]w_c = 2 *0.00126[/tex]
[tex]w_c = 0.00252 \ m[/tex]
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a point r < R1 is:
Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA = [tex]q_{int}[/tex] / ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁
A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of 36.4 m/s , blows a horn whose frequency is 123 Hz .what is its speed?
Answer:
51. 7m/s
Explanation:
Take speed of sound in air = 340 m/s
fp = fs (V + Vp)/(V + Vs)
128 = 123 (340 + Vp)/(340 + 36.4)
Vp = 51.7m/s
Explanation:
: A spaceship is traveling at the speed 2t 2 1 km/s (t is time in seconds). It is pointing directly away from earth and at time t 0 it is 1000 kilometers from earth. How far from earth is it at one minute from time t 0
Answer:
145060km
Explanation: Given that
speed = dx/dt = 2t^2 +1
integrate
x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)
given that at t=0, x = 1000
so 1000 = 2/3 X (0)^3 + 0 + c
or c = 1000
So x = 2/3t^3 + t + 1000
for t = 1 min = 60s
x = 2/3 X 60^3 + 60 + 1000
x = 2/3×216000+ 1060
x = 144000+1060
= 145060km
At one minute, it will be 145060km far from the earth
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
A square loop, length l on each side, is shot with velocity v0 into a uniform magnetic field B. The field is perpendicular to the plane of the loop. The loop has mass m and resistance R, and it enters the field at t = 0s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0m.
Required:
Find an expression for the loop's velocity as a function of time as it enters the magnetic field.
Answer:
v₀(1 + B²L²t/mR)
Explanation:
We know that the force on the loop is F = BIL where B = magnetic field strength, I = current and L = length of side of loop. Now the current in the loop I = ε/R where ε = induced e.m.f in the loop = BLv₀ where v₀ = velocity of loop and r = resistance of loop
F = BIL = B(BLv₀)L/R = B²L²v₀/R
Since F = ma where a = acceleration of loop and m = mass of loop
a = F/m = B²L²v₀/mR
Using v = u + at where u = initial velocity of loop = v₀, t = time after t = 0 and v = velocity of loop after time t = 0
Substituting the value of a and u into v, we have
v = v₀ + B²L²v₀t/mR
= v₀(1 + B²L²t/mR)
So the velocity of the loop after time t is v = v₀(1 + B²L²t/mR)
The expression for the loop's velocity as a function of time as it enters the magnetic field is v = v₀(1 + B²L²t/mR).
Calculation of the loop velocity:As we know that
Force on the loop
F = BIL
here
B = magnetic field strength,
I = current
and L = length of side of loop.
Now
the current in the loop I = ε/R
where
ε = induced e.m.f in the loop = BLv₀
where v₀ = velocity of loop
and r = resistance of loop
So,
F = BIL = B(BLv₀)L/R = B²L²v₀/R
Also, F = ma where a = acceleration of loop and m = mass of loop
Now
a = F/m = B²L²v₀/mR
We have to use
v = u + at
where
u = initial velocity of loop = v₀,
t = time after t = 0
and v = velocity of loop after time t = 0
So, it be like
v = v₀ + B²L²v₀t/mR
= v₀(1 + B²L²t/mR)
Learn more about velocity here: https://brainly.com/question/332163
g To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, you can
Answer:
Increase the distance by a factor of 6.
Explanation:
The intensity at a distance r is given by :
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
Here,
P is power emitted
r is distance from source
It means that the intensity is inversely proportional to the distance from the source.
To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, we can increase the distance by a factor of 6. Hence, this is the required solution.
g Assume you are a farsighted person who has a near point distance of 40 (cm). If you use a converging contact lens with focal length of 10 (cm). What is nearest distance you can vision with you contacts now?
Answer:
object distance p = 13.33 cm
Explanation:
For this problem of finding the image of an object we must use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distances to the object and the image, respectively.
In this case they indicate the focal length f = 10 cm, since the person has hyperopia, the image must be formed q = 40 cm, let's find where the object is (p)
1 / p = 1 / f - 1 / q
1 / p = 1/10 - 1/40
1 / p = 0.075
p = 13.33 cm
A clown 2 m tall looks at himself in a full-length mirror (floor-to-ceiling). Where in the mirror must he look to see his feet?
Answer:
Around the center of the mirror
An 1,820 W toaster, a 1,420 W electric frying pan, and a 55 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.)
Required:
a. What current is drawn by each device?
b. Will this combination blow the 15-A fuse?
Answer:
toaster- 15.1A
electric frying pan- 11.8 A
lamp- 0.5 A
b) The combination will blow the fuse.
Explanation:
When devices are connected in parallel, the potential difference across each of the devices is the same but the current through each is different. Hence;
V= 120 V
Power= IV
For the toaster;
I= 1820/120 = 15.1 A
For the electric frying pan;
I= 1420/120 = 11.8 A
For the lamp;
55/120 = 0.5 A
Total current = 15.1 +11.8 + 0.5 = 27.4 A
The combination will blow the fuse.
Explanation:
step one:
Given data
power of toaster= 1,820 W
power of electric frying pan= 1,420 W
power of lamp= 55 W
current of the outlet= 15 A
voltage of outlet = 120 V
step two
since all three appliances are connected in parallel to the socket outlet, they will use the same voltage of 120 V and the currents will be different across each appliance,
Hence the current across the Toaster will be I₁
using P=I₁V we have
I₁= P/V
I₁= 1820/120 = 15.16 A
A. The current drawn by each device
the current across the electric frying pan will be I₂
using P=I₂V we have
I₂= P/V
I₂= 1420/120 = 11.83 A
the current across the lamp will be I₃
using P=I₃V we have
I₃= P/V
I₃= 55/120 = 0.45 A
therefore the total current drawn by all appliances will be
Total current = I₁+I₂+I₃= 15.16 +11.83+ 0.45= 27.44
B. Will this combination blow the 15-A fuse?
27.44 A > 15 A by 45% ...and this will make fuse to blow
Complete each of the statements
A. Lines of force are lines used to represent ________ an ________ electric field
B. The intensity of an electric field is the coefficient between the _________ that in the field exerts on a test ___________ located at that point and the value of said charge
C. The electric field is uniform if at any point in the field its _________ and ________ is the same
D. The van der graff generator is a _________ machine which has two __________ that are driven by a _________ that generates a rotation
Answer:
A: magnitude and direction
B: Force that the field exerts on a test charge
C: its magnitude and direction is the same.
D: electrostatic machine
two rollers that are driven by a motor that generates a rotation
Explanation:
In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.
Answer:
1.24Mev
Explanation:
Using
E= hc/lambda
= (6.62x10^-19) x(3x10^8m/s)/(1x10^-12) x 1.602x10^-9
= 1.24Mev
Does the moon light originate from the moon only
Answer:
No
Explanation:
Moon has no light of its own. It just shines because its surface reflects light from the sun and that's what we see.
:-)
What type of energy conversion occurs when you place your feet near the fireplace and they become warm?
O Radiant to thermal
o Thermal to mechanical
O Mechanical to chemical
O Nuclear to thermal
Answer:
It is the first one RADIENT TO THERMAL
Explanation:
The heat emitted from the campfires is an an example of radiant energy and thermal energy is refers to the energy contained within a system that is responsible for its tempreture with in this case is the campfires and heat energy being reflected upon your feet.
Answer:
A
Explanation:
In the lab, you shoot an electron towards the south. As it moves through a magnetic field, you observe the electron curving upward toward the roof of the lab. You deduce that the magnetic field must be pointing:_______.
a. to the west.
b. upward.
c. to the north.
d. to the east.
e. downward.
Answer:
a. to the west.
Explanation:
An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), 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 (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.
An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase the magnitude of the induced emf? A. Straighten the wire out to be flat B. Reduce the resistance of the wire of which the loop is made C. Turning the plane of the loop to be parallel to the magnetic field D. Reducing the diameter of the loop
Answer:
changing the magnetic field more rapidly
Explanation:
According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.
Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.
A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor
Answer: R = 394.36ohm
Explanation: In a LR circuit, voltage for a resistor in function of time is given by:
[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]
ε is emf
L is indutance of inductor
R is resistance of resistor
After 4s, emf = 0.8*19, so:
[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]
[tex]0.8 = e^{-\frac{88}{R} }[/tex]
[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]
[tex]ln(0.8) = -\frac{88}{R}[/tex]
[tex]R = -\frac{88}{ln(0.8)}[/tex]
R = 394.36
In this LR circuit, the resistance of the resistor is 394.36ohms.
Unpolarized light is passed through three successive Polaroid filters, each with its transmission axis at 45.0° to the preceding filter. What percentage of light gets through?
Answer:
The percentage is [tex]k = 12.5 \%[/tex]
Explanation:
From the question we are told that
The axis is is at [tex]\theta = 45 ^o[/tex]
Generally the of intensity light emerging from the first polarizer is mathematically represented as
[tex]I_{1} = \frac{I_o}{ 2}[/tex]
Where [tex]I_o[/tex] is the intensity of unpolarized light
Now the light emerging from the second polarizer is mathematically represented as
[tex]I_2 = I_ 1 * cos ^2(\theta )[/tex]
[tex]I_2 = \frac{I_o}{2} * cos ^2(45 )[/tex]
[tex]I_2 = \frac{I_o}{2} * \frac{1}{2} = \frac{I_o}{4}[/tex]
Now the light emerging from the third polarizer is mathematically represented as
[tex]I_3 = I_ 2 * cos ^2(\theta )[/tex]
[tex]I_3 = \frac{I_o}{4} * cos ^2(45 )[/tex]
[tex]I_3 = \frac{I_o}{8}[/tex]
Now the percentage of the intensity of light that emerged with respect to the intensity of the unpolarized light is
[tex]k = \frac{\frac{I_o}{8} }{I_o } * 100[/tex]
[tex]k = 12.5 \%[/tex]
The percentage of light that gets through the three successive Polaroid filters is; 12.5%
We are given;
Angle of transmission axis; θ = 45°
Formula for intensity of light from first polarizer is;
I₁ = ¹/₂I₀
Formula for intensity of light from second polarizer is;
I₂ = I₁cos²θ
Formula for intensity of light from third polarizer is;
I₃ = I₂cos²(90 - θ)
Combining the 3 equations;
Put ¹/₂I₀ for I₁ in second formula to get;
I₂ = ¹/₂I₀cos²θ
Put ¹/₂I₀cos²θ for I₂ in third formula to get;
I₃ = ¹/₂I₀cos²θ*cos²(90 - θ)
Plugging in 45° for θ gives;
I₃ = ¹/₂I₀cos²45*cos²(90 - 45)
⇒ I₃ = ¹/₂I₀cos²45*cos²45
⇒ I₃ = ¹/₂I₀cos⁴45
Now, cos 45 in surd form is 1/√2. Thus;
I₃ = ¹/₂I₀(1/√2)⁴
I₃ = ¹/₂I₀(¹/₄)
I₃ = ¹/₈I₀
I₃/I₀ = ¹/₈
I₃/I₀ = 0.125
In percentage form, we have;
I₃/I₀ = 12.5%
Read more about unpolarized light at; https://brainly.com/question/1444040
Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of 0.860 eV from sodium metal. What is the binding energy of electrons to sodium metal?
Answer:
Binding Energy = 2.24 eV
Explanation:
First, we need to find the energy of the photon of light:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)
E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 3.1 eV
Now, from Einstein's Photoelectric Equation:
E = Binding Energy + Kinetic Energy
Binding Energy = E - Kinetic Energy
Binding Energy = 3.1 eV - 0.86 eV
Binding Energy = 2.24 eV
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
A hydraulic system is being used to lift a 1500-kg car. If the large piston under the car has a diameter of 50 cm, the small piston has a diameter of 4.0 cm, and the car is lifted a distance of 1.3 m, how much work is done on the car
Answer:
W = 122.3 J
Explanation:
First, we need to find out the force applied to the smaller piston. We know that the pressure applied to smaller piston must be equally transmitted to the larger piston. Therefore,
P₁ = P₂
F₁/A₁ = F₂/A₂
F₂ = F₁(A₂/A₁)
where,
F₁ = Force of Larger Piston = Weight of car = mg = (1500 kg)(9.8 m/s²)
F₁ = 14700 N
F₂ = Force applied to smaller piston = ?
A₁ = Area of larger piston = πd₁²/4
A₂ = Area of smaller piston = πd₂²/4
Therefore,
F₂ = (14700 N)[(πd₂²/4)/(πd₁²/4)]
F₂ = (14700 N)(d₂²/d₁²)
where,
d₁ = diameter of large piston = 50 cm
d₂ = diameter of small piston = 4 cm
Therefore,
F₂ = (14700 N)[(4 cm)²/(50 cm)²]
F₂ = 94.08 N
Now, for the work done on the car:
Work Done = W = F₂ d
where,
d = displacement of car = 1.3 m
Therefore,
W = (94.08 N)(1.3 m)
W = 122.3 J
The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1 % change in length.
A. Young's modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?
Answer:
a
[tex]k = 11600000 N/m[/tex]
b
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
c
[tex]F = 3750.28 \ N[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]E = 1.4 *10^{10} \ N/m^2[/tex]
The length is [tex]L = 0.35 \ m[/tex]
The area is [tex]2.9 \ cm^2 = 2.9 *10^{-4} \ m ^2[/tex]
Generally the force acting on the tibia is mathematically represented as
[tex]F = \frac{E * A * \Delta L }{L}[/tex] derived from young modulus equation
Now this force can also be mathematically represented as
[tex]F = k * \Delta L[/tex]
So
[tex]k = \frac{E * A }{L}[/tex]
substituting values
[tex]k = \frac{1.4 *10^{10} * 2.9 *10^{-4} }{ 0.35}[/tex]
[tex]k = 11600000 N/m[/tex]
Since the tibia support half the weight then the force experienced by the tibia is
[tex]F_k = \frac{750 }{2} = 375 \ N[/tex]
From the above equation the extension (compression) is mathematically represented as
[tex]\Delta L = \frac{ F_k * L }{ A * E }[/tex]
substituting values
[tex]\Delta L = \frac{ 375 * 0.35 }{ (2.9 *10^{-4}) * 1.4*10^{10} }[/tex]
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
From the above equation the maximum force is
[tex]F = \frac{1.4*10^{10} * (2.9*10^{-4}) * 3.233*10^{-5} }{ 0.35}[/tex]
[tex]F = 3750.28 \ N[/tex]
NASA is doing research on the concept of solar sailing. A solar sailing craft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion.
A) Should the sail be absorptive or reflective? Why?
B)The total power output of the sun is 3.90 × 1026 W . How large a sail is necessary to propel a 1.06 × 104 kg spacecraft against the gravitational force of the sun?
Answer:
A = 6.8 km²
Explanation:
A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.
B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;
F_rad = 2IA/c
I is given by the formula;
I = P/(4πr²)
Thus;
F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²
Where;
A is the area of the sail
r is the distance of the sail from the sun
c is the speed of light = 3 × 10^(8) m/s
P is total power output of the sun = 3.90 × 10^(26) W
Now,F_rad = F_g
Where F_g is gravitational force.
Thus;
PA/2cπr² = G•m•M_sun/r²
r² will cancel out to givw;
PA/2cπ = G•m•M_sun
Making A the subject, we have;
A = (2•c•π•G•m•M_sun)/P
Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg
G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²
Thus;
A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))
A = 6.8 × 10^(6) m² = 6.8 km²
A plano-convex glass lens of radius of curvature 1.4 m rests on an optically flat glass plate. The arrangement is illuminated from above with monochromatic light of 520-nm wavelength. The indexes of refraction of the lens and plate are 1.6. Determine the radii of the first and second bright fringes in the reflected light.
Given that,
Radius of curvature = 1.4 m
Wavelength = 520 nm
Refraction indexes = 1.6
We know tha,
The condition for constructive interference as,
[tex]t=(m+\dfrac{1}{2})\dfrac{\lambda}{2}[/tex]
Where, [tex]\lambda=wavelength[/tex]
We need to calculate the radius of first bright fringes
Using formula of radius
[tex]r_{1}=\sqrt{2tR}[/tex]
Put the value of t
[tex]r_{1}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(0+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=0.603\ mm[/tex]
We need to calculate the radius of second bright fringes
Using formula of radius
[tex]r_{2}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(1+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=1.04\ mm[/tex]
Hence, The radius of first bright fringe is 0.603 mm
The radius of second bright fringe is 1.04 mm.
A particle undergoes damped harmonic motion. The spring constant is 100 N/m, the damping constant is 8.0 x 10-3 kg.m/s, and the mass is 0.050 kg. If the particle starts at its maximum displacement, x = 1.5 m, at time t = 0. What is the amplitude of the motion at t = 5.0 s?
Answer:
The amplitude [tex]A(5) = 1 \ m[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 100 \ N/m[/tex]
The damping constant is [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]
The mass is [tex]m = 0.050 \ kg[/tex]
The maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]
The time considered is t = 5.0 s
Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as
[tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]
substituting values
[tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]
[tex]A(5) = 1 \ m[/tex]
Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, which is 300 K. (Use 300 K as the temperature of the hot reservoir of the engine). The heat capacity of the brick is C
Answer
Work done is 57.9KJ
Explanation
First solve the problem according to work done due to variation in temperature
So W= intergral Cu( 1-Tu/T). at Tu and T
So Given that
C = Heat capacity of the Brick
TEPc= Cold Temperature
TEPh = Hot Temperature
W = C ( TEPh-TEP) - TEPhCln ( TEPh/TEPc)
So
W= (1)-(300-150)-300 (1) ln 2
W= -57.9KJ
To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.68 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.
Answer:
1.8/0.61 =2.95 ft
Hope it helped u if yes mark me BRAINLIEST!
Tysm!
;)
How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled?
Answer:
a) the distance between the interference fringes is reduced by half
b) the distance between stripes is doubled
Explanation:
Interference experiments constructive interference is described by the expression
d sin θ = m λ
let's use trigonometry to find the distance between the interference fringes
tan θ= y / L
dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ / d L
a) it asks us when the screen doubles its distance
L ’= 2 L
subtitute in the equation
y ’= m λ / (d 2L)
y ’=( m λ / d L) /2
y ’= y / 2
the distance between the interference fringes is reduced by half
b) the density of the network doubles
if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half
d ’= d / 2
we substitute
y ’= m λ (L d / 2)
y ’= m λ / (L d) 2
y ’= y 2
the distance between stripes is doubled
Describe how, using a positively-charged rod and two neutral metal spheres, we canmake one sphere positive without touching it to the rod. You might want to draw adiagram to help you.
Answer:
se the principle of induction.
place the two metallic spheres together, now we bring the positively charged bar closer to the first sphere.
The charge that was induced in the sphere is distributed as infirm as possible,
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive
Explanation:
For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.
Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.
Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.
The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.
