Answer:
E. Coefficient of linear expansion
2. The nuclear model of the atom held that
a. electrons were randomly spread through "a sphere of uniform positive
electrification."
b. matter was made of tiny electrically charged particles that were smaller than the
atom
C. matter was made of tiny, indivisible particles.
d. the atom had a dense, positively charged nucleus.
Answer:
the atom had a dense, positively charged nucleus.
Explanation:
Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.
According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.
g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before interacting. From this information, find the time interval required for the strong interaction to occur.
Answer:
Time, [tex]t=1.07\times 10^{-23}\ s[/tex]
Explanation:
Given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel [tex]3.2\times 10^{-15}\ m[/tex] before interacting.
Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s[/tex]
So, the time interval is [tex]1.07\times 10^{-23}\ s[/tex]
A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and with mass 3.70 kg , while the balls each have mass 0.700 kg and can be treated as point masses.
Required:
Find the moment of inertia of this combination about an axis
a. perpendicular to the bar through its center.
b. perpendicular to the bar through one of the balls.
c. parallel to the bar through both balls.
d. parallel to the bar and 0.500 m from it.
Answer:
Explanation:
a )
moment of inertia in the first case will be sum of moment of inertia of two balls + moment of inertia of bar
= 2 x .700 x (2.1 / 2 )² + 3.7 x 2.1² / 12
= 1.5435 + 1.35975
= 2.90325 kg m²
b )
moment of inertia required
= moment of inertia of bar + moment of inertia of the other ball
= 3.70 x (2.1² / 3 ) + .7 x 2.1²
= 5.439 + 3.087
= 8.526 kg m²
c )
In this case moment of inertia of the combination = 0 as distance of masses from given axis is zero .
d )
masses = 3.7 + .7 = 4.4 kg
distance from axis = .5 m
moment of inertia about given axis
= 4.4 x .5²
= 1.1 kg m².
A heat engine operates between 200 K and 100 K. In each cycle it takes 100 J from the hot reservoir, loses 25 J to the cold reservoir, and does 75 J of work. This heat engine violates the second law but not the first law of thermodynamics. Why is this true?
Answer:
It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J
But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law
Explain why the two plates of a capacitor are charged to the same magnitude when a battery is connected to the capacitor.
Answer:
This is because the same electron removed from the positively charged plate is what is taken to the negatively charged plate, maintaining the same amount of electron according to the conservation of charge in an electric circuit.
Explanation:
In any circuit, electrons are neither created nor destroyed according to the laws of conservation of charge, but are transferred from one point to another on the circuit. When the plates of a capacitor are connected to battery, the battery pushes the electron to move due to its potential difference. Electrons are then moved from the positive plate, at a steady rate, to the negative plate. The removal of electrons from the positive plate is what leaves it positively charged from deficiency of electrons, and the addition of electrons at the negatively charged plate is what leaves the plate negatively charge from excess of electrons. From this, we can see that the same electrons removed from the positively charged plate are taken to the negatively charged plate.
which one is more powerful hydrogen bomb or atom bomb and why?
Hydrogen bomb is more powerful than atom bomb
Hydrogen has a calorie value of 150000KJ .It is very much than nuclear bomb or atom bombScientists also told that Hydrogen bomb is more powerful.But both bombs are destructive.12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and size of the image. (3)
Answer:
I think 9.5
Explanation:
............
A sample of gas is enclosed in a container of fixed volume. Identify which of the following statements are true. Check all that apply.If the container is heated, the gas particles will lose kinetic energy and temperature will increase.
Answer:
B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.
C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.
E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.
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In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than 540 nmnm . Part A What is the work function of this material
Answer:
Φ = 36.84 × 10^(-20) J
Explanation:
In the photoelectric effect, the energy of the incoming photon is usually used in part to extract the photoelectron from the material (work function) and then the rest is converted into kinetic energy of the photoelectron which is given by the formula;
K_max = hf - Φ
where;
hf represents the energy of the incoming photon
h is the Planck's constant
f is the light frequency
Φ is the work function of the material
K_max is the maximum kinetic energy of the photoelectrons.
From the question, we are told that no current flows unless the wavelength is less than 540 nm. This means that when the wavelength has this value, the maximum kinetic energy of the photoelectrons is zero i.e K_max = 0. Thus the energy of the incoming photons is just enough to extract the photoelectrons from the material.
Thus,
hf - Φ = 0
hf = Φ - - - (1)
We are given the wavelength as;
λ = 540 nm = 540 × 10^(-9) m
Now, let's find the frequency of the light by using the relationship between frequency and wavelength. The equation is;
f = c/λ
Where c is speed of light = 3 × 10^(8) m/s
f = (3 × 10^(8))/(540 × 10^(-9))
f = 5.56 × 10^(14) Hz
Thus, from equation 1,we can now find the work function;
Φ = hf
h is Planck's constant and has a value of 6.626 × 10^(-34) J.s
Thus;
Φ = 6.626 × 10^(-34) × 5.56 × 10^(14)
Φ = 36.84 × 10^(-20) J
(4) Use the preliminary observations to answer these questions; Compared to no polarizer or analyzer in the optical path, by what percent does the light intensity decrease when (a) The polarizer is introduced into the optical path? (b) The both polarizer and analyzer are introduced into the optical path?
Answer:
a) I = I₀/2, b) I = I₀/2 cos² θ
Explanation:
To answer these questions, let's analyze a little the way of working of a polarized
* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is
i = I₀ / 2
and is polarized in the direction of the polarizer
* when a polarized light reaches the analyzer it must comply with Malus's law
I = I₁ cos² θ
where the angle is between the polarized light and the analyzer.
With this, let's answer the questions
a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of
I = I₀/2
b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is
I = I₀/2 cos² θ
where the final value depends on the angle between the polarizer and the analyzer.
Let's look at two extreme cases
θ = 0 I = Io / 2
θ = 90º I = 0
How to do this question
Answer:
(a) 10 m/s
(b) 22.4 m/s
Explanation:
(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.
Sum of forces in the centripetal direction (towards the center):
∑F = ma
mg + N = mv²/r
At minimum speed, the normal force is 0.
mg = mv²/r
g = v²/r
v = √(gr)
v = √(10 m/s² × 10.0 m)
v = 10 m/s
(b) Energy is conserved.
Initial kinetic energy + initial potential energy = final kinetic energy
½ mv₀² + mgh = ½ mv²
v₀² + 2gh = v²
(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²
v = 22.4 m/s
An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -Imax, where Imax is the current amplitude. What is the unknown element?
Answer:
Capacitor, is the right answer.
Explanation:
The unknown element is a Capacitor.
Below is the calculation that proves that it is a capacitor.
We know that for the Capacitor
i = Imax × sin(wt+(pi/2)).
i = Imax × sin ((2 × pi/T) × (T/4) + (pi/2))
i = Imax × sin(3.142) = 0 A
at, t = T/2
wt = (2 × pi/T) × (T/2) = pi
wt + (pi/2) = pi + (pi/2) = ( 3 × pi/2) =
i = Imax × sin(3 × pi/2) = -Imax
Which is in a correct agreement with capacitor therefore, the answer is a Capacitor.
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
1. A 0.430kg baseball comes off a bar and goes straight up in the air. At a height of 10.0m, the baseball has a speed of 25.3m/s. Determine the mechanical energy at the height. Show all your work. 2. What is the baseball's mechanical energy when it is at a height of 8.0m? Explain?
Answer:
180 J
Explanation:
Mechanical energy = kinetic energy + potential energy
ME = KE + PE
ME = ½ mv² + mgh
ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)
ME = 180 J
Mechanical energy is conserved, so it is 180 J at all points of the trajectory.
The baseball's mechanical energy when it is at a height of 8.0m is 180 J.
What is mechanical energy?The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time. Mechanical energy is always conserved.
Mechanical energy = kinetic energy + potential energy
Given is the mass of baseball m= 0.430 kg, height h =10m, speed v= 25.3m/s.
ME = KE + PE
ME = ½ mv² + mgh
Substitute the values, we get
ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)
ME = 180 J
Thus, the baseball's mechanical energy when it is at a height of 8.0m is 180 J.
Learn more about mechanical energy.
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A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object and the screen. In terms of f, what is the smallest value d can have for an image to be in focus on the screen?
Answer:
2f
Explanation:
The formula for the object - image relationship of thin lens is given as;
1/s + 1/s' = 1/f
Where;
s is object distance from lens
s' is the image distance from the lens
f is the focal length of the lens
Total distance of the object and image from the lens is given as;
d = s + s'
We earlier said that; 1/s + 1/s' = 1/f
Making s' the subject, we have;
s' = sf/(s - f)
Since d = s + s'
Thus;
d = s + (sf/(s - f))
Expanding this, we have;
d = s²/(s - f)
The derivative of this with respect to d gives;
d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²
Equating to zero, we have;
(2s/(s - f)) - s²/(s - f)² = 0
(2s/(s - f)) = s²/(s - f)²
Thus;
2s = s²/(s - f)
s² = 2s(s - f)
s² = 2s² - 2sf
2s² - s² = 2sf
s² = 2sf
s = 2f
5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be changed to raise the wave speed to 180 m/s?
Answer:
The tension on string when the speed was raised is 134.53 N
Explanation:
Given;
Tension on the string, T = 120 N
initial speed of the transverse wave, v₁ = 170 m/s
final speed of the transverse wave, v₂ = 180 m/s
The speed of the wave is given as;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
μ is mass per unit length
[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]
The final tension T₂ will be calculated as;
[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]
Therefore, the tension on string when the speed was raised is 134.53 N
A 6.7 cm diameter circular loop of wire is in a 1.27 T magnetic field. The loop is removed from the field in 0.16 ss . Assume that the loop is perpendicular to the magnetic field.
Required:
What is the average induced emf?
Answer:
The induced emf is [tex]\epsilon = 0.0280 \ V[/tex]
Explanation:
From the question we are told
The diameter of the loop is [tex]d = 6.7 cm = 0.067 \ m[/tex]
The magnetic field is [tex]B = 1.27 \ T[/tex]
The time taken is [tex]dt = 0.16 \ s[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{\Delta \phi}{dt}[/tex]
Where N = 1 given that it is only a circular loop
[tex]\Delta \phi = \Delta B * A[/tex]
Where [tex]\Delta B = B_f - B_i[/tex]
where [tex]B_i[/tex] is 1.27 T given that the loop of wire was initially in the magnetic field
and [tex]B_f[/tex] is 0 T given that the loop was removed from the magnetic field
Now the area of the of the loop is evaluated as
[tex]A = \pi r^2[/tex]
Where r is the radius which is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.067}{2}[/tex]
[tex]r = 0.0335 \ m[/tex]
So
[tex]A = 3.142 * (0.0335)^2[/tex]
[tex]A = 0.00353 \ m^2[/tex]
So
[tex]\Delta \phi = (0- 127)* (0.00353)[/tex]
[tex]\Delta \phi = -0.00448 Weber[/tex]
=> [tex]\epsilon = - 1 * \frac{-0.00448}{0.16}[/tex]
=> [tex]\epsilon = 0.0280 \ V[/tex]
You measure the power delivered by a battery to be 4.26 W when it is connected in series with two equal resistors. How much power will the same battery deliver if the resistors are now connected in parallel across it
Answer:
The power delivered by the battery is 17.04 W
Explanation:
Power through a circuit is given as
P = IV ....1
where P is the power
I is the current through the circuit
V is the voltage through the circuit
The voltage in a circuit is given as
V = IR ....2
Let us take the value of each resistor as equal to R
when connected in series, the total resistance will be
[tex]R_{t}[/tex] = R + R = 2R
If we assume constant voltage through the circuit, then from equation 2, the current in this case is
I = V/2R
If the resistors are connected in parallel, then the total resistance will be
[tex]\frac{1}{R_{t} }[/tex] = [tex]\frac{1}{R}[/tex] +
[tex]R_{t}[/tex] = R/2
The current in this case will be increased since the resistance is reduced
I = 2V/R
comparing the two situations, we can see that the current increased when connected in parallel to a ratio of
[tex]\frac{2V}{R}[/tex] ÷ [tex]\frac{V}{2R}[/tex] =
This means that the current increased 4 times
From equation 1, we can see that electrical power is proportional to the current at a constant voltage, therefore, the power will also increase by four times to
P = 4 x 4.26 = 17.04 W
A stone is dropped from the upper observation deck of a tower, 50 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t. h(t) = (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) s (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) m/s (d) If the stone is thrown downward with a speed of 9 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)
Answer:
A. Using displacement =Ut + 1/2gt²
=> 0 + 1/2 (-9.8)t²
= -4.9t²
So
h(t) = 50+ displacement
= 50 - 4.9t²
B. To reach the ground
h(t) = 0
So
50-4.9t²= 0
t = √ (50/4.9)
= 3.2s
C. Using
V = u+ gt
U= 0
V= - 9.8(3.2)
= 31.4m/s
D. If u = -9m/s
Then s = ut + 1/2gt²
5t- 1/2gt²
But distance from the ground is
=.> 50-5t- 4.8t²= 0
So t solving the quadratic equation
t= 3.58s
(a) The distance of the stone above the ground level at time t is [tex]h(t) = 50 - 4.9t^2[/tex]
(b) The time taken for the stone to strike the ground is 3.19 s.
(c) The velocity of the stone when it strikes the ground is 31.4 m/s.
(d) The time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
The given parameters;
height above the ground, h₀ = 50 mThe distance of the stone above the ground level at time t is calculated as;
[tex]h(t) = h_0 - ut - \frac{1}{2} gt^2\\\\h(t) = 50 - 0 -0.5\times 9.8t^2\\\\h(t) = 50 - 4.9t^2[/tex]
The time taken for the stone to strike the ground is calculated as;
[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 50}{9.8} } \\\\t = 3.19 \ s[/tex]
The velocity of the stone when it strikes the ground is calculated as;
[tex]v =u + gt\\\\v = 0 + 3.2 \times 9.8\\\\v = 31.4 \ m/s[/tex]
The time taken for the stone to reach the ground when thrown at speed of 9 m/s is calculated as;
[tex]50 = 9t + \frac{1}{2} (9.8)t^2\\\\50 = 9t + 4.9t^2\\\\4.9t^2 + 9t - 50 = 0\\\\a = 4.9 \, \ b = 9, \ \ c = -50\\\\solve \ the \ quadratic \ equation\ using \ formula \ method\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-9 \ \ + /- \ \ \sqrt{(9)^2 - 4(4.9 \times -50)} }{2(4.9)} \\\\t = 2.41 \ s \ \ or \ \ - 4.24 \ s[/tex]
Thus, the time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
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At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to:_____
(a) 3 times that of the Celsius and
(b) 1/5 times that of the Celsius
Answer:
C = 26.67° and F = 80°C = -20° and F = -4°Explanation:
Find:
3 times that of the Celsius and 1/5 times that of the CelsiusComputation:
F = (9/5)C + 32
3 times that of the Celsius
If C = x
So F = 3x
So,
3x = (9/5)x + 32
15x = 9x +160
6x = 160
x = 26.67
So, C = 26.67° and F = 80°
1/5 times that of the Celsius
If C = x
So F = x/5
So,
x/5 = (9/5)x + 32
x = 9x + 160
x = -20
So, C = -20° and F = -4°
A car travels down the road for 535 m in 17.3 s. What is the velocity of the car in m/s and in km/h?
Answer:
30.92m/sExplanation:
[tex]Distance = 535m\\Time = 17.3s\\\\Velocity = \frac{Distane}{Time} \\\\V = \frac{535m}{17.3s} \\\\Velocity = 30.92m/s[/tex]
[tex]Distance = 535m\\\\535m \:to \: km=0.535km\\\\Time = 17.3s\\\\17.3s = 0.004805556hours\\\\Velocity = \frac{Distance}{Time}\\\\ V= \frac{0.535}{0.004805556} \\\\ V=111.329469472\\\\=111.33km/h[/tex]
What is the force that attracts objects with mass toward each other?
Explanation:
gravitional force attracts objects with mass toward each other.
A fan rotating with an initial angular velocity of 1500 rev/min is switched off. In 2.5 seconds, the angular velocity decreases to 400 rev/min. Assuming the angular acceleration is constant, answer the following questions.
How many revolutions does the blade undergo during this time?
A) 10
B) 20
C) 100
D) 125
E) 1200
Answer:
The blade undergoes 40 revolutions, so neither of the given options is correct!
Explanation:
The revolutions can be found using the following equation:
[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
Where:
α is the angular acceleration
t is the time = 2.5 s
[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min
First, we need to find the angular acceleration:
[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]
Now, the revolutions that the blade undergo are:
[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]
Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!
I hope it helps you!
Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.)
(a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?
Answer:
a
[tex]\lambda = 1.667 nm[/tex]
b
[tex]\theta = 0.8681^o[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]
The is distance of the screen from the slit is [tex]D = 2.60 \ m[/tex]
The distance between the central bright fringe and either of the adjacent bright [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]
Generally the condition for constructive interference is
[tex]d sin \tha(\theta ) = n \lambda[/tex]
From the question we are told that small-angle approximation is valid here.
So [tex]sin (\theta ) = \theta[/tex]
=> [tex]d \theta = n \lambda[/tex]
=> [tex]\theta = \frac{n * \lambda }{d }[/tex]
Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes
Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as
[tex]y = D * sin (\theta )[/tex]
From the question we are told that small-angle approximation is valid here.
So
[tex]y = D * \theta[/tex]
=> [tex]\theta = \frac{ y}{D}[/tex]
So
[tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]
[tex]\lambda =\frac{d * y }{n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]
[tex]\lambda = 1.667 *10^{-6}[/tex]
[tex]\lambda = 1.667 nm[/tex]
In the b part of the question we are considering the next set of bright fringe so n= 2
Hence
[tex]dsin (\theta ) = n \lambda[/tex]
[tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]
[tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]
[tex]\theta = 0.8681^o[/tex]
The Milky Way has a diameter (proper length) of about 1.2×105 light-years. According to an astronaut, how many years would it take to cross the Milky Way if the speed of the spacecraft is 0.890 c?
Answer:
t = 134834.31 years
Explanation:
First we find the speed of the ship:
v = 0.890 c
where,
v = speed of the ship = ?
c = speed of light = 3 x 10⁸ m/s
Therefore, using the values, we get:
v = (0.89)(3 x 10⁸ m/s)
v = 2.67 x 10⁸ m/s
Now, we find the distance in meters:
Distance = s = (1.2 x 10⁵ light years)(9.461 x 10¹⁵/1 light year)
s = 11.35 x 10²⁰ m
Now, for the time we use the following equation:
s = vt
t = s/v
t = (11.35 x 10²⁰ m)/(2.67 x 10⁸ m/s)
t = (4.25 x 10¹² s)(1 h/3600 s)(1 day/24 h)(1 year/365 days)
t = 134834.31 years
Question 2
A) A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released? (1 point)
The spring exerts a restoring force to the right and compresses even further
The spring exerts a restoring force to the left and returns to its equilibrium position
The spring exerts a restoring force to the right and returns to its equilibrium position
The spring exerts a restoring force to the left and stretches beyond its equilibrium position
1. Which example best describes a restoring force?
B) the force applied to restore a spring to its original length
2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?
C) The spring exerts a restoring force to the left and returns to its equilibrium position.
3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?
D) 1 m
4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?
D)It is a vector quantity.
5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?
A) It decreases in magnatude.
A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its trajectory, it explodes into two equal-mass fragments. One reaches the ground t1 = 2.71s after the explosion.When does the second reach the ground?t=?
Answer:
6.13 seconds
Explanation:
At the peak of the fireworks trajectory, the velocity of the firework would be zero. Using equation of motion, we have:
v² = u² + 2gh
0 = 40² - (2)(9.81)(h)
0 = 1600 - 19.62h
19.62h = 1600
h = 1600/19.62
h = 81.55 m
Now during the process of explosion, the two parts gained equal vertical momentum but in opposite directions.
We are told the first piece lands in a time of 2.71 s,
Using 3rd equation of motion, we have;
h = ut + ½gt²
81.55 = u(2.71) + ½(9.81 × 2.71²)
81.55 = 2.71u + 36.0228
2.71u = 81.55 - 36.0228
2.71u = 45.5272
u = 45.5272/2.71
u = 16.8 m/s
The time it takes a projectile to return back to its original launch point assuming the projectile was launched
vertically with speed u = 16.8 m/s is;
t = 2u/g
t = (2 × 16.8)/9.81
t = 3.43 s
Thus total time it takes the second mass to reach the ground = 3.43 + 2.71 = 6.13 seconds
Snell's Law: Light goes from material having a refractive index of n 1 into a material with refractive index n 2. If the refracted light is bent away from the normal, what can you conclude about the indices of refraction
Answer:
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
Explanation:
The expression for the angle of refraction is
n₁ sin θ₁ = n₂ sin θ₂
refractive index n₁ is for incident light and n₂ is for transmitted light.
We have two cases
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
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.
Peer assessment is a unique educational model. Think back to how you felt about peer assessment at the beginning of the term, and compare that to your feeling now. How have your feeling changed? Are you more comfortable with peer assessment? Have you learned something new while assessing your peer's work?
Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.