Answer:
The angle is [tex]\theta = 6.93 *10^{1}[/tex]
Explanation:
From the question we are told that
The intensity of light emerging from the second polarizer is [tex]I_2 = \frac{1}{8} * I_1[/tex]
Now the light emerging from the first polarizer is [tex]I_1 = \frac{I_o}{2}[/tex]
Where [tex]I_o[/tex] is the intensity of the unpolarized light
Now according to Malus law the intensity of light emerging from the second polarizer is mathematically represented as
[tex]I_2 = \frac{I_1}{8} = I_1 cos^2(\theta )[/tex]
=> [tex]I_2 = \frac{I_1}{8} = I_1 cos^2(\theta )[/tex]
=> [tex]\theta = cos^{-1}[0.3536][/tex]
=> [tex]\theta = 69.3^o[/tex]
=> [tex]\theta = 6.93 *10^{1}[/tex]
Marcel is helping his two children, Jacques and Gilles, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child, Jacques, simply sinking to the ground. Given that Jacques, whose weight is WWW, is sitting at distance LLL to the left of the pivot, at what distance L1L1L_1 should Marcel place Gilles, whose weight is www, to the right of the pivot to balance the seesaw
Answer:
L1 = WL/w
Explanation:
Jacques's weight = W
Jacques's distance from the pivot = L
Gilles's weight = w
Gilles's distance from the pivot must be L1
For the two boy to balance each other, they must generate equal amount of torques around the pivot
Torque = distance x weight
For Jacques, the torque generated = WL
For Gilles, the torque generated = wL1
Balancing the two torques, we have
WL = wL1
the distance L1 must be equal to
L1 = WL/w
Which of the following landforms would you NOT expect to see in the Inland South?
a) Swamplands
b) Flat, sandy plains
c) Rugged mountains
d) Rolling hills
e) Glaciers
Answer:
Option: e) Glaciers
Explanation:
Historically, the Inland South viewed as a backwater of the United States. The Inland South is known for its Appalachians mountains, swampland. Flatland, sandy soils, and meandering rivers are some of its features. The climate is hot and humid, which allow the area to grow as Pantanal floodplains. Glaciers will be one of the landforms that one will not expect to see in the Inland South.
Read the passage about the pygmy shrew.
The pygmy shrew is the smallest mammal in North America. However, when comparing the amount of food eaten to its body weight, the pygmy shrew eats more food than any other mammal. It will consume two to three times its own body weight in food daily. One explanation is that the pygmy shrew uses energy at a high rate. In fact, its heart beats over one thousand times per minute.
What is the best explanation for what happens to the food's mass and energy when it is consumed by the pygmy shrew?
Answer:
A very high metabolism and a very small size.
Explanation:
The pygmy shrew is a very small mammal, that forages day and night. The metabolism of the Pygmy shrew is so high that it must eat at least every 30 minutes or it might die. The best explanation for what happens to the food's mass and energy is that most of the food mass is rapidly used fro building up of the shrew due to its very high metabolism, and a bigger portion of the food is lost from the surface of the body of the shrew, due to its very small size. The combination of these two factors; a very high metabolism (rapidly uses up food material, and generates a large amount of heat in a very short time) and the very small size (makes heat loss due to surface area to volume ratio high) explains what happens to the food mass and energy.
A 1.70 kg block slides on a horizontal, frictionless surface until it encounters a spring with a force constant of 955 N/m. The block comes to rest after compressing the spring by a distance of 4.60 cm. The other end of the spring is attached to a wall. Find the initial speed of the block.
Answer:
The initial speed of the block is 1.09 m/s
Explanation:
Given;
mass of block, m = 1.7 kg
force constant of the spring, k = 955 N/m
compression of the spring, x = 4.6 cm = 0.046 m
From principle of conservation of energy
kinetic energy of the block = elastic potential energy of the spring
¹/₂mv² = ¹/₂kx²
mv² = kx²
[tex]v = \sqrt{\frac{kx^2}{m} }[/tex]
where;
v is the initial speed of the block
x is the compression of the spring
[tex]v = \sqrt{\frac{955*(0.046)^2}{1.7} } \\\\v = 1.09 \ m/s[/tex]
Therefore, the initial speed of the block is 1.09 m/s
The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8.00s, starting from rest?
Answer:
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 2.50 \ kg \cdot m^2[/tex]
The final angular speed is [tex]w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s[/tex]
The time taken is [tex]t = 8.0 s[/tex]
The initial angular speed is [tex]w_i = 0\ rad/s[/tex]
Generally the average angular acceleration is mathematically represented as
[tex]\alpha = \frac{w_f - w_i }{t}[/tex]
=> [tex]\alpha = \frac{41.89}{8}[/tex]
=> [tex]\alpha = 5.24 \ rad/s^2[/tex]
Generally the torque is mathematically represented as
[tex]\tau = I * \alpha[/tex]
=> [tex]\tau = 5.24 * 2.50[/tex]
=> [tex]\tau = 13.09 \ N \cdot m[/tex]
In a container of negligible mass, 020 kg of ice at an initial temperature of - 40.0 oC is mixed with a mass m of water that has an initial temperature of 80.0 oC. No heat is lost to the surroundings. If the final temperature of the system is 20.0 oC, what is the mass m of the water that was initially at 80.0 oC
Answer:
The mass is [tex]m_w = 0.599 \ kg[/tex]
Explanation:
From the question we are told that
The mass of ice is [tex]m_c = 0.20 \ kg[/tex]
The initial temperature of the ice is [tex]T_i = -40.0 ^oC[/tex]
The initial temperature of the water is [tex]T_{iw} = 80^o C[/tex]
The final temperature of the system is [tex]T_f = 20^oC[/tex]
Generally according to the law of energy conservation,
The total heat loss is = total heat gained
Now the total heat gain is mathematically represented as
[tex]H = H_1 + H_2 + H_3[/tex]
Here [tex]H_1[/tex] is the energy required to move the ice from [tex]-40^oC \to 0^oC[/tex]
And it mathematically evaluated as
[tex]H_1 = m_c * c_c * \Delta T[/tex]
Here the specific heat of ice is [tex]c_c = 2100 \ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]
So
[tex]H_1 = 0.20 * 2100 * (0-(-40))[/tex]
[tex]H_1 = 16800\ J[/tex]
[tex]H_2[/tex] is the energy to melt the ice
And it mathematically evaluated as
[tex]H_2 = m * H_L[/tex]
The latent heat of fusion of ice is [tex]H_L = 334 J/g = 334 *10^{3} J /kg[/tex]
So
[tex]H_2 = 0.20 * 334 *10^{3}[/tex]
[tex]H_2 = 66800 \ J[/tex]
[tex]H_3[/tex] is the energy to raise the melted ice to [tex]20^oC[/tex]
And it mathematically evaluated as
[tex]H_3 = m_c * c_w * \Delta T[/tex]
Here the specific heat of water is [tex]c_w= 4190\ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]
[tex]H_3 = 0.20 * 4190* (20-0))[/tex]
[tex]H_3 = 16744 \ J[/tex]
So
[tex]H = 16800 + 66800 + 16744[/tex]
[tex]H = 100344\ J[/tex]
The heat loss is mathematically evaluated as
[tex]H_d = m * c_h ( 80 - 20 )[/tex]
[tex]H_d = m_w * 4190 * ( 80 - 20 )[/tex]
[tex]H_d = 167600 m_w[/tex]
So
[tex]167600 m_w = 100344[/tex]
=> [tex]m_w = 0.599 \ kg[/tex]
Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
If she runs fhe same course again, what constant speed would let her finish in the same time as in the first race?
Answer:
The velocity is [tex]v = 4.76 \ m/s[/tex]
Explanation:
From the question we are told that
The first distance is [tex]d_1 = 4.0 \ km = 4000 \ m[/tex]
The first speed is [tex]v_1 = 5.0 \ m/s[/tex]
The second distance is [tex]d_2 = 1.0 \ km = 1000 \ m[/tex]
The second speed is [tex]v_2 = 4.0 \ m/s[/tex]
Generally the time taken for first distance is
[tex]t_1 = \frac{d_1 }{v_1 }[/tex]
[tex]t_1 = \frac{4000}{5}[/tex]
[tex]t_1 = 800 \ s[/tex]
The time taken for second distance is
[tex]t_1 = \frac{d_2 }{v_2 }[/tex]
[tex]t_1 = \frac{1000}{4}[/tex]
[tex]t_1 = 250 \ s[/tex]
The total time is mathematically represented as
[tex]t = t_1 + t_2[/tex]
=> [tex]t = 800 + 250[/tex]
=> [tex]t = 1050 \ s[/tex]
Generally the constant velocity that would let her finish at the same time is mathematically represented as
[tex]v = \frac{d_1 + d_2}{t }[/tex]
=> [tex]v = \frac{4000 + 1000}{1050 }[/tex]
=> [tex]v = 4.76 \ m/s[/tex]
The constant speed that will let her finish in the same time as in the first race is 4.76 m/s
Determination of the time taken for first 4 KmDistance = 4 Km = 4 × 1000 = 4000 mSpeed = 5 m/sTime 1 =?
Time 1 = distance / speed
Time 1 = 4000 / 5
Time 1 = 800 s
Determination of the time taken for the last 1 KmDistance = 1 Km = 1 × 1000 = 1000 mSpeed = 4 m/sTime 2 =?Time 2 = distance / speed
Time 2 = 1000 / 4
Time 2 = 250 s
Determination of the constant speedTotal distance = 4000 + 1000 = 5000 mTotal time = 800 + 250 = 1050 sConstant speed =?Constant speed = Total distance / total time
Constant speed = 5000 / 1050
Constant speed = 4.76 m/s
Learn more about average speed:
https://brainly.com/question/8819317
Let a metallic rod 20 cm in length be heated to a uniform temperature of 100°C Suppose that at t=0 the ends of the bar are plunged into an ice bath at 0°C and thereafter maintained at this temperature, but that no heat is allowed to escape through the lateral surface. Find the time that will elapse before the center of the bar cools to a temperature of 5°C if the bar is made of (a) silver (b) aluminum (c) cast iron. The α2 values for silver, aluminum and cast iron are 1.71, 0.86 and 0.12 respectively.
(Round your answer to two decimal places. Use two-term approximation of the series.)
For silver, it takes _____ seconds to cool the bar to a temperature of 5°C
For aluminum, it takes _____ seconds to cool the bar to a temperature of 5°C
For cast iron, it takes _____ seconds to cool the bar to a temperature of 5°C
Answer:
a) t = 59 s , b) t = 107 s , c) t = 466 s
Explanation:
This is an exercise in thermal conductivity. The power dissipated or transferred is
P = Q / Δt = k A dT/dx
where Q is the thermal energy of the bar, k the constant of thermal conductivity.
If we assume that we are in a stable regime
dT / dx = (T₀ - [tex]T_{f}[/tex]) / L
the energy in the bar is
Q = m [tex]c_{e}[/tex] T₀
we substitute
m c_{e} T₀ / t = k A (T₀ -T_{f}) / L
t = c_{e} / k m L / A T₀ / (T₀ -T_{f})
let's use the concept of density
ρ = m / V
V = A L
m = ρ AL
t = c_{e} / k (ρ A L) L / A T₀ / (T₀ -T_{f})
t = [tex]c_{e}[/tex] /k ρ L² T₀/(T₀ -T_{f})
In this exercise, the initial temperature is T₀ = 100ºC, the final temperature is T_{f }= 5ºC and the length
L = L₀ / 2 = 20/2 = 10cm = 0.1m
a) case of silver
c_{e} = 234 J / kg ºC
k = 437 W / m ºC
ρ = 10,490 10³ kg / m³
let's calculate
t = 234/437 10.49 10³ 0.1² 100 / (100 -5)
t = 59 s
b) case materials aluminum
c_{e} = 900 J / kg ºC
k = 238 W / m ºC
ρ = 2.70 10³ kg / m³
t = 900/238 2.70 10³ 0.1² 100 / (100-5)
t = 107 s
c) iron material
c_{e} = 448 J / kg ºC
k = 79.5 W / m ºC
ρ = 7.86 10³ kg / m³
t = 448 / 79.5 7.86 10³ 0.1² 100 / (100-5)
t = 466 s
Light from a 600 nm source goes through two slits 0.080 mm apart. What is the angular separation of the two first order maxima occurring on a screen 2.0 m from the slits
Answer:
The angular separation is [tex]k = 0.8594^o[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 600 \ nm = 600*10^{-9} \ m[/tex]
The distance of separation between the slit is [tex]d = 0.080 \ mm = 0.080 *10^{-3} \ m[/tex]
The distance from the screen is
Generally the condition for constructive interference is mathematically represented as
[tex]d \ sin(\theta) = n \lambda[/tex]
=> [tex]\theta = sin ^{-1} [ \frac{n * \lambda }{ d } ][/tex]
here [tex]\theta[/tex] is the angular separation between the central maxima and one side of the first order maxima
given that we are considering the first order of maxima n = 1
=> [tex]\theta = sin ^{-1} [ \frac{1 * 600*10^{-9} }{ 2.0 } ][/tex]
=> [tex]\theta = sin ^{-1} [ 0.0075 ][/tex]
=> [tex]\theta = 0.4297^o[/tex]
So the angular separation of the two first order maxima is
[tex]k = 2 * \theta[/tex]
[tex]k = 2 * 0.4297[/tex]
[tex]k = 0.8594^o[/tex]
Bailey wants to find out which frozen solid melts the fastest: soda, ice, or orange juice. She pours each of the three liquids into the empty cubes of an ice tray, and then places the ice tray in the freezer overnight. The next day, she pulls the ice tray out and sets each cube on its own plate. She then waits and watches for them to melt. When the last part of the frozen liquid melts, she records the time.
Answer:
its 45 over 6
Explanation:the answer is in the question
Answer: Only the melted cube's shape changed.
Explanation:
The center of the galaxy is filled with low-density hydrogen gas that scatters light rays. An astronomer wants to take a picture of the center of the galaxy. Will the view be better using ultra violet light, visible light, or infrared light? Explain.
Answer:
Infrared light
Explanation:
Infrared light is the spectrum of electromagnetic wave given off by a body possessing thermal energy. Infrared light is preferred over visible light in this region of space because visible light is easily scattered in the presence of fine particles. Infrared ray makes it easy for us to observe Cold, dark molecular clouds of gas and dust in our galaxy that glows when irradiated by the stars . Infrared can also be used to detect young forming stars, even before they begin to emit visible light. Stars emit a smaller portion of their energy in the infrared spectrum, so nearby cool objects such as planets can be more readily detected with infrared light which won't be possible with an ultraviolet or visible light.
Find the work done by the gas during the following stages. (a) A gas is expanded from a volume of 1.000 L to 4.000 L at a constant pressure of 2.000 atm. (b) The gas is then cooled at constant volume until the pressure falls to 1.500 atm
Answer:
a) 607.95 J
b) 0 J
Explanation:
a) Initial volume = 1 L = 0.001 m^3
final volume = 4 L = 0.004 m^3
pressure = 2 atm = 202650 Pa (1 atm = 101325 Pa)
work done by the gas on the environment = PΔV
P is the pressure = 101325 Pa
ΔV is the change in volume from the initial volume to the final volume
ΔV = 0.004 m^3 - 0.001 m^3 = 0.003 m^3
work done by the gas = 202650 x 0.003 = 607.95 J
b) If the gas is cooled at constant volume, then the gas does no work. For a gas to do work, there must be a change in its volume.
Therefore the work done in cooling at constant volume until pressure falls to 1.5 atm = 0 J
Light of wavelength 575 nm passes through a double-slit and the third order bright fringe is seen at an angle of 6.5^degree away from the central fringe. What is the separation between the double slits? a) 5.0 mu m b) 10 mu m c) 15 mu m d) 20 mu m e) 25 mu m
Answer:
The correct option is C
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 575 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 6.5^o[/tex]
The order of maxima is n = 3
Generally for constructive interference
[tex]dsin \theta = n * \lambda[/tex]
=> [tex]d = \frac{n * \lambda }{ sin \theta }[/tex]
=> [tex]d = \frac{3 * 575 *10^{-9} }{ sin 6.5 }[/tex]
=> [tex]d = 15.24 *10^{-6} \ m[/tex]
=> [tex]d = 15 \mu m[/tex]
If the prism is surrounded by a fluid, what is the maximum index of refraction of the fluid that will still cause total internal reflection
Answer:
n₁ > n₂.
prisms are made of glass with refractive index n₂ = 1.50, so the fluid that surrounds the prism must have an index n₁> 1.50
Explanation:
Total internal reflection occurs when the refractive index of the incident medium the light is greater than the medium to which the light is refracted, let's use the refraction equation
n₁ sin θ₁ = n₂ sin θ₂
the incident medium is 1, at the limit point where refraction occurs is when the angle in the refracted medium is 90º, so sin θ₂ = 1
n₁ sin θ₁ = n₂
sin θ₁ = n₂ / n₁
We mean that this equation is defined only for n₁ > n₂.
In our case, for the total internal reflection to occur, the refractive incidence of the medium must be greater than the index of refraction of the prism.
In general, prisms are made of glass with refractive index n₂ = 1.50, so the fluid that surrounds the prism must have an index n₁> 1.50
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An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the
speed of sound as 1100 km/h at the aircraft's altitude, how long will it take to reach the
Sound barrier'?
Answer:
50 seconds
Explanation:
Acceleration = change in velocity / change in time
a = Δv / Δt
10 km/h/s = (1100 km/h − 600 km/h) / t
t = 50 s
A ranger in a national park is driving at 15.0 m/s when a deer jumps into the road 60 m ahead of the vehicle. After a reaction time, t, the ranger applies the brakes to produce an acceleration of a = -3.00 m/s2. What is the maximum reaction time allowed if she is to avoid hitting the deer?
Answer:
t = 5 s
Explanation:
In uniform rectilinear movement, the equation for final speed is:
vf = v₀ + a*t
In this case we need that the car stops just before 60 m after applied the brakes, then
vf = v₀ - a*t
vf = final speed = 0
v₀ = initial speed = 15 m/s
And negative acceleration is 3 m/s²
0 = 15 (m/s) - 3 ( m/s²)*t
t = 15 / 3 (m/s /m/s²)
t = 5 s
The point is that with that value ranger will hit the deer so in order to not to hit the deer that time should be smaller than 5 seconds
Can someone tell me a very very simple physics experiment topic that links to biology?
Explanation:
One idea would be to investigate the correlation between your pulse pressure and your pulse rate. To do this, you'll need a blood pressure monitor.
First, measure your resting pressure and rate. Then exercise for 30 seconds. Measure your new blood pressure and pulse rate. Wait for your pressure and rate to return to normal, then repeat the trial for 1 minute, 1.5 minutes, 2 minutes, etc.
List the results in a table. This should include the amount of exercise time, your pulse rate, your systolic pressure (the high number, which is your blood pressure during contraction of your heart muscle), and your diastolic pressure (the low number, which is your blood pressure between heartbeats). Calculate your pulse pressure (systolic minus diastolic) for each trial. Graph the pulse pressure on the x-axis, and your pulse rate (beats per minute) on the y-axis.
What do you hypothesize will be the shape of the graph? Consider Bernoulli's formula, which relates fluid pressure and flow. How close do the results match your hypothesis? What might explain any differences?
How many electrons would have to be removed from a coin to leave it with a charge of +1.0 10-7 C?
Answer:
[tex]n=6.25\times 10^{11}[/tex]
Explanation:
We need to find the number of electrons that would have to be removed from a coin to leave it with a charge of [tex]+10^{-7}\ C[/tex]. Then the number of electrons be n. Using quantization of electric charge as :
q = ne
e is charge on an electron
[tex]n=\dfrac{q}{e}\\\\n=\dfrac{10^{-7}}{1.6\times 10^{-19}}\\\\n=6.25\times 10^{11}[/tex]
So, the number of electrons are [tex]6.25\times 10^{11}[/tex].
A 0.145 kg baseball pitched at 33.m/s is hit on a horizontal line drive straight back at the pitcher at 46.0 m/s. If the contact time between bat and ball is 5.70×10−3 s, calculate the magnitude of the force (assumed to be constant) exerted on the ball by the bat.
Answer:
F = 2009.64 N
Explanation:
It is given that,
Mass of a baseball, m = 0.145 kg
Initial speed if the baseball, u = 33 m/s
It hit on a horizontal line drive straight back at the pitcher at 46.0 m/s, final velocity, v = -46 m/s
Time of contact between the bat and the ball is [tex]t=5.7\times 10^{-3}\ s[/tex]
We need to find the magnitude of the force exerted by the ball on the bat. It can be calculated using impulse-momentum theorem. So,
[tex]Ft=m(v-u)\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.145\times (-46-33)}{5.7\times 10^{-3}}\\\\F=-2009.64\ N[/tex]
So, the magnitude of force exerted on the ball by the bat is 2009.64 N.
An ideal gas is at a temperature of 320 K. What is the average translational kinetic energy of one of its molecules
Answer:
6.624 x 10^-21 J
Explanation:
The temperature of the ideal gas = 320 K
The average translational energy of an ideal gas is gotten as
[tex]K_{ave}[/tex] = [tex]\frac{3}{2}K_{b}T[/tex]
where
[tex]K_{ave}[/tex] is the average translational energy of the molecules
[tex]K_{b}[/tex] = Boltzmann constant = 1.38 × 10^-23 m^2 kg s^-2 K^-1
T is the temperature of the gas = 320 K
substituting value, we have
[tex]K_{ave}[/tex] = [tex]\frac{3}{2} * 1.38*10^{-23} * 320[/tex] = 6.624 x 10^-21 J
A person is lying on a diving board 3.00 m above the surface of the water in a swimming pool. She looks at a penny that is on the bottom of the pool directly below her. To her, the penny appears to be a distance of 8.00 m from her.
Required:
What is the depth of the water at this point?
Answer:
The depth of water at the point is [tex]d_A = 6.55 \ m[/tex]
Explanation:
From the question we are told that
The height of the person above water is [tex]d = 3.00 \ m[/tex]
The distance of the coin as seen by the person is [tex]d' = 8.00 \ m[/tex]
Generally the apparent depth is mathematically represented as
[tex]d_a = \frac{d_A}{n}[/tex]
Here [tex]d_A[/tex] is the actual depth of water while n is the refractive index of water with a constant value [tex]n = 1.33[/tex]
Now from the point the person is the apparent depth is evaluated as
[tex]d_a = d'-d[/tex]
=> [tex]d_a = 8 - 3[/tex]
=> [tex]d_a = 5 \ m[/tex]
So
[tex]5 = \frac{d_A}{1.33}[/tex]
=> [tex]d_A = 5 * 1.33[/tex]
=> [tex]d_A = 6.55 \ m[/tex]
Find the position of the center of mass of two bodies points of masses m1 and m2 joined by a rod of mass negligible in length d. Find the acceleration of the center of mass for the case that m1 = 1 [kg] and m2 = 3 [kg] and the applied forces of the figure with d = 2 [m]
Explanation:
If m₁ is at the origin, then the center of mass is at:
x = (m₁ × 0 m + m₂ × d m) / (m₁ + m₂)
x = m₂ d / (m₁ + m₂)
If m₁ = 1 kg, m₂ = 3 kg, and d = 2 m:
x = (3 kg) (2 m) / (1 kg + 3 kg)
x = 1.5 m
Sum of forces in the x direction:
∑F = ma
16 N = (1 kg + 3 kg) aₓ
aₓ = 4 m/s²
Sum of forces in the y direction:
∑F = ma
20 N = (1 kg + 3 kg) aᵧ
aᵧ = 5 m/s²
In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separation between adjacent bright fringes on a screen 5 m from the slits is: CONVERT FIRST
Answer:
The value is [tex]y = 0.0025 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 500 \ nm[/tex]
The distance of separation is [tex]d = 1\ mm = 0.001 \ m[/tex]
The distance from the screen is [tex]D = 5 \ m[/tex]
Generally the separation between the adjacent bright fringe is mathematically represented as
[tex]y = \frac{D * \lambda }{ d}[/tex]
=> [tex]y = \frac{5 * 500 *10^{-9}}{0.001}[/tex]
=> [tex]y = 0.0025 \ m[/tex]
If the person generates another pulse like the first but the rope is tightened, the pulse will move:_________.
A. at the same rate
B. faster
C. slower
Answer:
the correct answer is B
Explanation:
The speed of a pulse on a string is described by the expression
v =√T/μ
where v is the speed of the pulse, t is the tension of the string and μ the linear density of the string
When applying this equation to our case, if the string is taut, it implies that the tension has increased so that the pulse speed is FASTER
the correct answer is B
A double-slit experiment is performed with light of wavelength 640 nm. The bright interference fringes are spaced 1.6 mm apart on the viewing screen.What will the fringe spacing be if the light is changed to a wavelength of 360nm?
Answer:
1.44*10^-3m
Explanation:
Given that distance BTW two bright fringes is
DetaY = lambda* L/d
So for second wavelength
Deta Y2= Lambda 2* L/d
=lambda 2 x deta y1/ lambda1
So substituting
= 360 x 10^-9 x (1.6*10^-3/640*10^-9)
1.44*10^ -3m
Calculate the radius of curvatuire of the concave lens based on the measured focal length.
Answer:
R₁ = (n -1) f
Explanation:
In geometric optics the focal length and the radius of curvature are related, for the case of a lens
1 / f = (n₂-n₀) (1 / R₁ - 1 / R₂)
where f is the focal length, n₂ is the refractive index of the material, n₀ is the refractive index of the medium surrounding the material, R₁ and R₂ are the radius of curvature of each of the material's
In our case, the most common is that the lens is in the air, so n1 = 1, in many cases one of the surfaces is flat, so its radius of curvature R₂ = ∞.
1 / f = (n-1) (1 / R₁)
we look for the radius of curvature R₁
1 / R₁ = 1 / f (n-1)
R₁ = (n -1) f
With this expression we can find the radius of curvature of a concave-plane lens
How are period and frequency related to each other?
You push a box across the floor with a force of 20 N. You push it 10 meters in 5 seconds. How much work did you do? How much power did you use? Enter your answer in the space provided.
Explanation:
ur answer is in attachment.
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Explanation:
Force applied on the box = 20 N
Displacement, s = 10 m
Time taken = 5 sec
According to first condition of the question, we could find the value of work done
i.e
Work done = force × displacement
= 20 × 10
= 200 Joule
According to second condition of the question, we could find the value of power
i.e
Power = work done/Time taken
= 200/5
= 40 watt.
Hope it helpful!!!!!!!!!!!!
A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?
Answer:
The thickness of the film is 4.32 μm.
Explanation:
Given;
index of refraction of the thin film on one beam, n₂ = 1.5
number of bright fringes shift in the pattern produced by light, ΔN = 8
wavelength of the Michelson interferometer, λ = 540 nm
The thickness of the film will be calculated as;
[tex]\delta N = \frac{2L}{\lambda} (n_2 -n_1)[/tex]
where;
n₁ and n₂ are the index of refraction on the beam
L is the thickness of the film
[tex]\delta N = \frac{2L}{\lambda} (n_2 -n_1)\\\\L = \frac{\lambda}{2} (\frac{N}{n_2-n_1} )\\\\L = \frac{540*10^{-9}}{2} (\frac{8}{1.5-1} )\\\\L = 4.32*10^{-6} \ m\\\\L = 4.32 \mu m[/tex]
Therefore, the thickness of the film is 4.32 μm.
On a day that the temperature is 10.0°C, a concrete walk is poured in such a way that the ends of the walk are unable to move. Take Young's modulus for concrete to be 7.00 109 N/m2 and the compressive strength to be 2.00 109 N/m2. (The coefficient of linear expansion of concrete is 1.2 10-5(°C−1).)
What is the stress in the cement on a hot day of 42.0°C? N/m2
Answer:
The stress is [tex]stress = 2688000 \ N[/tex]
Explanation:
From the question we are told that
The first temperature is [tex]T_1 = 10 ^o \ C[/tex]
The young modulus is [tex]Y = 7.00 *10^9\ N/m^2[/tex]
The compressive strength is [tex]\sigma = 2.00 *10^{9} \ N/m^2[/tex]
The coefficient of linear expansion is [tex]\alpha = 1.2 *10^{-5} \ ^o C ^{-1}[/tex]
The second temperature is [tex]T_2 = 42.0^o \ C[/tex]
Generally the change in length of the concrete is mathematically represented as
[tex]\Delta L = \alpha * L * [T_2 - T_1 ][/tex]
=> [tex]\frac{\Delta L}{L} = \alpha * [T_2 - T_1 ][/tex]
=> [tex]strain = \alpha * [T_2 - T_1 ][/tex]
Now the young modulus is mathematically represented as
[tex]Y = \frac{stress}{strain}[/tex]
=> [tex]7.00 *10^9 = \frac{stress}{\alpha(T_2 - T_1 ) }[/tex]
=> [tex]stress = \alpha (T_2 - T_1 ) * 7.00 *10^9[/tex]
=> [tex]stress = 1.2* 10^{-5} (42 - 10 ) * 7.00 *10^9[/tex]
=> [tex]stress = 2688000 \ N[/tex]