Answer: I0*0.853
Explanation:
Ok, the Malus's law says that:
If you have light polarized along a given line with an intensity I0, and it passes through a polaroid which axis of polarization forms an angle θ with respect to the polarization of the light, then the intensity of the resulting beam is:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the axis of polarization of the light beam that will impact it, then we have θ = 0°, and the equation above says that the intensity of the beam will not change.
In this particular case, we have that the intensity of the light is I0, and the angle is θ = 22.5°
Then:
I(22.5°) = I0*cos^2(22.5°) = I0*0.853
(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
Which statement accurately describes the inner planets? Uranus is one of the inner planets. The inner planets formed when the solar system cooled. The inner planets are also called terrestrial planets. The inner planets are larger than the outer planets.
The correct answer is C. The inner planets are also called terrestrial planets.
Explanation:
Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".
Answer:
The answer is c.) The inner planets are also called terrestrial planets.
Explanation:
n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How fast does the electron move away from the metal
Answer:
The speed of the electron is 1.371 x 10⁶ m/s.
Explanation:
Given;
wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m
the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f = c / λ
[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J[/tex]
Photo electric effect equation is given by;
E = W₀ + K.E
Where;
K.E is the kinetic energy of the emitted electron
K.E = E - W₀
K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J
K.E = 8.563 x 10⁻¹⁹ J
Kinetic energy of the emitted electron is given by;
K.E = ¹/₂mv²
where;
m is mass of the electron = 9.11 x 10⁻³¹ kg
v is the speed of the electron
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s[/tex]
Therefore, the speed of the electron is 1.371 x 10⁶ m/s.
You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?
Answer:
Here,
v (final velocity) = 0
u (initial velocity) = u
a = ?
s = 1m
t = 0.4s
using the first equation of motion,
0 = u + 0.4a
= -0.4a = u
using the second equation of motion:
1 = 0.4u + 0.08a
from the bold equation
1 = 0.4(-0.4a) + 0.08a
1 = -0.16a + 0.08a
1 = -0.08a
a = -1/0.08
a = -100/8
a = -12.5 m/s/s
please make me brainly, i am 1 brainly away from the next rank
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?
Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.
Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m
Explanation: A speed of wave on a string under a tension force can be calculated as:
[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]
[tex]F_{T}[/tex] is tension force (N)
μ is linear density (kg/m)
Determining velocity:
[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{0.00874 }[/tex]
[tex]|v| =[/tex] 0.0935 m/s
The displacement a pulse traveled in 1.23ms:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]
Δx = 11.5×[tex]10^{-6}[/tex]
With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex] m.
Doubling Tension:
[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{2.0.00874 }[/tex]
[tex]|v| = \sqrt{0.01568}[/tex]
|v| = 0.1252 m/s
Displacement for same time:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]
[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]
With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m
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
A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy
Answer: her rotational kinetic energy increases
A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses in diopters he must use in his contacts to allow him to focus on distant objects?
Answer:
-0.24diopters
Explanation:
The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)
So 1/do +1/di = 1/f
1/infinity + 1/-4.2 = 1/f
1/f = 1/-4.2 = -0.24diopters
What happens to the magnetic field when you reverse the direction of current by sliding the battery voltage bar past 0 volts
Answer:
The polarity of the magnetic field changes
Explanation:
This because The magnetic field generated is always perpendicular to the direction of the current and parallel to the solonoid. Hence if we reverse the current the direction of magnetism also reverses. In other words the magnetic poles gets reversed (North pole becomes south pole and the south pole becomes the north pole)
A viewing screen is separated from a double slit by 5.20 m. The distance between the two slits is 0.0300 mm. Monochromatic light is directed toward the double slit and forms an interference pattern on the screen. The first dark fringe is 3.70 cm from the center line on the screen.
Required:
a. Determine the wavelength of light.
b. Calculate the distance between the adjacent bright fringes.
Answer:
The wavelength of this light is approximately [tex]427\; \rm nm[/tex] ([tex]4.27\times 10^{-7}\; \rm m[/tex].)The distance between the first and central maxima is approximately [tex]7.40\; \rm cm[/tex] (about twice the distance between the first dark fringe and the central maximum.)Explanation:
WavelengthConvert all lengths to meters:
Separation of the two slits: [tex]0.0300\; \rm mm = 3.00\times 10^{-5}\; \rm m[/tex].Distance between the first dark fringe and the center of the screen: [tex]3.70\; \rm cm = 3.70\times 10^{-2}\; \rm m[/tex].Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:
The angle between the filter and the beam of light from the lower slit, andThe angle between the screen and that same beam of light.These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].
The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)
Therefore:
[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].
On the other hand:
[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].
Because the cotangent of [tex]\theta[/tex] is very close to zero,
[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].
[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].
[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].
Distance between two adjacent maximaIf the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.
The path difference required for the central maximum is [tex]0[/tex].The path difference required for the first maximum is [tex]\lambda[/tex].The path difference required for the second maximum is [tex]2\,\lambda[/tex].On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:
[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].
Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].
Why was Bohr's atomic model replaced by the
modern atomic model?
Answer:
Explanation:
Bohr's atomic model was replaced by the modern atomic model because of its limitations, which included :
(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )
(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).
(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.
If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed
Answer:
The frequency will change by a factor of 0.4
Explanation:
T = 2(pi)*sqrt(L/g)
Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.
A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.
(a) Is she nearsighted or farsighted?
(b) What type of lens (converging or diverging) is needed to correct her vision?
(c) What focal length contact lens is needed, and what is its power in diopters?
Answer:
(a) nearsighted
(b) diverging
(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D
Explanation:
(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.
(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)
(c) the absolute value of the focal length (f) is given by the formula:
[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]
So it would normally be written with a negative signs in front indicating a divergent lens.
In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to easily count them. To spread out the fringe pattern, one could
Answer:
halve the slit separation
Explanation:
As we know that
In YDS experiment, the equation of fringe width is as follows
[tex]\beta = \frac{\lambda D}{d}[/tex]
where,
D denotes the separation in the middle of screen and slits
d denotes the distance in the middle of two slits
And to increase the Δx we have to decrease the d i.e, the distance between the two slits
Hence, the first option is correct
You plan to take your hair blower to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The blower puts out 1700 W at 120 V.Required:a. What could you do to operate your blower via the 240V line in Europe? which one is it?b. What current will your blower draw from a European outlet?c. What resistance will your blower appear to have when operated at 240 ?
Answer:
a) Connect a series resistance of 8,47 ohms
b)14,16 [A]
c) r = 10,96 ohms
Explanation:
My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower
In America
P = V*I
1700 (w) = 120*I
I = 1700/120 [A]
I = 14,16 [A] current needed for the blower
In Europe
120 (v) (the drop of voltage I need) when a current of 14,16 passes through to series resistor is
V = I*R 120 = 14,16* R R = 8,47 ohms
c) P = I*r²
1700 (w) = 14,16 (A) * r²
r² = 120,06
r = 10,96 ohms
Suppose you exert a force of 185 N tangential to the outer edge of a 1.73-m radius 76-kg grindstone (which is a solid disk).
Required:
a. What torque is exerted?
b. What is the angular acceleration assuming negligible opposing friction?
c. What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?
Answer:
a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².
Explanation:
a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.
τ = Frsinθ
= 185 N × 1.73 m × sin90°
= 320.05 Nm
So, the torque τ = 320.05 Nm
b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m
α = angular acceleration of grindstone
τ = Iα
α = τ/I = τ/(MR²/2) = 2τ/MR²
substituting the values of the variables, we have
α = 2τ/MR²
= 2 × 320.05 Nm/[76 kg × (1.73 m)²]
= 640.1 Nm/227.4604 kgm²
= 2.814 rad/s²
So, the angular acceleration α = 2.814 rad/s²
c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.
So τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm
The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm
Since τ'' = Iα
α' = τ''/I where α' = its new angular acceleration
α' = 2τ/MR²
= 2 × 319.75 Nm/[76 kg × (1.73 m)²]
= 639.5 Nm/227.4604 kgm²
= 2.811 rad/s²
So, the angular acceleration α' = 2.811 rad/s²
In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.497 m away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in what direction (toward or away from the slits) should the screen be moved, so that the centers of the first-order bright orange fringes just appear on the screen
Answer:
0.5639m
Explanation:
For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ
=sin⁻¹(mλ/d)
mλ /d =y/L
for the first order,
y= mλL/d
For ratio separation y₀/yD=1 and d= 1
y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]
1=λ ₀L₀/λD.LD.
λD.LD= λ ₀L₀
L₀= λD.LD/ λ ₀..............(1)
Then substitute the given values into (1) we have
L₀=471 *0.497/611
= 0.3831m
Distance by which the screen has to be moved towards the slit is
LD- Lo
0.947-0.3831= 0.5639m
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.
Answer:
a) k = 95.54 N / m, c = 19.55 , b) m₃ = 0.9078 kg
Explanation:
In a simple harmonic movement with friction, we can assume that this is provided by the speed
fr = -c v
when solving the system the angular value remains
w² = w₀² + (c / 2m)²
They give two conditions
1) m₁ = 1 kg
f₁ = 1.1 Hz
the angular velocity is related to frequency
w = 2π f₁
Let's find the angular velocity without friction is
w₂ = k / m₁
we substitute
(2π f₁)² = k / m₁ + (c / 2m₁)²
2) m₂ = 2 kg
f₂ = 0.8 Hz
(2π f₂)² = k / m₂ + (c / 2m₂)²
we have a system of two equations with two unknowns, so we can solve it
we solve (c / 2m)² is we equalize the expression
(2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁
k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)
k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)
a) Let's calculate
k = 4 π² (0.8² -1.1²) / (½ -1/1)
k = 39.4784 (1.21) / (-0.5)
k = 95.54 N / m
now we can find the constant of friction
(2π f₁) 2 = k / m₁ + (c / 2m₁)²
c2 = ((2π f₁)² - k / m₁) 4m₁²
c2 = (4ππ² f₁² - k / m₁) 4 m₁²
let's calculate
c² = (4π² 1,1² - 95,54 / 1) 4 1²
c² = (47.768885 - 95.54) 8
c² = -382.1689
c = 19.55
b) f₃ = 0.2 Hz
m₃ =?
(2πf₃)² = k / m₃ + (c / 2m₃) 2
we substitute the values
(4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²
1.579 = 95.54 / m₃ + 95.542225 / m₃²
let's call
x = 1 / m₃
x² = 1 / m₃²
- 1.579 + 95.54 x + 95.542225 x² = 0
60.5080 x² + 60.5080 x -1 = 0
x² + x - 1.65 10⁻² = 0
x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2
x = [1 ± 1.03] / 2
x₁ = 1.015 kg
x₂ = -0.015 kg
Since the mass must be positive we eliminate the second results
x₁ = 1 / m₃
m₃ = 1 / x₁
m₃ = 1 / 1.1015
A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium. a) What is the total mechanical energy of the motion
Answer:
Explanation:i think this would help u
A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the wavelength (in cm) of the first harmonic?
Answer:
200cm
Explanation:
Answer:
100cm
Explanation:
Using
F= ( N/2L)(√T/u)
F1 will now be (0.5*2)( √600/0.015)
=> L( wavelength)= 200/2cm = 100cm
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.
an ideal gas is confined to a container with adjustable volume. the number of moles, n, and temperature, t, are constant. by what factor will the volume change if pressure increase by a factor of 5.1
Answer:
The volume will decrease by a factor of 10/51.
Explanation:
Hello,
In this case, since both moles and temperature remain constant, we can use the Boyle's law that relates the volume and pressure as an inversely proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
Thus, since the pressure increases by a factor of 5.1 (statement), we have:
[tex]P_2=5.1P_1[/tex]
Thus, the final volume is:
[tex]V_2=\frac{P_1V_1}{P_2} =\frac{P_1V_1}{5.1P_1}\\\\V_2=\frac{10}{51}V_1[/tex]
It means that the volume will decrease by a factor of 10/51.
Regards.
Somebody please help it’s urgent!!!!
In the tug of war game, none of the teams won. What can you conclude about the forces of the two teams ? Write all the evidence to support your answer.
Answer:
Explanation:
We can conclude that the forces of the two teams are equal and opposite and hence they cancel each other. Therefore none of the teams won as the rope did not move.
hope this helps
plz mark as brainliest!!!!!!!
A concrete slab shown in Figure 5 is being lifted by using three cables connected to the slab at points A, B and C. The slab is in the xy plane. The vertical force required to lift this slab is 60 kN (F 60 kN). Find the tensions in cables DA, DB and DC (show all your workings that you do to find these)
Answer:
Fad = 28.8 kN
Fbd = 16.4 kN
Fcd = 28.1 kN
Explanation:
First, find the length of each cable.
AD = √((2 m)² + (0.5 m)² + (2.5 m)²)
AD = √10.5 m
AD ≈ 3.24 m
BD = √((1.5 m)² + (1 m)² + (2.5 m)²)
BD = √9.5 m
BD ≈ 3.08 m
CD = √((1 m)² + (1 m)² + (2.5 m)²)
CD = √8.25 m
CD ≈ 2.87 m
Next, use similar triangles to find the x, y, and z components of each tension force.
Fadx = 2/3.24 Fad = 0.617 Fad
Fady = 0.5/3.24 Fad = 0.154 Fad
Fadz = 2.5/3.24 Fad = 0.772 Fad
Fbdx = 1.5/3.08 Fbd = 0.487 Fbd
Fbdy = 1/3.08 Fbd = 0.324 Fbd
Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd
Fcdx = 1/2.87 Fcd = 0.348 Fcd
Fcdy = 1/2.87 Fcd = 0.348 Fcd
Fcdz = 2.5/2.87 Fcd = 0.870 Fcd
Now sum the forces in the x, y, and z directions:
∑Fx = ma
-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0
∑Fy = ma
-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0
∑Fz = ma
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.
-0.463 Fad + 0.811 Fbd = 0
0.811 Fbd = 0.463 Fad
Fbd = 0.571 Fad
Substitute into either of the first two equations:
-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0
-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0
-0.339 Fad + 0.348 Fcd = 0
0.348 Fcd = 0.339 Fad
Fcd = 0.975 Fad
Now substituting into the third equation:
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0
60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0
60 kN − 2.083 Fad = 0
Fad = 28.8 kN
Solving for the other two tension forces:
Fbd = 0.571 Fad = 16.4 kN
Fcd = 0.975 Fad = 28.1 kN
Answer:
Tensions of:
DA = 28.81 KN
DB = 16.45 KN
DC = 28.07 KN
Explanation:
see attached
Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration
Answer:
The acceleration of the refrigerator together with the objects decreases.
Explanation:
If the mass of the refrigerator is increased by stacking more masses (objects) on it,
and the force applied remains constant, then we know from
F = ma
where
F is the applied force
m is the total mass of the refrigerator and the objects
a is the acceleration of the masses.
If F is constant, and m is increased, the acceleration will decrease
Answer:
The acceleration decreases.
Explanation:
its right
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to
Answer:
Ok, the question is incomplete buy ill try to answer this in a general way.
Suppose that you have no-polarized light.
When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.
Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°
and:
I(0°) = I0*cos^2(0°) = I0
So the intensity does not change.
Now, knowing the initial intensity, you can find the angle needed to get a given intensity.
For example, if the question was:
"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"
We should solve:
I(θ) = A = I0*cos^2(θ)
(A/i0) = cos^2(θ)
√(A/I0) = cos(θ)
Acos(√(A/I0)) = θ
A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the circle from which light escapes from the liquid into the air above the surface
Answer:
The radius is [tex]r = 3.1905 \ m[/tex]
Explanation:
From the question we are told that
The distance beneath the liquid is [tex]d = 2.70 \ m[/tex]
The refractive index of the liquid is [tex]n_i = 1.31[/tex]
Now the critical value is mathematically represented as
[tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]
substituting values
[tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]
[tex]\theta = 49.76^o[/tex]
Using SOHCAHTOA rule we have that
[tex]tan \theta = \frac{ r}{d}[/tex]
=> [tex]r = d * tan \theta[/tex]
substituting values
[tex]r = 2.7 * tan (49.76)[/tex]
[tex]r = 3.1905 \ m[/tex]
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 580 nm
Answer:
Explanation:
In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is
2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .
2 x 1.34 x t = 1 x 580
t = 216.42 nm .
Thickness must be 216.42 nm .
A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 10-7 N/A2 .) A. 0.02219 m B. 327 m C. 52 m D. 0.00199 m
Answer:
A. 2.2*10^-2m
Explanation:
Using
Area = length x L/ uo xN²
So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*
3000²)
A = 17.5*10^-3/ 1.13*10^-5
= 15.5*10^-2m²
Area= π r ²
15.5E-2/3.142 = r²
2.2*10^2m
Explanation:
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. Calculate: a. the electric field between the plates b. the surface charge density c. the capacitance d. the charge on each plate.
Answer:
(a) 1.47 x 10⁴ V/m
(b) 1.28 x 10⁻⁷C/m²
(c) 3.9 x 10⁻¹²F
(d) 9.75 x 10⁻¹¹C
Explanation:
(a) For a parallel plate capacitor, the electric field E between the plates is given by;
E = V / d -----------(i)
Where;
V = potential difference applied to the plates
d = distance between these plates
From the question;
V = 25.0V
d = 1.70mm = 0.0017m
Substitute these values into equation (i) as follows;
E = 25.0 / 0.0017
E = 1.47 x 10⁴ V/m
(c) The capacitance of the capacitor is given by
C = Aε₀ / d
Where
C = capacitance
A = Area of the plates = 7.60cm² = 0.00076m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m
d = 1.70mm = 0.0017m
C = 0.00076 x 8.85 x 10⁻¹² / 0.0017
C = 3.9 x 10⁻¹²F
(d) The charge, Q, on each plate can be found as follows;
Q = C V
Q = 3.9 x 10⁻¹² x 25.0
Q = 9.75 x 10⁻¹¹C
Now since we have found other quantities, it is way easier to find the surface charge density.
(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e
σ = Q / A
σ = 9.75 x 10⁻¹¹ / 0.00076
σ = 1.28 x 10⁻⁷C/m²
