Answer:
The electric field is [tex]E = 2.2625 *10^{6} \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the inner sphere is [tex]r_1 = 0.008\ m[/tex]
The radius of the outer sphere is [tex]r _2 = 0.018 \ m[/tex]
The charge on the inner sphere is [tex]q_1 = 3.62 *10^{-8} \ C[/tex]
The charge on the outer sphere is [tex]q_2 = 1.62 *10^{-8} \ C[/tex]
The position from the origin is [tex]d = 0.012 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{k (q_1 )}{ r^2}[/tex]
The reason for using [tex]q_1[/tex] for the calculation is due to the fact that the position considered is greater than the [tex]r_1[/tex] but less than [tex]r_2[/tex]
Here k is the Coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A{-2}[/tex]
So
[tex]E = \frac{9*10^9 (3.62 *10^{-8}}{0.012^2}[/tex]
[tex]E = 2.2625 *10^{6} \ N/C[/tex]
12. Rainbows occur at __________. A. 40–42 degrees above the horizon B. at or below 25 degrees above the horizon C. the horizon D. 60–75 degrees above the horizon.
Answer:
the right answer is option C
Explanation:
at the horizon
The position of rainbows depends on the some factors like wavelength of light and position of sun. So, it can occur at the horizon or above the horizon at the 40 - 42 Degrees. Hence, option (A) and (C) are correct.
The given problem is based on the concepts and fundamentals of Rainbow. The rainbow is the band of seven of seven colors namely, violet, indigo, blue, green, yellow, orange and red.
Rainbows are caused by the sunlight and always appears in the section of sky, directly opposite to the sun.The position of rainbow depends on the wavelength of light such that the light leaves the collection of water droplets (or raindrops) to extend and angle in front of the observer. (Generally between 40 - 42 Degrees).But in many general cases, the rainbows occurs when the sun is exactly on the horizon and most of the time we can only observe the smaller segment of an arc.Thus, we can conclude that the position of rainbows depends on the some factors like wavelength of light and position of sun. So, it can occur at the horizon or above the horizon at the 40 - 42 Degrees. Hence, option (A) and (C) are correct.
Learn more about the Rainbows here:
https://brainly.com/question/15758504
A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
Answer:
If the rod is a conductor, the electrons are free to move within. So when the negative charge is brought towards the rod, the negatively charged electrons near the ball are repelled away towards the other end of the rod, leaving a net positive charge on the end of the rod near the ball. This causes the rod to be attracted towards the negatively charged ball and move closer.
Explanation:
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of what, independent of the direction thrown.
Complete question is;
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of 31.1 m/s, independent of the direction thrown
Answer:
It is proved that the final speed is truly 31.1 m/s
Explanation:
From energy - conservation principle;
E_i = Initial potential energy + Initial Kinetic Energy
Or
E_i = U_i + K_i
Similarly, for final energy
E_f = U_f + K_f
So, expressing the formulas for potential and kinetic energies, we now have;
E_i = (m × g × y_i) + (½ × m × v_i²)
Similarly,
E_f = (m × g × y_f) + (½ × m × v_f²)
We are given;
y_i = 25 m
y_f = 0 m
v_i = 20 m/s
v_f = 31.1 m/s
So, plugging in relevant values;
E_i = m((9.8 × 25) + (½ × 20²))
E_i = 485m
Similarly;
E_f = m((9.8 × 0) + (½ × v_f²)
E_f ≈ ½m•v_f²
From energy conservation principle, E_i = E_f.
Thus;
485m = ½m•v_f²
m will cancel out to give;
½v_f² = 485
v_f² = 485 × 2
v_f² = 970
v_f = √970
v_f ≈ 31.1 m/s
Two 10-cm-diameter charged rings face each other, 25 cm apart. The left ring is charged to ? 25 nC and the right ring is charged to + 25 nC.A) What is the magnitude of the electric field E? at the midpoint between the two rings?B) What is the direction of the electric field E? at the midpoint between the two rings?C) What is the magnitude of the force on a proton at the midpoint?D) What is the direction of the force F? on a proton at the midpoint?
Answer:
A) E = 0N/C
B) 0i + 0^^j
C) F = 0N
D) 0^i + 0^j
Explanation:
You assume that the rings are in the zy plane but in different positions.
Furthermore, you can consider that the origin of coordinates is at the midway between the rings.
A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.
You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:
[tex]E=k\frac{rQ}{(r+R^2)^{3/2}}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C
Q: charge of the ring
r: perpendicular distance to the center of the ring
R: radius of the ring
You use the equation (1) to calculate the net electric field at the midpoint between the rings:
[tex]E=k\frac{rQ}{(r^2+R^2)^{3/2}}-k\frac{rQ}{(r^2+R^2)^{3/2}}=0\frac{N}{C}[/tex]
The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.
B) The direction of the electric field is 0^i + 0^j
C) The magnitude of the force on a proton at the midpoint between the rings is:
[tex]F=qE=q(0N/C)=0N[/tex]
D) The direction of the force is 0^i + 0^j
Part A: The magnitude of the electric field generated at the midpoint between two rings is equal that is 0 N/C.
Part B: The direction of the electric field at the midpoint is opposite.
Part C: The magnitude of the force generated on a proton at the midpoint between two rings is equal that is 0 N.
Part D: The direction of the force on a proton at the midpoint is opposite.
Electric fieldAn electric field is defined as the region that surrounds electrically charged particles and exerts a force on all other charged particles within the region, either attracting or repelling them.
Given that diameter of the ring is 10 m and they are 25 m apart from each other. The charge on the left ring is -25nC and on the right ring is 25nC. The electric field can be given as below.
[tex]E = \dfrac {kQ}{(r+R)^2}\\ [/tex]
Where Q is the charge, r is the radius of the ring, R is the mid-point distance and k is the constant.
Part A
The electric field at the mid-point will be the sum of the electric field generated by both the rings. Substituting the values in the above equation,
[tex]E = \dfrac {8.9\times 10^9\times 25}{(10 +12.5)^2}+\dfrac {8.9\times 10^9\times (-25)}{(10 +12.5)^2}[/tex]
[tex]E = \dfrac {222.5\times 10^9}{506.25} - \dfrac {222.5\times 10^9}{506.25}\\ [/tex]
[tex]E = 0\;\rm N/C[/tex]
Hence we can conclude that both the rings generate the electric field with the same magnitude but they are opposite in direction.
Part B
The electric field at the mid-point is 0 N/C. In the vector form, the electric field can be given as below.
[tex]E = 0i+0j[/tex]
The vector form shows that the electric field at the mid-point between the two rings has the same magnitude but is opposite in direction.
Part C
The force can be given as below.
[tex]F = qE[/tex]
[tex]F = 0 \;\rm N[/tex]
If the electric field at the mid-point is zero, then the force at the mid-point will be zero.
Part D
The vector form of the force at the midpoint is given below.
[tex]F = 0i+0j[/tex]
Hence we can conclude that at the midpoint of two rings, the electric field generates an equal force on the proton but in opposite direction. Hence the net force will be zero.
To know more about the electric field, follow the link given below.
https://brainly.com/question/4440057.
Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Upon using Thomas Young’s double-slit experiment to obtain measurements, the following data were obtained. Use these data to determine the wavelength of light being used to create the interference pattern. Do this using three different methods.
The angle to the eighth maximum is 1.12°.
The distance from the slits to the screen is 302.0 cm.
The distance from the central maximum to the fifth minimum is 3.33 cm.
The distance between the slits is 0.000250 m.
The 3 equations I used were 1). d sin θ_m =(m)λ 2). delta x =λL/d and 3.) d(x_n)/L=(n-1/2)λ
but all my answers are different.
DID I DO SOMETHING WRONG!!!!!!!
Given info
d = 0.000250 meters = distance between slits
L = 302 cm = 0.302 meters = distance from slits to screen
[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])
[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )
---------------
Method 1
[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]
Make sure your calculator is in degree mode.
-----------------
Method 2
[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]
-----------------
Method 3
[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]
There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by
Answer:
the two beams must travel paths that differ by one-half of a wavelength.
A 30 L electrical radiator containing heating oil is placed in a 50 m3room. Both the roomand the oil in the radiator are initially at 10◦C. The radiator with a rating of 1.8 kW is nowturned on. At the same time, heat is lost from the room at an average rate of 0.35 kJ/s.After some time, the average temperature is measured to be 20◦C for the air in the room,and 50◦C for the oil in the radiator. Taking the density and the specific heat of the oil to be950 kg/m3and 2.2 kJ/kg◦C, respectively, determine how long the heater is kept on. Assumethe room is well sealed so that there are no air leaks.
Answer:
Explanation:
Heat absorbed by oil
= mass x specific heat x rise in temperature
= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )
= 25.08 x 10⁵ J
Heat absorbed by air
= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )
= 6.03 x 10⁵ J
Total heat absorbed = 31.11 x 10⁵ J
If time required = t
heat lost from room
= .35 x 10³ t
Total heat generated in time t
= 1.8 x 10³ t
Heat generated = heat used
1.8 x 10³ t = .35 x 10³ t + 31.11 x 10⁵
1.45 x 10³ t = 31.11 x 10⁵
t = 31.11 x 10⁵ / 1.45 x 10³
t = 2145.5 s
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge ( e). Express your answer in vector form. Do not enter units in your expression.
Required:
a. If a positron is accelerated by a constant electric field, find the acceleration of the positron.
b. Assuming the positron started from rest, find the velocity of the positron
Answer:
a) a = + 1,758 10¹¹ m / s , b) = √ (2 1,758 10¹¹ E x)
Explanation:
a) A charged particle is an electric field undergoing force given by the expression
F = qE
where q is the charge of the paticle and E electric field.
In this case we are told that the particle is positron
q = + 1.6 10⁻¹⁹ C
let's calculate the force
F = + 1.6 10⁻¹⁹ E
we write the positive sign, to show that the particle accelerates in the same direction of the electric field
let's write Newton's second law to find the acceleration
F = ma
a = F / m
a = + 1.6 10-19 / 9.1 10-31 E
a = + 1,758 10¹¹ m / s
b) the velocity of the particle starting from rest
v² = v₀² + 2 a x
v = √ (2 1,758 10¹¹ E x)
James drove his car down a mountain road at an average rate of 30 mph and retunred over the same road at an average rate of 20 mph. How far did he drive down the road before he turned around and drove back if his trip took 5 hours?
A. 1B. 2C. 3D. 4E. 5
Answer:
150 miles
Explanation:
5 times 30 is 150
1. The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere
Complete Question
The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere? Convert the result in the still popular non-SI unit Gauss (1 Tesla = 10^4 Gauss).
B = N*mue*I/(2*r)
# of loops = 140
radius of the coil = 0.14m
Answer:
The magnetic induction is [tex]B = 2.639 \ Gauss[/tex]
Explanation:
From the question we are told that
The coil current is [tex]I = 1.5 \ A[/tex]
The number of loops is [tex]N = 140[/tex]
The magnetic field due to the current is mathematically represented as
[tex]B = \mu_o * N * I[/tex]
[tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting value
[tex]B = 4\pi * 10^{-7} * 140 * 1.5[/tex]
[tex]B = 2.639*19^{-4} \ T[/tex]
From question
(1 Tesla = [tex]10^4 \ Gauss[/tex]).
=> [tex]B = 2.693 *10^{-4} *10^4 = 2.63 \ Gauss[/tex]
=> [tex]B = 2.639 \ Gauss[/tex]
When a potential difference of 12 V is applied to a wire 7.2 m long and 0.35 mm in diameter the result is an electric current of 2.1 A. What is the resistivity of the wire?
Answer:
7.63 x 10^-8ohmm
Explanation:
resistivity of the wire = 7.63 x 10^-8ohmm
The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of uniform magnetic field that is oriented at 43.0° to the vertical. The radius of the plate is 8.50 cm. Determine the strength of the magnetic field. mT A circular plate of radius r is lying flat. A field of arrows labeled vector B rising up and to the right pass through the plate.
Answer:
B = 4.1*10^-3 T = 4.1mT
Explanation:
In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:
[tex]\Phi_B=S\cdot B=SBcos\alpha[/tex] (1)
ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2
S: surface area of the circular plate = π.r^2
r: radius of the circular plate = 8.50cm = 0.085m
B: magnitude of the magnetic field = ?
α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°
You solve the equation (1) for B, and replace the values of the other parameters:
[tex]B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT[/tex]
The strength of the magntetic field is 4.1mT
How far apart (in mm) must two point charges of 90.0 nC (typical of static electricity) be to have a force of 3.80 N between them
Answer:
The distance between the two charges is =4.4mm
A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?
Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following statements are true of the resulting interference pattern? (There could be more than one correct choice.)
a. The distance between the maxima decreases.
b. The distance between the minima decreases.
c. The distance between the maxima stays the same.
d. The distance between the minima increases.
e. The distance between the minima stays the same.
Answer:
he correct answers are a, b
Explanation:
In the two-slit interference phenomenon, the expression for interference is
d sin θ= m λ constructive interference
d sin θ = (m + ½) λ destructive interference
in general this phenomenon occurs for small angles, for which we can write
tanθ = y / L
tan te = sin tea / cos tea = sin tea
sin θ = y / La
un
derestimate the first two equations.
Let's do the calculation for constructive interference
d y / L = m λ
the distance between maximum clos is and
y = (me / d) λ
this is the position of each maximum, the distance between two consecutive maximums
y₂-y₁ = (L 2/d) λ - (L 1 / d) λ₁ y₂ -y₁ = L / d λ
examining this equation if the wavelength decreases the value of y also decreases
the same calculation for destructive interference
d y / L = (m + ½) κ
y = [(m + ½) L / d] λ
again when it decreases the decrease the distance
the correct answers are a, b
From their observations, Darcel and Chandra found that the speed of an object at the bottom of an incline depends on how the mass is distributed in the object. Now they consider the moments of inertia of the objects in the simulation, which are related to their mass distributions. Darcel asks Chandra what effect the moment of inertia has on the final speed. Which is the correct response?
Answer:
The larger the moment of inertia of the object, the slower it will be moving at the bottom of the incline
Explanation:
This is because Rotational kinetic energy varies proportionally to the moment oThe larger the moment of inertia of the object,therefore the slower it will be moving at the result is a slower speed
An erect object is placed on the central axis of a thin lens, further from the lens than the magnitude of its focal length. The magnification is +0.4. This means:
A) the image is real and erect and the lens is a converging lens
B) the image is real and inverted and the lens is a converging lens
C) the image is virtual and erect, and the lens is a diverging lens
D) the image is virtual and erect, and the lens is a converging lens
E) the image is virtual and inverted and the lens is a diverging lens
Answer:
the image is virtual and erect and the lens divergent; therefore the correct answer is C
Explanation:
In a thin lens the magnification given by
m = h '/ h = - q / p
where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.
It indicates that the object is straight and is placed at a distance p> f
analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.
Consequently the lens must be divergent
The magnification value is
0.4 = h ’/ h
h ’= 0.4 h
therefore the erect images
therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C
You’re driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 m/s2.
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?
Answer:
(a) Distance between deer and car = 5 m
(b) Vmax = 21.92 m/s
Explanation:
a.
First we calculate distance covered during response time:
s₁ = vt --------- equation 1
where,
s₁ = distance covered during response time = ?
v = speed of car = 20 m/s
t = response time = 0.5 s
Therefore,
s₁ = (20 m/s)(0.5 s)
s₁ = 10 m
Now, we calculate the distance covered by the car during deceleration. Using 3rd equation of motion:
2as₂ = Vf² - Vi²
s₂ = (Vf² - Vi²)/2a ------ eqation 2
where,
a = deceleration = - 10 m/s²
s₂ = Distance covered during deceleration = ?
Vf = Final Velocity = 0 m/s (since car finally stops)
Vi = Initial Velocity = 20 m/s
Therefore,
s₂ = [(0 m/s)² - (20 m/s)²]/2(-10 m/s²)
s₂ = (400 m²/s²)/(20 m/s²)
s₂ = 20 m
thus, the total distance covered by the car before coming to rest is given as:
s = s₁ + s₂
s = 10 m + 20 m
s = 30 m
Now, the distance between deer and car, when it comes to rest, can be calculated as:
Distance between deer and car = 35 m - s = 35 m - 30 m
Distance between deer and car = 5 m
b.
Since, the distance covered by the car in total must be equal to 35 m at maximum. Therefore,
s₁ + s₂ = 35 m
using equation 1 and equation 2 from previous part:
Vi t + (Vf² - Vi²)/2a = 35 m
Vi(0.5 s) + [(0 m/s)² - Vi²]/2(-10 m/s²) = 35 m
0.5 Vi + 0.05 Vi² = 35
0.05 Vi² + 0.5 Vi - 35 = 0
solving this quadratic equation, we get:
Vi = - 31.92 m/s (OR) Vi = 21.92 m/s
For maximum velocity:
Vmax = 21.92 m/s
A hard drive disk rotates at 7200 rpm. The disk has a diameter of 5.1 in (13 cm). What is the speed of a point 6.0 cm from the center axle
Answer:
The speed will be "3.4×10⁴ m/s²".
Explanation:
The given values are:
Angular speed,
w = 7200 rpm
i.e.,
= [tex]7200 \times \frac{2 \pi}{60}[/tex]
= [tex]753.6 \ rad/s[/tex]
Speed from the center,
r = 6.0 cm
As we know,
⇒ Linear speed, [tex]v=wr[/tex]
On putting the estimated values, we get
[tex]=753.6\times 0.06[/tex]
[tex]=45.216 \ m[/tex]
Now,
Acceleration on disk will be:
⇒ [tex]a=\frac{v^2}{r}[/tex]
[tex]=34074 \ m/s^2[/tex]
[tex]=3.4\times 10^4 \ m/s^2[/tex]
Radio waves transmitted through space at 3.00 ✕ 108 m/s by the Voyager spacecraft have a wavelength of 0.137 m. What is their frequency (in Hz)?
Answer:
Frequency, [tex]f=2.18\times 10^9\ Hz[/tex]
Explanation:
We have,
Speed of radio waves is [tex]3\times 10^8\ m/s[/tex]
Wavelength of radio waves is [tex]\lambda=0.137\ m[/tex]
It is required to find the frequency of the radio waves. The speed of a wave is given by :
[tex]v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.137}\\\\f=2.18\times 10^9\ Hz[/tex]
So, the frequency of the radio wave is [tex]2.18\times 10^9\ Hz[/tex].
A simple pendulum of length 1.62 m has a mass of 117 g attached. It is drawn back 38.0 degrees and then released. What is the maximum speed of the mass
Answer:
The maximum speed of the mass is 4.437 m/s.
Explanation:
Given;
length of pendulum, L = 1.62 m
attached mass, m = 117 g
angle of inclination, θ = 38°
This mass was raised to a height of
h = 1.62 - cos38° = 1.0043 m
Apply the principle of conservation of mechanical energy
PE = KE
mgh = ¹/₂mv²
v = √(2gh)
v = √(2 * 9.8 * 1.0043)
v = 4.437 m/s.
Therefore, the maximum speed of the mass is 4.437 m/s.
What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3
Answer:
3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²
- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²
Explanation:
Complete Question
3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?
- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.
The image of this setup attached to this question as obtained from online is attached to this solution.
Solution
3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.
Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²
The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²
- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.
I₂ = I₁ cos² θ
where
I₂ = intensity of light that passes through the second polarizer = ?
I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²
θ = angle between the major axis of the first and second polarizer = 30°
I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²
In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through
I₃ = I₂ cos² θ
I₃ = intensity of light that passes through the third/additional polarizer = ?
I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²
θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)
I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²
Hope this Helps!!!
A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use g=9.8m/sec2 as the acceleration of gravity.
Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
θ₀ = 5.22°
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
θ₀ = 84.78°
A commuter train passes a passenger platform at a constant speed of 40.0 m/s. The train horn is sounded at its characteristic frequency of 320 Hz.
(a) What overall change in requency is detected by a person on the platform as the train moves from approaching to receding?
(b) What wavelength is detected by a person on the platform as the train approaches?
Answer:
a) -75 Hz
b) 0.11 [tex]m^{-1}[/tex]
Explanation:
a) Let us first find the frequency detected by the person on the platform.
We have to find the frequency observed by the person when the train was approaching and when the train was receding.
When the train was approaching:
[tex]f_o = \frac{v}{v - v_s} f_s[/tex]
where fo = frequency observed
fs = frequency from the source = 320 Hz
v = speed of sound = 343 m/s
vs = speed of the train = 40 m/s
Therefore:
[tex]f_o = \frac{343}{343 - 40} * 320\\\\f_o = \frac{343}{303} * 320\\\\f_o = 362 Hz[/tex]
The person on the platform heard the sound at a frequency of 362 Hz when the train was approaching.
When the train was receding:
[tex]f_o = \frac{v}{v + v_s} f_s[/tex]
[tex]f_o = \frac{343}{343 + 40} * 320\\\\f_o = \frac{343}{383} * 320\\\\f_o = 287 Hz[/tex]
The person on the platform heard the sound at a frequency of 287 Hz when the train was receding.
Therefore, the frequency change is given as:
Δf = 287 - 362 = -75 Hz
b) We can find the wavelength detected by the person on the platform as the train approaches by using the formula for speed:
[tex]v = \lambda f[/tex]
where λ = wavelength
f = frequency of the train as it approaches = 362 Hz
v = speed of train = 40 m/s
Therefore, the wavelength detected is:
40 = λ * 362
λ = 40 / 362 = 0.11 [tex]m^{-1}[/tex]
A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What will be the wavelength if the tension is decreased to 600 N and the frequency is kept constant
Answer:
The wavelength will be 33.9 cm
Explanation:
Given;
frequency of the wave, F = 1200 Hz
Tension on the wire, T = 800 N
wavelength, λ = 39.1 cm
[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]
Where;
F is the frequency of the wave
T is tension on the string
μ is mass per unit length of the string
λ is wavelength
[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]
when the tension is decreased to 600 N, that is T₂ = 600 N
[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]
Therefore, the wavelength will be 33.9 cm
Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70.0-kg pilot at that point?
Answer:
The apparent weight of the pilot is 3311 N
Explanation:
Given;
radius of the vertical circle, r = 600 m
speed of the plane, v = 150 m/s
mass of the pilot, m = 70 kg
Weight of the pilot due to his circular motion;
[tex]W= F_v\\\\F_v = \frac{mv^2}{r} \\\\F_v = \frac{70*150^2}{600} \\\\F_v = 2625 \ N[/tex]
Real weight of the pilot;
[tex]W_R = mg\\\\W_R = 70 *9.8\\\\W_R = 686 \ N[/tex]
Apparent weight - Real weight of pilot = weight due to centripetal force
[tex]F_N - mg = \frac{mv^2}{r} \\\\F_N = \frac{mv^2}{r} + mg\\\\F_N = 2625 \ N + 686 \ N\\\\F_N = 3311\ N[/tex]
Therefore, the apparent weight of the pilot is 3311 N
The center of gravity of an ax is on the centerline of the handle, close to the head. Assume you saw across the handle through the center of gravity and weigh the two parts. What will you discover?
Answer:
I believe it is they will weigh the same
Explanation:
Center of gravity is the axis on which the mass rotates evenly if I remember correctly from AP Physics
The head side is heavier than the handle side. - this will be discovered.
What is center of gravity of a object?Theoretically, the body's center of gravity is where all of the weight is believed to be concentrated. Knowing the centre of gravity is crucial because it may be used to forecast how a moving object will behave when subjected to the effects of gravity. In designing immobile constructions like buildings and bridges, it is also helpful.
We know that center of gravity is close to some particular point refers the mass of the point is greater then others. It is given that: The center of gravity of an ax is on the centerline of the handle, close to the head.
So, we can conclude that the head side of the ax is heavier than the handle side of it.
Learn more about center of gravity here:
https://brainly.com/question/17409320
#SPJ5
A 16.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 65.0° angle with the horizontal.
(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.20 m along the ladder from the bottom. Horizontal Force Magnitude = Direction = Vertical Force Magnitude = Direction =
(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?
Answer:
we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.
Coefficient of static friction = 0.26
Explanation:
Given that:
length of the ladder = 16.0 m
weight of the ladder = 520 N
angle θ = 65.0°
(a) We are to find the horizontal and vertical forces the ground exerts on the base of the ladder when an :
Force = 850 N
distance of the climber from the base of the ladder = 4.20 m
The diagrammatic illustration representing what the given information entails can be seen from the attached file below.
Let consider the Ladder being at point A with the horizontal layer of the ground.
From the whole system; the condition for the equilibrium at the point A can be computed as :
[tex]N_2 (16 \ Sin\ 65) = 850(4.2 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]
[tex]N_2 (14.50) = 850(1.7749 )+ 520 (8) \times 0.4226[/tex]
[tex]N_2 (14.50) = 1508.665+1758.016[/tex]
[tex]N_2 (14.50) = 3266.681[/tex]
[tex]N_2 =\dfrac{ 3266.681}{14.50}[/tex]
[tex]N_2 =225.28 \ N[/tex]
[tex]N_1 = mg+F\\[/tex]
where ;
w =mg
[tex]N_1 = 520+850[/tex]
[tex]N_1 = 1370 \ N[/tex]
Therefore; we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.
(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?
the coefficient of static friction between ladder and ground when the firefighter is 9.40 m from the bottom can be calculated as:
[tex]N_2 (16 \ Sin\ 65) = 850(9.4 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]
[tex]N_2 (14.50) = 850(3.9726 )+ 520 (8) \times 0.4226[/tex]
[tex]N_2 (14.50) =3376.71+1758.016[/tex]
[tex]N_2 (14.50) =5134.726[/tex]
[tex]N_2 =\dfrac{5134.726}{14.50}[/tex]
[tex]N_2 =354.12 \ N[/tex]
Therefore; the coefficient of the static friction is;
[tex]\mu = \dfrac{f_s}{N_1}[/tex]
[tex]\mu = \dfrac{354.12}{1370}[/tex]
[tex]\mu[/tex] = 0.26
Coefficient of static friction = 0.26
A 3-liter container has a pressure of 4 atmospheres. The container is sent underground, with resulting compression into 2 L. Applying Boyle's Law, what will the new pressure be? choices: 2.3 atm 8 atm 6 atm 1.5 atm
Answer:
6 atm
Explanation:
PV = PV
(4 atm) (3 L) = P (2 L)
P = 6 atm
An 88.0 kg spacewalking astronaut pushes off a 645 kg satellite, exerting a 110 N force for the 0.450 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 1.40 min?
Answer:
The astronaut and the satellite are 53.718 m apart.
Explanation:
Given;
mass of spacewalking astronaut, = 88 kg
mass of satellite, = 645 kg
force exerts by the satellite, F = 110N
time for this action, t = 0.45 s
Determine the acceleration of the satellite after the push
F = ma
a = F / m
a = 110 / 645
a = 0.171 m/s²
Determine the final velocity of the satellite;
v = u + at
where;
u is the initial velocity of the satellite = 0
v = 0 + 0.171 x 0.45
v = 0.077 m/s
Determine the displacement of the satellite after 1.4 m
d₁ = vt
d₁ = 0.077 x (1.4 x 60)
d₁ = 6.468 m
According to Newton's third law of motion, action and reaction are equal and opposite;
Determine the backward acceleration of the astronaut after the push;
F = ma
a = F / m
a = 110 / 88
a = 1.25 m/s²
Determine the final velocity of the astronaut
v = u + at
The initial velocity of the astronaut = 0
v = 1.25 x 0.45
v = 0.5625 m/s
Determine the displacement of the astronaut after 1.4 min
d₂ = vt
d₂ = 0.5625 x (1.4 x 60)
d₂ = 47.25 m
Finally, determine the total separation between the astronaut and the satellite;
total separation = d₁ + d₂
total separation = 6.468 m + 47.25 m
total separation = 53.718 m
Therefore, the astronaut and the satellite are 53.718 m apart.