Answer:
The unknown quantities are:
E and F
The final velocity of the proton is:
√(8/3) k e^2/(m*r)
Explanation:
Hello!
We can solve this problem using conservation of energy and momentum.
Since both particles are at rest at the beginning, the initial energy and momentum are:
Ei = k (q1q2)/r
pi = 0
where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)
and q1 = e and q2 = 2e
When the distance between the particles doubles, the energy and momentum are:
Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2
pf = m1v1 + m2v2
with m1 = m, m2 = 4m, v1=vf_p, v2 = vf_alpha
The conservation momentum states that:
pi = pf
Therefore:
m1v1 + m2v2 = 0
That is:
v2 = (1/4) v1
The conservation of energy states that:
Ei = Ef
Therefore:
k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2
Replacing
m1 = m, m2 = 4m, q1 = e, q2 = 2e
and v2 = (1/4)v1
We get:
(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 = k e^2/r + (1/8)mv1^2
(3/8) mv1^2 = k e^2/r
v1^2 = (8/3) k e^2/(m*r)
A horizontal rod 0.300 m long carries a current through a uniform horizontal magnetic field of magnitude 6.40×10−2 T that points perpendicular to the rod. Part A If the magnetic force on this rod is measured to be 0.140 N , what is the current flowing through the rod?
Answer:
Current, 7.29 A
Explanation:
It is given that,
Length of the horizontal rod, L = 0.3 m
Magnetic field through a horizontal rod, [tex]B=6.4\times 10^{-2}\ T[/tex]
The magnetic force acting on the rod, F = 0.14 N
Let the current flowing through the rod is given by I. The magnetic force acting on an object in the uniform magnetic field is given by :
[tex]F=ILB\ sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=ILB[/tex]
[tex]I=\dfrac{F}{LB}[/tex]
[tex]I=\dfrac{0.14\ N}{0.3\ m\times 6.4\times 10^{-2}\ T}[/tex]
I = 7.29 A
So, the current flowing through the rod 7.29 A.
Final answer:
The current flowing in the rod is approximately 7.292 A when it is subject to a 0.140 N magnetic force within a 6.40 x [tex]10^-^2[/tex] T magnetic field. The calculation uses the formula F = I * L * B, and since the rod is perpendicular to the magnetic field, the angle θ is 90°, which simplifies the calculation.
Explanation:
The question asks for the current flowing through a rod that is experiencing a magnetic force due to an external magnetic field.
The force on a current-carrying conductor in a magnetic field is given by the equation F = I * L * B * sin(θ), where F is the force in newtons, I is the current in amperes, L is the length of the conductor in meters, B is the magnetic field in teslas, and θ is the angle between the direction of the current and the magnetic field. In this case, the rod is perpendicular to the magnetic field, so the angle θ is 90°, making sin(θ) equal to 1.
To find the current I, we rearrange the formula to be I = F / (L * B). Substituting the given values:
Force F = 0.140 N
Length L = 0.300 m
Magnetic field B = 6.40×[tex]10^-^2[/tex] T
The current I can thus be calculated as I = 0.140 N / (0.300 m * 6.40×[tex]10^-^2[/tex] T).
Performing the calculation, I equals approximately 7.292 A.
Part APart complete If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, what is α, the magnitude of the angular acceleration of the CD, as it spins to a stop?
Answer:
20.13841 rad/s²
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = [tex]500\times \frac{2\pi}{60}\ rad/s[/tex]
[tex]\omega_f[/tex] = Final angular velocity = 0
t = Time taken = 2.6 s
[tex]\alpha[/tex] = Angular acceleration
Equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-500\times \frac{2\pi}{60}}{2.6}\\\Rightarrow \alpha=-20.13841\ rad/s^2[/tex]
The magnitude of the angular acceleration of the CD, as it spins to a stop is 20.13841 rad/s²
Case 1: A 0.780-kg silver pellet with a temperature of 85 oC is added to 0.150 kg of water in a copper cup of unknown mass. The initial temperature of the water and the copper cup is 14 oC. The equilibrium temperature of the system (silver water copper cup) is measured to be 26.0 °C. Assume no heat is exchanged with the surroundings. The specific heats of silver, water and copper are: 234 J/(kg oC), 4186 J/(kg oC) and 387 J/(kg oC) , respectively. (a) Which substance releases heat
Answer: The silver pellet will release heat
Explanation:
Based on the case scenario, the silver pellet has a higher temperature that the system of water and copper cup and is thereby added to the system. Because of the higher kinetic energy of the molecules of silver in the silver pellet, some of energy will be released to the water and copper cup system because the system will aim to achieve thermal equilibrium.
The diffusion constant for oxygen diffusing through tissue is 1.0 × 10-11 m2/s. In a certain sample oxygen flows through the tissue at 2.0 × 10-6 kg/s. If the thickness of the tissue is doubled, then what is the rate of oxygen flow through the tissue?
Answer:
m' = 1 x 10⁻⁶ kg/s
Explanation:
Given that
Diffussion constant = 1 x 10⁻¹¹
Mass flow rate ,m = 2 x 10⁻⁶ kg/s
The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.
So new flow rate m'
[tex]m'=\dfrac{m}{2}[/tex]
[tex]m'=\dfrac{2\times 10^{-6}}{2}\ kg/s[/tex]
m' = 1 x 10⁻⁶ kg/s
A liquid of density 1290 kg/m 3 1290 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.83 m/s 9.83 m/s and the pipe diameter d 1 d1 is 12.1 cm 12.1 cm . At Location 2, the pipe diameter d 2 d2 is 17.7 cm 17.7 cm . At Location 1, the pipe is 8.35 m higher than it is at location 2. Ignoring viscosity, calculate the difference between fluid pressure at location 2 and the fluid pressure at location 1.
Answer:
[tex]\Delta P=1060184.8946\ Pa[/tex]
[tex]P_1=124651.2383\ Pa[/tex]
Explanation:
Given:
density of liquid, [tex]\rho=1290\ kg.m^{-3}[/tex]speed of flow at location 1, [tex]v_1=9.83\ m.s^{-1}[/tex]diameter of pipe at location 1, [tex]d_1=0.121\ m[/tex]diameter of pipe at location 2, [tex]d_2=0.177\ m[/tex]height of pipe at location 1, [tex]z_1=8.35\ m[/tex]We know the Bernoulli's equation of in-compressible flow:
[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} + z=constant[/tex] ........................(1)
Cross sectional area of pipe at location 2:
[tex]A_2=\pi \frac{d_2^2}{4}[/tex]
[tex]A_2=\pi\times \frac{0.177^2}{4}[/tex]
[tex]A_2=0.0246\ m^2[/tex]
Cross sectional area of pipe at location 1:
[tex]A_1=\pi \frac{d_1^2}{4}[/tex]
[tex]A_1=\pi\times \frac{0.121^2}{4}[/tex]
[tex]A_1=0.0115\ m^2[/tex]
Using continuity equation:
[tex]A_1.v_1=A_2.v_2[/tex]
[tex]0.0115\times 9.83=0.0246\times v_2[/tex]
[tex]v_2=4.5953\ m.s^{-1}[/tex]
Now apply continuity eq. on both the locations:
[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} + z_1= \frac{P_2}{\rho.g} +\frac{v_2^2}{2g} + z_2[/tex]
[tex](P_2-P_1) = \rho.g [\frac{v_1^2}{2g} + z_1-\frac{v_2^2}{2g} ][/tex]
[tex]\Delta P=1290\times 9.8 [\frac{9.83^2}{19.6} + 8.35-\frac{4.5953^2}{19.6} ][/tex]
[tex]\Delta P=154266.016\ Pa[/tex]...................................Ans (a)
Now the mass flow rate at location 1:
[tex]\dot{m_1}=\rho\times \dot{V}[/tex]
[tex]\dot{m_1}=1290\times (0.0115\times 9.83)[/tex]
[tex]\dot{m_1}=145.828\ kg.s^{-1}[/tex]
Now pressure at location 1:
[tex]P_1=\frac{\dot{m_1}\times v_1}{A_1}[/tex]
[tex]P_1=\frac{145.828\times 9.83}{0.0115}[/tex]
[tex]P_1=124651.2383\ Pa[/tex] ...................................Ans (b)
The difference between fluid pressure at location 2 and fluid pressure at location 1 is mathematically given as
dP = 114 kPa
What is the difference between fluid pressure at location 2 and fluid pressure at location 1.?Question Parameter(s):
Generally, the Bernoulli's equation is mathematically given as
P + ρ*g*y + v² =pipe constant
Where
A1*v1 = A2*v2
π*(0.105/2)²*9.91 = π*(0.167/2)²*v2
v2 = 3.9 m/s
Therefore
P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²
dP = 1290*9.8*9.01 + 9.91² - 3.9²
dP = 114 kPa
In conclusion, difference between fluid pressure is
dP = 114 kPa
Read more about Pressure
https://brainly.com/question/25688500
A thin flashlight beam traveling in air strikes a glass plate at an angle of 52° with the plane of the surface of the plate. If the index of refraction of the glass is 1.4, what angle will the beam make with the normal in the glass?
To solve this problem it is necessary to apply Snell's law and thus be able to calculate the angle of refraction.
From Snell's law we know that
[tex]n_1sin\theta_1 = n_2 sin\theta_2[/tex]
Where,
n_i = Refractive indices of each material
[tex]\theta_1[/tex] = Angle of incidence
[tex]\theta_2[/tex] = Refraction angle
Our values are given as,
[tex]\theta_1 = 38\°[/tex]
[tex]n_1 = 1[/tex]
[tex]n_2 = 1.4[/tex]
Replacing
[tex]1*sin38 = 1.4*sin\theta_2[/tex]
Re-arrange to find [tex]\theta_2[/tex]
[tex]\theta_2 = sin^{-1} \frac{sin38}{1.4}[/tex]
[tex]\theta_2 = 26.088°[/tex]
Therefore the angle will the beam make with the normal in the glass is 26°
(7%) Problem 5: A thermos contains m1 = 0.73 kg of tea at T1 = 31° C. Ice (m2 = 0.095 kg, T2 = 0° C) is added to it. The heat capacity of both water and tea is c = 4186 J/(kg⋅K), and the latent heat of fusion for water is Lf = 33.5 × 104 J/kg. dho32@student.mtsac.edu
The final temperature of the mixture is approximately 29.91°C.
The heat exchange between the tea and the ice. The temperature of the final mixture will be somewhere between 0°C and 31°C, and we need to determine that final temperature. The heat transfer can be calculated using the principle of conservation of energy:
Qin = Qout
The heat gained by the ice as it melts is given by:
Qice = m2 ⋅ Lf
The heat gained by the tea as it cools down is given by:
Qtea = m1 ⋅ c ⋅ (T1 - Tfinal)
The negative sign is used because the tea is losing heat. Setting these equal to each other and solving for Tfinal, we get:
m1 ⋅ c ⋅ (T1 - Tfinal) = m2 ⋅ Lf
Now, let's plug in the given values:
0.73 kg ⋅ 4186 J/(kg ⋅ K) ⋅ (31°C - Tfinal) = 0.095 kg ⋅ 33.5 × 10^4 J/kg
Now, solve for Tfinal:
0.73 ⋅ 4186 ⋅ (31 - Tfinal) = 0.095 ⋅ (33.5 × 10^4)
3050.78 ⋅ (31 - Tfinal) = 3182.5
94602.78 - 3050.78 ⋅ Tfinal = 3182.5
-3050.78 ⋅ Tfinal = -91420.28
Tfinal = 91420.28/3050.78
Tfinal ≈ 29.91°C
So, the final temperature of the mixture is approximately 29.91°C.
A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction.
Part A
If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 3.40V/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?
Part B
What is the direction of the magnetic field?
Answer:
a) 1.13 10-8 T. b) +y direction
Explanation:
a)
For an electromagnetic wave propagating in a vacuum, the wave speed is c = 3. 108 m/s.
At a long distance from the source, the components of the wave (electric and magnetic fields) can be considered as plane waves, so the equations for them can be written as follows:
E(z,t) = Emax cos (kz-ωt-φ) +x
B(z,t) = Bmax cos (kz-ωt-φ) +y
In an electromagnetic wave, the magnetic field and the electric field, at any time, and at any point in space, as the perturbation is propagating at a speed equal to c (light speed in vacuum), are related by this expression:
Bmax = Emax/c
So, solving for Bmax:
Bmax = 3.4 V/m / 3 108 m/s = 1.13 10-8 T.
b) As we have already said, in an electromagnetic wave, the electric field and the magnetic field are perpendicular each other and to the propagation direction, so in this case, the magnetic field propagates in the +y direction.
The hydraulic oil in a car lift has a density of 8.81 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 5.07 x 10-3 m and 0.150 m, respectively. What input force F is needed to support the 27800-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.20 m above that of the input plunger?
Answer:
a. [tex]F_2=31.76N[/tex]
b. [tex]F_2=185.86N[/tex]
Explanation:
Given:
[tex]F_1=27800N[/tex]
[tex]r_1=5.07x10^{-3}m[/tex]
[tex]r_2=0.150 m[/tex]
[tex]p=8.81x10^2 kg/m^3[/tex]
Using the equation to find the force so replacing
a.
[tex]F_1*A_2=F_2*A_1[/tex]
[tex]A=\pi*r^2[/tex]
[tex]F_2=F_1*\frac{A_2}{A_1}=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}[/tex]
[tex]F_2=31.76N[/tex]
b.
[tex]F_2=F_1+F_p[/tex]
[tex]F_2=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}+(8.81x10^2kg/m^3*9.8m/s^2*1.20m*\pi*(5.07x10^{-3})m^2)[/tex]
[tex]F_2=185.86N[/tex]
Suppose you are selling apple cider for two dollars a gallon when the temperature is 3.3 degree C. The coefficient of volume expansion of the cider is 280*10^-6(C degree)^-1. How much more money (in pennies) would you make per gallon be refilling the container on a day when the temperature is 32 degrees C? Ignore the expansion of the container. Round your answer to 0.1 penny.
Answer:
1.6 penny
Explanation:
[tex]V_0[/tex] = Original volume = 1 gal (Assumed)
[tex]\Delta T[/tex] = Change in temperature
[tex]\beta[/tex] = Coefficient of volume expansion = [tex]280\times 10^{-6}\ /^{\circ}[/tex]
Change in volume is given by
[tex]\Delta_V=\beta V_0\Delta T\\\Rightarrow \Delta_V=280\times 10^{-6}\times 1\times (32-3.3)\\\Rightarrow \Delta_V=0.008036[/tex]
New volume would be
[tex]1+0.008036=1.008036\ gal[/tex]
The amount of money earned extra would be
[tex]0.008036\times 2=0.016072\ \$[/tex]
1.6 penny more would be earned if the temperature is 32°C
Final answer:
By refilling a container of apple cider at 32 degrees C instead of 3.3 degrees C, you would make approximately 1.6 pennies more per gallon due to thermal expansion of the cider.
Explanation:
To calculate how much more money you would make per gallon by refilling the container of apple cider when the temperature is 32 degrees C, as opposed to 3.3 degrees C, you need to determine the change in volume due to thermal expansion.
The formula for volume expansion is ΔV = βV₀ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V₀ is the initial volume, and ΔT is the change in temperature.
The initial temperature T1 is 3.3°C, and the final temperature T2 is 32°C, thus ΔT = T2 - T1 = 32°C - 3.3°C = 28.7°C. The coefficient of volume expansion of the cider, given as β, is 280 x 10^-6 (C°)^-1.
Assuming that the initial volume V₀ of the cider is 1 gallon, the change in volume ΔV would be:
ΔV = 280 x 10^-6 x 1 x 28.7 = 0.008036 gallons
To convert gallons to liters, we use the fact that 1 gallon is approximately 3.78541 liters. So, the increase in volume in liters would be:
ΔV (liters) = 0.008036 x 3.78541 = 0.0304 liters
Since there are approximately 3.78541 liters in a gallon, and knowing that the price for one gallon is two dollars, we can calculate the additional revenue (in pennies) as follows:
Extra revenue = ΔV (liters) / 3.78541 x 200 pennies = 0.0304 / 3.78541 x 200 ≈ 1.6 pennies
Therefore, you would make approximately 1.6 pennies more per gallon by refilling the container at 32°C compared to 3.3°C.
A planet is in an elliptical orbit around a distant star. At periastron (the point of closest approach to the star), the planet is rp=4.50×108 km from the star and is moving with a speed of vp=18.5 km/s . When the planet is at apastron (the point of greatest distance from the star), it is ra=9.10×108 km from the star. How fast is the planet moving at apastron? va=? km/s
Answer:
9.15 km/s
Explanation:
rp = 4.5 x 10^8 km
vp = 18.5 km/s
ra = 9.10 x 10^8 km
va = ?
According to the conservation of angular momentum constant.
Let m be the mass of planet
m x rp x vp = m x ra x va
4.5 x 10^8 x 18.5 = 9.10 x 10^8 x va
va = 9.15 km/s
To calculate the speed of the planet at apastron, we can use Kepler's second law and the given values of rp, vp, and ra. Plugging in the values, we find that the planet is moving at a speed of 0.92 km/s at apastron.
Explanation:To calculate the speed of the planet at apastron, we can use Kepler's second law, which states that the area swept out by a planet in equal time intervals is constant. At periastron, the planet is moving fastest, so we can use the equation:
A1 = A2
where A1 is the area swept out at periastron and A2 is the area swept out at apastron.
Since the areas are equal, we can set up the following equation:
0.5 * rp * vp = 0.5 * ra * va
where rp is the distance at the periastron, vp is the velocity at the periastron, ra is the distance at the apastron, and VA is the velocity at the apastron. We can rearrange this equation to solve for va:
va = (rp * vp) / ra
Plugging in the given values, we get:
va = (4.50 x 10^8 km * 18.5 km/s) / (9.10 x 10^8 km)
va = 0.92 km/s
Learn more about Speed of Planet at Apastron here:https://brainly.com/question/13999216
#SPJ11
Two horizontal curves on a bobsled run are banked at the same angle, but one has twice the radius of the other. The safe speed (no friction needed to stay on the run) for the smaller radius curve is v. What is the safe speed on the larger radius curve?
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, [tex]\sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}[/tex]................ii
form i and ii we can write
[tex]v^2= \frac{1}{2} u^2[/tex]
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static friction µstat beltween the block and the slab is 0.70, whereas their kinetic friction coefficient µkin is 0.40. The 10 kg block is pulled by a horizontal force with a magnitude of 100 N.
Answer:
[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]
Explanation:
Normal reaction from 40 kg slab on 10 kg block
M × g = 10 × 9.8 = 98 N
Static frictional force = 98 × 0.7 N
Static frictional force = 68.6 N is less than 100 N applied
10 kg block will slide on 40 kg slab and net force on it
= 100 N - kinetic friction
[tex]=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)[/tex]
= 100 - 39.2
= 60.8 N
[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}[/tex]
[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}[/tex]
[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}[/tex]
[tex]\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }[/tex]
Frictional force on 40 kg slab by 10 kg block = 98 × 0.4
Frictional force on 40 kg slab by 10 kg block = 39.2 N
[tex]40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}[/tex]
[tex]40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}[/tex]
40 kg slab will move with = [tex]0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]
[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]
A 45.0-kg girl is standing on a 166-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.48 m/s to the right relative to the plank. (Let the direction the girl is moving in be positive. Indicate the direction with the sign of your answer.)
1. What is her velocity relative to the surface of the ice?
2. What is the velocity of the plank relative to the surface of ice?
Answer:
-0.31563 m/s
1.16437 m/s
Explanation:
[tex]m_1[/tex] = Mass of girl = 45 kg
[tex]m_2[/tex] = Mass of plank = 166 kg
[tex]v_1[/tex] = Velocity of girl relative to plank = 1.48 m/s
[tex]v_2[/tex] = Velocity of the plank relative to ice surface
In this system the linear momentum is conserved
[tex](m_1+m_2)v_2+m_1v_1=0\\\Rightarrow v_2=-\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=-\frac{45\times 1.48}{45+166}\\\Rightarrow v_2=-0.31563\ m/s[/tex]
Velocity of the plank relative to ice surface is -0.31563 m/s
Velocity of the girl relative to the ice surface is
[tex]v_1+v_2=1.48-0.31563=1.16437\ m/s[/tex]
The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of the calculations, treat the Pacific Ocean as a point like object (obviously a very rough approximation). 1) Determine the gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon. (Express your answer to three significant figures.) Answer in Joules 2) Repeat the calculation when Earth has rotated so that the Pacific Ocean faces toward the Moon. (Express your answer to three significant figures.) Answer in Joules
The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.
By definition the gravitational potential energy is given by,
[tex]PE=\frac{GMm}{r}[/tex]
Where,
m = Mass of Moon
G = Gravitational Universal Constant
M = Mass of Ocean
r = Radius
First we calculate the mass through the ratio given by density.
[tex]m = \rho V[/tex]
[tex]m = (1030Kg/m^3)(7*10^8m^3)[/tex]
[tex]m = 7.210*10^{11}Kg[/tex]
PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon
Now we define the radius at the most distant point
[tex]r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m[/tex]
Then the potential energy at this point would be,
[tex]PE_1 = \frac{GMm}{r_1}[/tex]
[tex]PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}[/tex]
[tex]PE_1 = 9.05*10^{15}J[/tex]
PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.
At the nearest point we perform the same as the previous process, we calculate the radius
[tex]r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m[/tex]
The we calculate the Potential gravitational energy,
[tex]PE_2 = \frac{GMm}{r_2}[/tex]
[tex]PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}[/tex]
[tex]PE_2 = 9.361*10^{15}J[/tex]
A student pushes a 21-kg box initially at rest, horizontally along a frictionless surface for 10.0 m and then releases the box to continue sliding. If the student pushes with a constant 10 N force, what is the box's speed when it is released?
Answer:v=3.08 m/s
Explanation:
Given
mass of student [tex]m=21 kg[/tex]
distance moved [tex]d=10 m[/tex]
Force applied [tex]F=10 N[/tex]
acceleration of system during application of force is a
[tex]a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2[/tex]
using [tex]v^2-u^2=2 as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex]v^2-0=2\times 0.476\times 10[/tex]
[tex]v=\sqrt{9.52}[/tex]
[tex]v=3.08 m/s[/tex]
Unpolarized light is passed through an optical filter that is oriented in the vertical direction.
If the incident intensity of the light is 46 W/m2 , what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)
In order to solve this problem it is necessary to apply the concepts related to intensity and specifically described in Malus's law.
Malus's law warns that
[tex]I = I_0 cos^2\theta[/tex]
Where,
[tex]\theta=[/tex] Angle between the analyzer axis and the polarization axis
[tex]I_0 =[/tex]Intensity of the light before passing through the polarizer
The intensity of the beam from the first polarizer is equal to the half of the initial intensity
[tex]I = \frac{I_0}{2}[/tex]
Replacing with our the numerical values we get
[tex]I = \frac{46}{2}[/tex]
[tex]I = 23W/m^2[/tex]
Therefore the intensity of the light that emerges from the filter is [tex]23W/m^2[/tex]
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 1.30×104 kg/m3 at the center and 2100 kg/m3 at the surface. Part A What is the acceleration due to gravity at the surface of this planet?
The acceleration due to gravity at the surface of a planet depends on its mass and radius, and assumes a uniform density. Since your model has a density that decreases linearly from the center to the surface, the exact value for gravity would require integration over the volume of the planet to account for mass distribution. This arrangement involves advanced calculus.
Explanation:The acceleration due to gravity at the surface of any planet, including Earth, is determined by a constant (G), the mass of the planet (M), and the radius of the planet (R). The formula is g = GM/R². However, this calculation assumes a uniform density throughout the planet, which is often not the case. In reality, like in your model where the density decreases linearly from the center to the surface, finding the precise acceleration due to gravity at the surface becomes more complicated and involves integration over the entire volume of the planet to account for how the mass is distributed.
Given that you provided the densities at the center and surface of the modeled planet, and these densities decrease linearly, one can utilize the formula for the linear density ρ(r) = ρ_center - r(ρ_center - ρ_surface)/R, where R is the radius of the planet, r is the distance from the center, and ρ_center and ρ_surface are the density at the center and surface, respectively. Then, integrate over the volume of the planet to find the total mass.
Once you have the mass, you can use the formula g = GM/R² again to find the acceleration due to gravity at the surface. However, this calculation goes beyond a basic understanding of gravity and requires knowledge of calculus. Without specific numbers for the mass and the integration result, I cannot provide the exact value for surface gravity in this case.
Learn more about Gravity here:https://brainly.com/question/31321801
#SPJ12
The acceleration due to gravity at a planet's surface depends on the planet's radius, mass and the linear decrease of density from center to surface. The formula of this acceleration is G×M/r², considering that M is the planet's mass obtained by the product of volume and average density. However, as the density changes linearly, the force of gravity also decreases linearly from the center to the surface.
Explanation:To calculate the acceleration due to gravity at the surface of the planet, we have to consider the planet's radius, mass and density. Given the density at the center and surface, we can calculate the average density which is the total mass of the planet divided by the total volume. In this spherically symmetric planet model, we can use the formula for the volume of a sphere, which is 4/3πr³, with r being the Earth's radius. We consider that mass (M) equals density (ρ) times volume (V), and the force of gravity (F) is G×(M1×M2)/r², where G is the gravitational constant. In this case, M1 is the mass of the planet and M2 is the mass of the object where we want to know the acceleration, and r is the distance between the centers of the two masses, or in this case the radius of the planet. As force is also mass times acceleration, we can replace F in the formula with M2 times a (acceleration), and find that acceleration is G×M1/r². However, as the density changes linearly from the center to the surface, the force of gravity will also decrease linearly, affecting the acceleration.
Learn more about Acceleration due to gravity here:https://brainly.com/question/21775164
#SPJ11
A beam of x-rays with wavelength λ = 0.300 nm is directed toward a sample in which the x-rays scatter off of electrons that are effectively free. The wavelength of the outgoing electrons is measured as a function of scattering angle, where a scattering angle of 0 means the direction of the x-rays was unchanged when passing through the sample. When looking at all possible scattering angles, what are the longest and shortest wavelengths that the scattered x-rays can have?
Answer:
Explanation:
The problem relates to Compton Effect in which electrons are scattered due to external radiation . The electron is scattered out and photons relating to radiation also undergo scattering at angle θ .
The formula relating to Compton Effect is as follows
[tex]\lambda_f-\lambda_i=\frac{h}{m_0c} (1-cos\theta)[/tex]
Here [tex]\lambda_i[/tex] = 3 0 x 10⁻¹¹
For longest [tex]\lambda_f[/tex] θ =180°
[tex]\lambda_f[/tex] = [tex]\lambda_i + \frac{2\times h}{m_0c}[/tex]
= .3 x 10⁻⁹ + [tex]\frac{2\times6.6\times 11^{-34}}{9\times10^{-31}\times3\times10^8}[/tex]
= .348 nm
For shortest wavelength θ = 0
Putting this value in the given formula
[tex]\lambda_f=\lambda_i[/tex]
[tex]\lambda_f[/tex] = .3 nm
Model the concrete slab as being surrounded on both sides (contact area 24 m2) with a 2.1-m-thick layer of air in contact with a surface that is 5.0 ∘C cooler than the concrete. At sunset, what is the rate at which the concrete loses thermal energy by conduction through the air layer?
Final answer:
The rate at which the concrete loses thermal energy by conduction through the air layer can be calculated using Fourier's Law of Heat Conduction. The formula involves the thermal conductivity, area, temperature difference, and thickness of the air layer. However, without the thermal conductivity value for air, the calculation cannot be completed.
Explanation:
To calculate the rate at which the concrete slab loses thermal energy by conduction through the surrounding air layer at sunset, we can apply Fourier's Law of Heat Conduction. This law states that the heat transfer rate (Q) through a material is directly proportional to the temperature difference across the material (ΔT), the area through which heat is being transferred (A), and the thermal conductivity (k), and inversely proportional to the thickness of the material (L).
The formula to calculate the rate of heat loss is given by Q = k*A*(ΔT/L), where ΔT is the temperature difference between the two sides of the material, A is the contact area, k is the thermal conductivity of the material, and L is the thickness of the material.
Unfortunately, without the thermal conductivity value for air in the provided data, we cannot calculate the exact rate of heat loss for this specific scenario. Thermal conductivity is required to solve this problem, and it's typically obtained from tables in textbooks or scientific references.
A 60-kg woman stands on the very end of a uniform board, which is supported one-quarter of the way from one end and is balanced. What is the mass of the board?
a. 60 kg
b. 30 kg
c. 20 kg
d. 15 kg
e. 120 kg
The correct option can be seen in Option A.
The diagrammatic expression of the question can be seen in the image attached below.
From the given question, we are being informed that the uniform board is balanced. As a result, the torque(i.e. a measurement about how significantly a force acts on a body for it to spin about an axis) acting on the right-hand side of the balance point should be equal to that of the left-hand side.
Mathematically;
[tex]\mathbf{\tau_{_{right}}= \tau_{_{left}}}[/tex]
Given that the mass of the woman = 60 kg
[tex]\mathbf{\tau =\dfrac{m\times g \times l}{\mu}}[/tex]
[tex]\mathbf{\tau_{left} =\dfrac{m\times g \times l}{\mu}}---(1)[/tex]
[tex]\mathbf{\tau_{_{right}} =\dfrac{60 \times g \times l}{\mu}}---(2)[/tex]
Equating both (1) and (2) together, we have:
[tex]\mathbf{\dfrac{m\times g \times l}{\mu} =\dfrac{60 \times g \times l}{\mu} }[/tex]
Dividing like terms on both side
mass (m) = 60 kg
As such, the correct option can be seen in Option A.
Thus, we can conclude that from the 60-kg woman who stands on the very end of a uniform board, the mass of the board on the other end is also 60 kg.
Learn more about mass here:
https://brainly.com/question/6240825?referrer=searchResults
A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvature with magnitudes of |R1|=10cm and |R2|=15cm. The lens is made of glass with index of refraction nglass=1.5. We will employ the convention that R1 refers to the radius of curvature of the surface through which light will enter the lens, and R2 refers to the radius of curvature of the surface from which light will exit the lens.Part AIs this lens converging or diverging?Part BWhat is the focal length f of this lens in air (index of refraction for air is nair=1)?Express your answer in centimeters to two significant figures or as a fraction.
A biconvex lens with the given parameters is a converging lens. Using the Lens Maker's Equation with the radii of curvature and index of refraction for glass and air, the focal length of the lens is calculated to be approximately 12 cm.
Explanation:A biconvex lens, where both surfaces of the lens bulge outwards, will bend light rays such that they converge at a focal point. With the parameters given (|R1|=10cm, |R2|=15cm, and nglass=1.5), we can deduce that this lens is a converging lens.
Part A: Since a biconvex lens makes parallel rays of light converge at a point after passing through the lens, it is classified as a converging lens.
Part B: To calculate the focal length (f) of the lens, we use the Lens Maker's Equation:
First, we convert the radii of curvature to the appropriate signs as per the lensmaker's convention (positive for convex surfaces when the outside medium is air). R1 = +10cm and R2 = -15cm, since the light exits from the second surface.Next, we plug the values into the equation (1/f) = (nglass - nair) ((1/R1) - (1/R2)) to get the reciprocal of the focal length.Carrying out the calculation with the data given (nglass=1.5, nair=1, R1=+10cm, and R2=-15cm), we get:
(1/f) = (1.5 - 1) ((1/10cm) - (1/(-15cm)))
(1/f) = 0.5 * (0.1cm⁻¹ + 0.0667cm⁻¹)
(1/f) = 0.5 * 0.1667cm⁻¹
(1/f) = 0.08335cm⁻¹
Therefore, the focal length f is the reciprocal of 0.08335cm⁻¹ which is approximately:
f ≈ 12cm
Water flows through a horiztonal pipe at a rate of 94 ft3/min. A pressure gauge placed on a 3.3 inch diameter section of the pipe reads 15 psi.
What is the gauge pressure in a section of pipe where the diameter is 5.2 inches?
Answer:
The gauge pressure is 1511.11 psi.
Explanation:
Given that,
Flow rate = 94 ft³/min
Diameter d₁=3.3 inch
Diameter d₂ = 5.2 inch
Pressure P₁= 15 psi
We need to calculate the pressure on other side
Using Bernoulli equation
[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2[/tex]
We know that,
[tex]V=Av[/tex]
[tex]v=\dfrac{V}{A}[/tex]
Where, V = volume
v = velocity
A = area
Put the value of v into the formula
[tex]P_{1}+\dfrac{1}{2}\rho (\dfrac{V}{A_{1}})^2=P_{2}+\dfrac{1}{2}\rho (\dfrac{V}{A_{2}})^2[/tex]
Put the value into the formula
[tex]15+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2=P_{2}+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]
[tex]P_{2}=15+\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2-\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]
[tex]P_{2}=1525.8\ psi[/tex]
We need to calculate the gauge pressure
Using formula of gauge pressure
[tex]P_{g}=P_{ab}-P_{atm}[/tex]
Put the value into the formula
[tex]P_{g}=1525.8-14.69[/tex]
[tex]P_{g}=1511.11\ psi[/tex]
Hence, The gauge pressure is 1511.11 psi.
A parallel plate capacitor is connected to a battery that maintains a constant potential difference between the plates. If the plates are pulled away from each other, increasing their separation, what happens to the amount of charge on the plates?
a. The amount of the charge decreases, because the capacitance increases.
b. Nothing happens; the amount of charge stays the same.
c. The amount of the charge increases, because the capacitance increases.
d. The amount of the charge increases, because the capacitance decreases.
e. The amount of the charge decreases, because the capacitance decreases.
When the separation between the plates of a parallel plate capacitor is increased, the amount of charge on the plates decreases due to the decrease in capacitance (option e), with the voltage remaining constant.
When parallel plate capacitor plates are pulled away from each other while connected to a battery maintaining a constant potential difference, the capacitance decreases. This is because the capacitance is inversely proportional to the distance between the plates. As the capacitance decreases, the charge on the plates also decreases since the voltage (V) remains constant, and the relation between charge (Q), capacitance (C), and voltage (V) is given by Q = CV. Therefore, the amount of charge on the plates decreases because the capacitance decreases (option e).
The return-air ventilation duct in a home has a cross-sectional area of 900 cm^2. The air in a room that has dimensions 5.0 m x 11.0 m ×x 2.4 m is to be completely circulated in a 50-min cycle.
1) What is the speed of the air in the duct? (Express your answer to two significant figures.)
To solve the problem it is necessary to apply the concepts related to the flow rate of a fluid.
The flow rate is defined as
[tex]Q = Av[/tex]
Where,
[tex]Q = Discharge (m^3/s)[/tex]
[tex]A = Area (m^2)[/tex]
v = Average speed (m / s)
And also as
[tex]Q = \frac{V}{t}[/tex]
Where,
V = Volume
t = time
Let's start by finding the total volume according to the given dimensions, that is to say
[tex]V = 5*11*2.4[/tex]
[tex]V = 132m^3[/tex]
The entire cycle must be completed in 50 min = 3000s
In this way we know that the [tex]132m ^ 3[/tex] must be filled in 3000s, that is to say that there should be a flow of
[tex]Q = \frac{V}{t}[/tex]
[tex]Q = \frac{132}{3000}[/tex]
[tex]Q = 0.044m^3/s[/tex]
Using the relationship to find the speed we have to
[tex]Q = Av[/tex]
[tex]v = \frac{Q}{A}[/tex]
Replacing with our values,
[tex]v = \frac{0.044}{900*10^{-4}m^2}[/tex]
[tex]v = 0.488m/s[/tex]
Therefore the air speed in the duct must be 4.88m/s
A truck horn emits a sound with a frequency of 238 Hz. The truck is moving on a straight road with a constant speed. If a person standing on the side of the road hears the horn at a frequency of 220 Hz, then what is the speed of the truck? Use 340 m/s for the speed of the sound.
Answer:
[tex]v_s=27.8m/s[/tex]
Explanation:
If the person hearing the sound is at rest, then the equation for the frequency heard [tex]f[/tex] given the emitted frequency [tex]f_0[/tex], the speed of the truck [tex]v_s[/tex] and the speed of sound [tex]c[/tex] will be:
[tex]f=f_0\frac{c}{c+v_s}[/tex]
Where [tex]v_s[/tex] will be positive if the truck is moving away from the person, and negative otherwise. We then do:
[tex]\frac{f}{f_0}=\frac{c}{c+v_s}[/tex]
[tex]\frac{f_0}{f}=\frac{c+v_s}{c}=1+\frac{v_s}{c}[/tex]
[tex]v_s=c(\frac{f_0}{f}-1)=(340m/s)(\frac{238Hz}{220Hz}-1)=27.8m/s[/tex]
Steam is accelerated by a nozzle steadily from zero velocity to a velocity of 280 m/s at a rate of 2.5 kg/s. If the temperature and pressure of the steam at the nozzle exit are 400°C and 2 MPa, determine the exit area of the nozzle. Solve using appropriate software.
Final answer:
To determine the exit area of the nozzle, use the principle of conservation of mass and the equation for mass flow rate. Calculate the density using the Ideal Gas Law and substitute it into the equation for area.
Explanation:
To determine the exit area of the nozzle, we can use the principle of conservation of mass and the equation for mass flow rate:
Mass flow rate = density x velocity x area
Given that the mass flow rate is 2.5 kg/s and the velocity is 280 m/s, we can rearrange the equation to solve for the area:
Area = mass flow rate / (density x velocity)
However, we need to find the density of the steam at the nozzle exit. To do this, we can use the Ideal Gas Law:
Pressure x Volume = n x R x Temperature
Where pressure = 2 MPa, volume can be assumed to be the volume of the nozzle exit, R is the gas constant, and temperature is 400°C converted to Kelvin.
Once we have the density, we can substitute it into the equation for the area to find the exit area of the nozzle.
The exit area of the nozzle is approximately [tex]\( 0.00140 \text{ m}^2 \) or \( 1.40 \text{ mm}^2 \)[/tex].
The continuity equation for a steady-state flow is given by:
[tex]\[ \dot{m} = \rho \cdot A \cdot v \][/tex]
To find the density [tex]\( \rho \)[/tex], we need to use the ideal gas law, which is a good approximation for steam under these conditions:
[tex]\[ P = \rho \cdot R \cdot T \][/tex]
where:
- P is the absolute pressure at the nozzle exit (2 MPa or 2000 kPa),
- R is the specific gas constant for steam (0.4615 kJ/kg·K),
- T is the absolute temperature at the nozzle exit (400°C + 273.15 = 673.15 K).
Rearranging the ideal gas law to solve for [tex]\( \rho \)[/tex]:
[tex]\[ \rho = \frac{P}{R \cdot T} \][/tex]
Now, we can substitute the density [tex]\( \rho \)[/tex] back into the continuity equation to solve for the exit area A:
[tex]\[ A = \frac{\dot{m}}{\rho \cdot v} \][/tex]
Substituting the values we have:
[tex]\[ \rho = \frac{2000 \text{ kPa}}{0.4615 \text{ kJ/kg·K} \cdot 673.15 \text{ K}} \] \[ \rho = \frac{2000}{310.56} \text{ kg/m}^3 \] \[ \rho \approx 6.44 \text{ kg/m}^3 \][/tex]
Now, we can find the exit area A:
[tex]\[ A = \frac{2.5 \text{ kg/s}}{6.44 \text{ kg/m}^3 \cdot 280 \text{ m/s}} \] \[ A = \frac{2.5}{1787.2} \text{ m}^2 \] \[ A \approx 0.00140 \text{ m}^2 \][/tex]
Why is fusion an appealing energy source?
Fusion products are generally not radioactive.
Extremely high temperatures are required.
The reaction can be confined by available structural materials.
Extremely high pressures are required.
To take place the process of nuclear fusion basically seeks to reach heavy nuclei through light nuclei. Reaching this process implies a release of energy that is what makes this process attractive because it is possible to obtain significant volumes of energy. The procedure to arrive at this process also implies a high cost concerning high temperatures and exorbitant pressures as it is necessary to be able to overcome the barrier of electrostatic repulsion.
This process does not generate any type of radioactive waste like other processes, therefore it is not as dangerous as nuclear fission. For this reason the correct answer is A. Fusion products are generally not radioactive.
Which one of the following phrases best describes the electric potential of a charged particle?
A) the total force exerted on or by the charged particle
B) the force per unit charge
C) the potential energy of the particle relative to infinity
D) the potential energy per unit charge
E) the potential energy per unit force on the particle
Answer:
D.The potential energy per unit charge
Explanation:
Electric potential of a charged particle:
It is scalar quantity because it has magnitude but it does not have direction.
It is the amount of work done required to move a unit positive charge from reference point to specific point in the electric field without producing any acceleration.
Mathematical representation:
[tex]V=\frac{W}{Q_0}[/tex]
Where W= Work done
[tex]Q_0[/tex]= Unit positive charge
Other formula to calculate electric field:
[tex]V=\frac{KQ}{r}[/tex]
Where K=[tex]\frac{1}{4\pi \epsilon_0}[/tex]
It can be defined as potential energy per unit charge.
Hence, option D is true.
A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. You may want to review (Pages 1080 - 1086) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Reflection and refraction.
a. What is the angle between the reflected part of the beam and the surface of the glass? θθ = nothing ∘
b. What is the angle between the refracted beam and the surface of the glass? θθ = nothing ∘
a. The angle between the reflected part of the beam and the surface of the glass is [tex]\(43.5^\circ\).[/tex]
b. The angle between the refracted beam and the surface of the glass is [tex]\(64.5^\circ\).[/tex]
To solve this problem, we need to apply the laws of reflection and refraction. Let's address each part separately.
Part (a): Angle Between the Reflected Beam and the Surface of the Glass
The law of reflection states that the angle of incidence is equal to the angle of reflection. The angle of incidence is given as 43.5° with respect to the surface of the glass. However, angles in optics are typically measured with respect to the normal (a line perpendicular to the surface).
So, the angle of incidence with respect to the normal (which we'll call [tex]\(\theta_i\)[/tex] ) is:
[tex]\[ \theta_i = 90^\circ - 43.5^\circ = 46.5^\circ \][/tex]
Since the angle of incidence equals the angle of reflection:
[tex]\[ \theta_r = \theta_i = 46.5^\circ \][/tex]
Therefore, the angle between the reflected part of the beam and the surface of the glass is:
[tex]\[ 90^\circ - \theta_r = 90^\circ - 46.5^\circ = 43.5^\circ \][/tex]
So, the angle between the reflected beam and the surface of the glass is:
[tex]\[ 43.5^\circ \][/tex]
Part (b): Angle Between the Refracted Beam and the Surface of the Glass
For the refracted beam, we need to apply Snell's Law, which is:
[tex]\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_t) \][/tex]
Where:
- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air), [tex]\( n_1 = 1.00 \)[/tex],
- [tex]\( \theta_i \)[/tex] is the angle of incidence with respect to the normal, [tex]\( \theta_i = 46.5^\circ \),[/tex]
- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass), [tex]\( n_2 = 1.68 \)[/tex],
- [tex]\( \theta_t \)[/tex] is the angle of refraction with respect to the normal.
Using Snell's Law, we can solve for [tex]\(\theta_t\):[/tex]
[tex]\[ 1.00 \sin(46.5^\circ) = 1.68 \sin(\theta_t) \][/tex]
[tex]\[ \sin(\theta_t) = \frac{\sin(46.5^\circ)}{1.68} \][/tex]
Calculating [tex]\(\sin(46.5^\circ)\):[/tex]
[tex]\[ \sin(46.5^\circ) \approx 0.723 \][/tex]
So,
[tex]\[ \sin(\theta_t) = \frac{0.723}{1.68} \approx 0.430 \][/tex]
Now we find [tex]\(\theta_t\):[/tex]
[tex]\[ \theta_t = \sin^{-1}(0.430) \approx 25.5^\circ \][/tex]
The angle between the refracted beam and the surface of the glass is:
[tex]\[ 90^\circ - \theta_t = 90^\circ - 25.5^\circ = 64.5^\circ \][/tex]
So, the angle between the refracted beam and the surface of the glass is:
[tex]\[ 64.5^\circ \][/tex]