Answer:
a) [tex]v_{max} = 2\ \textup{m/s}[/tex]
b) [tex]v_{avg} = 1\ \textup{m/s}[/tex]
c) Q = 1.256 × 10⁻³ m³/s
Explanation:
Given:
The velocity profile as:
[tex]u(r) = 2(1-\frac{r^2}{R^2} )[/tex]
Now, the maximum velocity of the flow is obtained at the center of the pipe
i.e r = 0
thus,
[tex]v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )[/tex]
or
[tex]v_{max} = 2\ \textup{m/s}[/tex]
Now,
[tex]v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}[/tex]
or
[tex]v_{avg} = \frac{2}}{2}\ \textup{m/s}[/tex]
or
[tex]v_{avg} = 1\ \textup{m/s}[/tex]
Now, the flow rate is given as:
Q = Area of cross-section of pipe × [tex]v_{avg}[/tex]
or
Q = [tex]\frac{\pi D^2}{4}\times v_{avg}[/tex]
or
Q = [tex]\frac{\pi 0.04^2}{4}\times 1[/tex]
or
Q = 1.256 × 10⁻³ m³/s
For fully developed laminar pipe flow in a circular pipe, the maximum velocity is 2 m/s, the average velocity is (4/3) m/s, and the volume flow rate is (16/3)π cm^3/s. The velocity profile equation and the formulas for maximum velocity, average velocity, and volume flow rate are explained in detail.
Explanation:For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by the equation u(r) = 2(1-r^2/R^2) in m/s, where R is the inner radius of the pipe. To find the maximum velocity, we need to substitute r = R into the equation. The maximum velocity u_max is then equal to 2(1-1^2/R^2), which simplifies to 2 m/s.
The average velocity can be found by integrating the velocity profile equation over the entire cross-sectional area of the pipe. The cross-sectional area is given by A = πR^2, so the average velocity v_avg is equal to (1/A) times the integral of 2(1-r^2/R^2) over the range r = 0 to r = R. Simplifying the integral and dividing by A, we get v_avg = (4/3) m/s.
The volume flow rate Q can be calculated by multiplying the cross-sectional area A by the average velocity v_avg. Using the given diameter of the pipe, we can find the radius R = 2 cm. Substituting the values into the equation Q = (πR^2)(4/3), we get Q = (16/3)π cm^3/s.
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A velocity selector has an electric field of magnitude 2170 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 5.45 × 103 m/s, enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of +4.10 × 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.54 × 10-9 N, pointing directly upward. What is the speed of this particle?
Answer:
[tex]v = 4.51 \times 10^3 m/s[/tex]
Explanation:
electric field = 2170 N/C
now the speed of the charge particle is given as
[tex]v = 5.45 \times 10^3 m/s[/tex]
here we know that charge particle moves without any deviation
so we will have
[tex]qvB = qE[/tex]
now magnetic field in this region is given as
[tex]B = \frac{E}{v}[/tex]
[tex]B = \frac{2170}{5.45 \times 10^3}[/tex]
[tex]B = 0.398 T[/tex]
Now another charge particle enters the region with different speed and experience the force upwards
[tex]F = qE - qvB[/tex]
[tex]1.54 \times 10^{-9} = (4.10\times 10^{-12})[2170 - v(0.398)][/tex]
[tex]375.6 = 2170 - v(0.398)[/tex]
[tex]v = 4.51 \times 10^3 m/s[/tex]
The answer is: [tex]9.43 \times 10^5 \, \text{m/s}[/tex].
To determine the speed of the different particle that enters the velocity selector, we need to consider the forces acting on it due to the electric and magnetic fields. The net force acting on a charged particle in an electric field [tex]\( E \)[/tex] and a magnetic field [tex]\( B \)[/tex] is given by the Lorentz force equation:
[tex]\[ \vec{F} = q\vec{E} + q(\vec{v} \times \vec{B}) \][/tex]
where [tex]\( q \)[/tex] is the charge of the particle, [tex]\( \vec{v} \)[/tex] is the velocity of the particle, [tex]\( \vec{E} \)[/tex] is the electric field, and [tex]\( \vec{B} \)[/tex] is the magnetic field.
For the particles that pass through the velocity selector without being deflected, the forces due to the electric and magnetic fields must cancel each other out. This means that:
[tex]\[ qE = qvB \][/tex]
where [tex]\( v \)[/tex] is the speed of the particles that pass through undeflected.
Given that the electric field [tex]\( E \)[/tex] is 2170 N/C directed vertically upward, and the particles are traveling east, the magnetic field [tex]\( B \)[/tex] must be directed south to produce a force that cancels the electric force. The speed [tex]\( v \)[/tex] of the undeflected particles is given as [tex]\( 5.45 \times 10^3 \)[/tex] m/s.
Now, for the different particle with a charge of [tex]\( +4.10 \times 10^{-12} \)[/tex] C, the net force [tex]\( F \)[/tex] acting on it is [tex]\( 1.54 \times 10^{-9} \)[/tex] N, pointing directly upward. This means that the magnetic force is not sufficient to cancel the electric force, and the net force is due to the electric force only:
[tex]\[ F = qE \][/tex]
We can use this equation to find the speed [tex]\( v' \)[/tex] of the different particle. Since the net force is equal to the electric force, the magnetic force must be zero. This implies that the velocity of the particle is such that the magnetic force component is equal and opposite to the electric force component, but since the net force is not zero, the particle is not moving at the correct speed to pass through undeflected.
Let's solve for the speed [tex]\( v' \)[/tex] of the different particle:
[tex]\[ F = qE \][/tex]
[tex]\[ 1.54 \times 10^{-9} \, \text{N} = (4.10 \times 10^{-12} \, \text{C})(2170 \, \text{N/C}) \][/tex]
[tex]\[ v' = \frac{F}{qB} \][/tex]
We know [tex]\( F \), \( q \), and \( E \)[/tex], but we need to find [tex]\( B \)[/tex] from the information given for the undeflected particles:
[tex]\[ qvB = qE \][/tex]
[tex]\[ vB = E \][/tex]
[tex]\[ B = \frac{E}{v} \][/tex]
[tex]\[ B = \frac{2170 \, \text{N/C}}{5.45 \times 10^3 \, \text{m/s}} \][/tex]
[tex]\[ B = 3.98 \times 10^{-4} \, \text{T} \][/tex]
Now we can find [tex]\( v' \)[/tex]:
[tex]\[ v' = \frac{1.54 \times 10^{-9} \, \text{N}}{(4.10 \times 10^{-12} \, \text{C})(3.98 \times 10^{-4} \, \text{T})} \][/tex]
[tex]\[ v' = \frac{1.54 \times 10^{-9}}{1.6322 \times 10^{-15}} \][/tex]
[tex]\[ v' = 9.43 \times 10^5 \, \text{m/s} \][/tex]
Therefore, the speed of the different particle is [tex]\( 9.43 \times 10^5 \) m/s[/tex].
An electron is to be accelerated in a uniform electric field having a strength of 2.00 × 106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?
Answer:
a) 800 keV
b) 25 km
Explanation:
[tex]Strength\ of\ Electric\ field=2\times 10^6\ V/m\\a)\ Potential\ Difference=Strength\ of\ Electric\ field\times Distance\\\Rightarrow Potential\ Difference=Kinetic\ Energy\ =2\times 10^6\times 0.4\\\therefore Energy=0.8\times 10^6\ eV=800\ keV\\[/tex]
[tex]b)\ Potential\ difference=50\ GeV=50\times 10^9\ eV\\Distance=\frac{Potential\ difference}{electric\ field}\\\Rightarrow Distance=\frac{50\times 10^9}{2\times 10^6}\\\Rightarrow Distance=25\times 10^3\ m\\\therefore Distance=25\ km[/tex]
Answer:
a) 800 keV
b) 24.996 km.
Explanation:
(a) we have
[tex]\large \Delta K.E=q\Delta V[/tex] .............(1)
where,
[tex]\large \Delta K.E[/tex] = Change in kinetic energy
[tex]q[/tex] = charge of an electron
[tex]\Delta V[/tex] = Potential difference
also
[tex]\large E=\frac{V}{d}[/tex] .......(2)
E = electric field
d = distance traveled
Now from (1) and (2) we have,
[tex]\large \Delta K. E=qV=qEd[/tex]
substituting the values in the above equation, we get
[tex]\large \Delta K. E=(1.6\times 10^{-19}C)(2\times 10^6V/m)(0.400m)(\frac{1eV}{1.6\times 10^{-19}J})(\frac{1keV}{1000eV})[/tex]
[tex]\large \Delta K. E=800keV[/tex]
Thus, the energy gained by the electron is 800 keV if it is accelerated over a distance of 0.400 m.
(b) Using the equation (1), we have
[tex]\large d=\frac{\Delta K.E}{qE}[/tex]
[tex]\large d=\frac{(50\times 10^9eV)}{(1.6\times 10^{-19C})(2\times 10^6V/m)}(\frac{1.6\times 10^{-19}J}{1eV})[/tex]
or
[tex]\large d=2.4996\times 10^4m[/tex]
or
[tex]\large d=24.996\times 10^3m=24.996km[/tex]
Thus, to gain 50.0 GeV of energy the electron must be accelerated over a distance of 24.996 km.
What is the distance between adjacent crests of ocean waves that have a frequency of 0.20 Hz if the waves have a speed of 2.4 m/s?
Answer:
12 m
Explanation:
f = 0.2 Hz, v = 2.4 m/s
v = f x λ
Where, λ is the wavelength
λ = v / f = 2.4 / 0.2 = 12 m
The wavelength is defined as the distance between two consecutive crests or troughs.
So, the distance between two consecutive crests is 12 m.
Water is flowing in a straight horizontal pipe of variable cross section. Where the cross-sectional area of the pipe is 3.70·10-2 m2, the pressure is 6.10·105 Pa and the velocity is 0.260 m/s. In a constricted region where the area is 9.50·10-4 m2, what is the velocity?
Answer:
v = 10.1 m/s
Explanation:
As we know that by the law of conservation of volume the rate of volume flowing through the pipe will remain conserved
so here we have flow rate given as
[tex]Q = Area\times velocity[/tex]
now we have
[tex]A_1 v_1 = A_2 v_2[/tex]
now we have
[tex]A_1 = 3.70 \times 10^{-2} m^2[/tex]
[tex]v_1 = 0.260 m/s[/tex]
[tex]A_2 = 9.50 \times 10^{-4} m^2[/tex]
now from above equation we have
[tex]v_2 = \frac{A_1}{A_2} v_1[/tex]
[tex]v_2 = \frac{3.70\times 10^{-2}}{9.50\times 10^{-4}}(0.260)[/tex]
[tex]v_2 = 10.1 m/s[/tex]
The small piston of a hydraulic lift has a diameter of 8.0 cm, and its large piston has a diameter of 40 cm. The lift raises a load of 15,000 N. Assume the pistons each have negligible weight. (a) Determine the force that must be applied to the small piston. (b) Determine the pressure applied to the fluid in the lift.
Answer:
a)
600 N
b)
1.2 x 10⁵ Pa
Explanation:
(a)
d₁ = diameter of small piston = 8 cm = 0.08 m
d₂ = diameter of large piston = 40 cm = 0.40 m
F₂ = force applied to large piston = 15000 N
F₁ = force applied to small piston = ?
Using pascal's law
[tex]\frac{F_{1}}{(0.25)\pi d_{1}^{2}} = \frac{F_{2}}{(0.25)\pi d_{2}^{2}}[/tex]
Inserting the values
[tex]\frac{F_{1}}{(0.08)^{2}} = \frac{15000}{(0.40)^{2}}[/tex]
F₁ = 600 N
b)
Pressure applied is given as
[tex]P = \frac{F_{1}}{(0.25)\pi d_{1}^{2}}[/tex]
[tex]P = \frac{(600)}{(0.25)(3.14) (0.08)^{2}}[/tex]
P = 1.2 x 10⁵ Pa
To determine the force needed on the small piston of a hydraulic lift, the ratios of the pistons' areas are used in combination with the load on the large piston, applying Pascal's Principle. The pressure in the hydraulic fluid is then found by dividing the force by the area of the piston where the force is applied.
Force applied to small piston = (Force on large piston) × (Area of small piston / Area of large piston)Pressure in the hydraulic fluid = Force / AreaGiven the diameters of the pistons, we can find the areas by using the formula for the area of a circle, A = πr², where r is the radius of the piston.
For the provided diameters of 8.0 cm and 40 cm, the areas of the pistons are:
Area of small piston = π × (4 cm)²Area of large piston = π × (20 cm)²With a load of 15,000 N on the large piston:
Force applied to small piston = 15,000 N × (Area of small piston / Area of large piston)Pressure = 15,000 N / (Area of large piston)Once we perform these calculations, we can determine the required force to apply to the small piston and the pressure applied to the fluid within the lift.
A 62.0-kg woman runs up a flight of stairs having a rise of 4.28 m in a time of 4.20 s. What average power did she supply? a) 63.2 W b) 619 W c) 596 W d) 629 W e) 609 W
Answer:
The average power is 619 W.
(B) is correct option
Explanation:
Given that,
Weight = 62.0 kg
Height = 4.28 m
Time t = 4.20 s
We need to calculate the work done
Work done by woman
[tex]W=mgh[/tex]
Where,
m = mass
g = acceleration due to gravity
t = time
Put the value into the formula
[tex]W=62\times9.8\times 4.28[/tex]
[tex]W=2600.528 J[/tex]
We need to calculate the power
Using formula of power
[tex]P=\dfrac{W}{t}[/tex]
Where,
W = work done
t = time
Put the value in to the formula
[tex]P=\dfrac{2600.528}{4.20}[/tex]
[tex]P=619\ W[/tex]
Hence, The average power is 619 W.
The average power supplied by the woman is calculated by first determining the work done against gravity to move up the stairs, and then dividing this by the time taken. After performing these calculations, the most accurate answer is approximately 629 W.
Explanation:The question is asking to calculate the average power supplied by a woman while running up the stairs. Power is defined as the rate at which work is done. In this scenario, the work is the woman's movement against the gravitational force, which is calculated using the formula W = mgh, where m is mass, g is the acceleration due to gravity and h is the height of the stairs. Substituting the given values, we get W = 62.0 Kg * 9.8 m/s² * 4.28 m = 2650.368 Joules. Power is calculated by dividing the work done by the time taken, so P = W/t = 2650.368 J / 4.20 s = 630.8 Watts, which is closest to option (d) 629 W.
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Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3. (Assume that acceleration due to gravity is g = 9.8 m/s2. Round your answer to one decimal place.)
Answer:
Work done = 35467.278 J
Explanation:
Given:
Height of the cone = 4m
radius (r) of the cone = 1.2m
Density of the cone = 600kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Now,
The total mass of the cone (m) = Density of the cone × volume of the cone
Volume of the cone = [tex]\frac{1}{3}\pi r^2 h[/tex]
thus,
volume of the cone = [tex]\frac{1}{3}\pi 1.2^2\times 4[/tex] = 6.03 m³
therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg
The center of mass for the cone lies at the [tex]\frac{1}{4}[/tex]times the total height
thus,
center of mass lies at, h' = [tex]\frac{1}{4}\times4=1m[/tex]
Now, the work gone (W) against gravity is given as:
W = mgh'
W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J
The work against gravity required to build a cone of height 4 m and base of radius 1.2 m out of a material of density 600 kg/m3 is 35,467.278 J.
Given to us
Height of the cone = 4m
The radius of the cone = 1.2 m
Density of the material = 600 kg/m³
We know that the work gone against the gravity is given as,
[tex]W = mgh'[/tex]
where W is the work, m is the mass, g is the acceleration due to gravity, and h' is the center of mass.
Also, the mass of an object is the product of its volume and density, therefore,
[tex]m = v \times \rho[/tex]
We know the value of the volume of a cone,
[tex]m = \dfrac{1}{3}\pi r^2 h \times \rho[/tex]
The center of mass of a cone lies at the center of its base at 1/4 of the total height from the base,
[tex]h' = \dfrac{h}{4}[/tex]
Substitute all the values we will get,
[tex]W = mgh'\\\\W = (v \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi r^2h \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi (1.2)^2 \times4 \times 600) \times 9.81 \times \dfrac{4}{4}\\\\W = 35467.278\rm\ J[/tex]
Hence, the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3 is 35,467.278 J.
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A 15,000 N truck starts from rest and moves down a 15∘ hill with the engine providing a 8,000 N force in the direction of the motion. Assume the rolling friction force between the truck and the road is very small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill?
Answer:
[tex]v_f = 27.9 m/s[/tex]
Explanation:
Component of the weight of the truck along the inclined plane is given as
[tex]F_1 = W sin\theta[/tex]
[tex]F_1 = 15000 sin15[/tex]
[tex]F_1 = 3882.3 N[/tex]
also the engine is providing the constant force to it as
[tex]F_2 = 8000 N[/tex]
now the net force along the the plane is given as
[tex]F_{net} = 8000 + 3882.3[/tex]
[tex]F = 11882.3 N[/tex]
mass of the truck is given as
[tex]m = \frac{w}{g} = 1529 kg[/tex]
now the acceleration is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = 7.77 m/s^2[/tex]
now the speed of the truck after travelling distance of d = 50 m is given as
[tex]v_f^2 = v_i^2 + 2 a d[/tex]
[tex]v_f^2 = 0 + 2(7.77)(50)[/tex]
[tex]v_f = 27.9 m/s[/tex]
When a comet comes close to the Sun, what happens to it that makes it bright and easier to see?
Answer:
Explanation:
Comet is made by dust particles, icy particles, gases etc.
A comet has a fixed time to complete a revolution around the sun.
As a comet comes nearer to the sun, due to the heat of the sun the vapour and the icy particles becomes gases and due to the radial pressure of energy od sun, we observe a tail of comet which has vapours mainly. SI the comet is visible easily.
A baseball player wants to hit a home run over the wall of a stadium. He hits the ball 0.751 m above the ground with at an angle of 30 degrees above the horizontal and a speed of 37.5 m/s. What is the tallest wall the player can clear if the wall is 150 m away?
Answer:
The baseball does not reach the wall, because the ball falls at 125.33 meters and the wall is at 150 meters
Explanation:
V= 37.5 m/s
α= 30º
g= 9.8 m/s²
Vx= V * cos(30º) = 32.47 m/s
Vy= V * sin(30º) = 18.75 m/s
flytime of the baseball:
t= 2 * Vy/g
t= 3.86 sec
distance of baseball fall:
d= Vx * t
d= 125.33 m
Answer: 18.67 m wall
Explanation:this is projectile.
For Max height,
H = v²sin²©/2g
H= (37.5)²sin²(30)/(2*9.81)
H = 17.919 m
But he is standing 0.751 m above ground, so total height = 17.919+0.751= 18.67m
The temperature of air decreases as it is compressed by an adiabatic compressor. a)-True b)-False
Answer: false
Explanation: the air will increase if it is compressed by an adiabatic compressor
A satellite orbits the earth in a circle with constant speed. Which of the following is true? The net force on the satellite is zero because the satellite is not accelerating. The net force on the satellite is directed forward, in the direction of travel. The net force on the satellite is directed straight down, toward the Earth. The net force on the satellite is directed outward, away from the Earth.
Explanation:
A satellite orbiting the Earth is a characteristic example of the uniform circular motion, where the direction of the velocity vector [tex]\vec{V}[/tex] is perpendicular to the radius [tex]r[/tex] of the trajectory. Hence, the velocity is changing although the speed is constant.
On the other hand, acceleration [tex]\vec{a}[/tex] is directed toward the center of the circumference (that is why it is called centripetal acceleration).
Now, according to Newton's 2nd law, the force [tex]\vec{F}[/tex] is directly proportional and in the same direction as the acceleration:
[tex]\vec{F}=m.\vec{a}[/tex]
Therefore, the net force on the satellite resulting from its circular motion points towards the center of the circle (where the Earth is in the Earth-satellite system).
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is spun at 646.1 rad/s646.1 rad/s while it is being read, and then is allowed to come to rest over 0.234 seconds0.234 seconds , what is the magnitude of the average angular acceleration of the disk?
Answer:
The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².
Explanation:
Given that,
Angular velocity of optical disk= 1045 rad/s
Angular velocity of particular disk = 646.1 rad/s
Time = 0.234 sec
We need to calculate the average angular acceleration
Using formula of angular acceleration
[tex]\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}[/tex]
Where, [tex]\omega_{f}[/tex] = final angular velocity
[tex]\omega_{i}[/tex] = Initial angular velocity
t = time
Put the value in the equation
[tex]\alpha = \dfrac{0-646.1}{0.234}[/tex]
[tex]\alpha= -2761.1\ rad/s^2[/tex]
Negative sign shows the angular deceleration.
Hence, The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².
The force between two electrical charges is 0.3 N. What would be the force if the charges are doubled, and the distance between them increased by 100%?
(Please show clear steps/explanation)
Answer:
The new force F' will be same of the original force F.
Explanation:
Given that,
Charges = 0.3
We need to calculate the force between the charges
Suppose that the distance between the charges is r.
The force between the charges
[tex]F =\dfrac{kq_{1}q_{2}}{r^2}[/tex]
Put the value into the formula
[tex]F=\dfrac{k\times0.3\times0.3}{r^2}[/tex]
[tex]F=\dfrac{k\times(0.09)}{r^2}[/tex]
If the charges are doubled, and the distance between them increased by 100%.
So, The charges are 0.6 and the distance is 2r.
Then,
The force between the charges
[tex]F'=\dfrac{k\times0.6\times0.6}{(2r)^2}[/tex]
[tex]F'=\dfrac{k\times0.09}{r^2}[/tex]
[tex]F'=F[/tex]
Hence, The new force F' will be same of the original force F.
A pump is used to empty a 5200 L wading pool. The water exits the 3.0-cm-diameter hose at a speed of 2.2 m/s .How long will it take to empty the pool? Express your answer using two significant figures.
Answer:
55.79 minutes
Explanation:
Volume of water = 5200 L = 5.2 m^3
Diameter = 3 cm
radius, r = 1.5 cm = 0.015 m
v = 2.2 m/s
The volume of water coming out in 1 second = velocity x area
= 2.2 x 3.14 x 0.015 x 0.015
= 1.55 x 10^-3 m^3
1.55 x 10^-3 m^3 water flows = 1 second
5.2 m^3 water flows = 5.2 / (1.55 x 10^-3) = 3345.57 second
= 55.79 minutes
A book slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m. Calculate the coefficient of kinetic friction ????k between the book and the carpet. Assume the only forces acting on the book are friction, weight, and the normal force.
A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.
A book slides across a level, carpeted floor with an initial speed (u) of 3.25 m/s. It comes to rest (final speed = v = 0m/s) after 3.25 m (s).
Since this is a uniformly decelerated rectilinear motion, we can calculate the acceleration (a) using the following kinematic expression.
[tex]v^{2} = u^{2} + 2as\\\\(0m/s)^{2} = (3.25m/s)^{2} + 2a(3.25m)\\\\a = -1.63 m/s^{2}[/tex]
The negative sign only explains that the friction opposes the movement. We can calculate the force exerted by the friction (F) using Newton's second law of motion.
[tex]F = m \times a[/tex] [1]
where,
m: mass of the bookWe can also calculate the force of friction using the following expression.
[tex]F = k \times N = k \times m \times g[/tex] [2]
where,
k: coefficient of kinetic frictionN: normal forceg: gravity (9.81 m/s²)Given [1] = [2], we get,
[tex]m \times a = k \times m \times g\\\\k = \frac{1.63 m/s^{2} }{9.81 m/s^{2}} = 0.166[/tex]
A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.
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This question involves the concepts of the equations of motion, Newton's Second Law, and Frictional Force.
The coefficient of kinetic friction is found to be "0.15".
First, we will use the third equation of motion to find out the acceleration of the book:
[tex]2as = v_f^2-v_i^2\\[/tex]
where,
a = acceleration = ?
s = distance covered = 3.25 m
vf = final speed = 0 m/s
vi = initial speed = 3.25 m/s
Therefore,
[tex]2a(3.25\ m) = (0\ m/s)^2-(3.25\ m/s)^2\\\\a=\frac{-10.56\ m^2/s^2}{7\ m/s}[/tex]
a = - 1.51 m/s² (negative sign indicates deceleration)
Now the force can be calculated using Newton's Second Law of motion:
F = ma
This force is also equal to the frictional force:
F = μmg
comparing both forces, we get:
ma = μmg
a = μg
[tex]\mu = \frac{a}{g} = \frac{1.51\ m/s^2}{9.81\ m/s^2}[/tex]
μ = 0.15
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The attached picture shows the equations of motion.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0; (5.0 m, 5.0 m) at t = 2.0 s; and (14.0 m, 12.0 m) at t = 5.0 s. (a) Find the magnitude of the average velocity from t = 0 to t = 2 s. (b) Find the magnitude of the average velocity from t = 0 to t = 5 s.
Answer:
a) Magnitude of the average velocity from t = 0 to t = 2 s = 2.24 m/s
b) Magnitude of the average velocity from t = 0 to t = 5 s = 3.16 m/s
Explanation:
a) Velocity is rate of change of position.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0
A particle's position coordinates (x, y) are (5.0 m, 5.0 m) at t = 2.0 s
Displacement = (5-1)i + (5-3)j = 4i + 2j
Change in time = 2s
Velocity
[tex]v=\frac{4i+2j}{2}=2i+j[/tex]
Magnitude of velocity
[tex]v=\sqrt{2^2+1^2}=2.24m/s[/tex]
b) Velocity is rate of change of position.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0
A particle's position coordinates (x, y) are (14.0 m, 12.0 m) at t = 5.0 s
Displacement = (14-1)i + (12-3)j = 13i + 9j
Change in time = 5s
Velocity
[tex]v=\frac{13i+9j}{5}=2.6i+1.8j[/tex]
Magnitude of velocity
[tex]v=\sqrt{2.6^2+1.8^2}=3.16m/s[/tex]
horizontal block–spring system with the block on a frictionless surface has total mechanical energy E 5 47.0 J and a maximum displacement from equilibrium of 0.240 m. (a) What is the spring constant? (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is 3.45 m/s, what is its mass? (d) What is the speed of the block when its displacement is 0.160 m? (e) Find the kinetic energy of the block at x 5 0.160 m. (f) Find the potential energy stored in the spring when x 5 0.160 m. (g) Suppose the same system is released from rest at x 5 0.240 m on a rough surface so that it loses
Answer:
Part a)
[tex]k = 1632 J[/tex]
Part b)
[tex]KE = 47 J[/tex]
Part c)
[tex]m = 7.9 kg[/tex]
Part d)
[tex]v = 2.57 m/s[/tex]
Part e)
[tex]KE = 26.1 J[/tex]
Part f)
[tex]PE = 20.9 J[/tex]
Explanation:
Total Mechanical energy is given as
[tex]E = 47.0 J[/tex]
Its maximum displacement from mean position is given as
[tex]A = 0.240 m[/tex]
Part a)
Now from the formula of energy we know that
[tex]E = \frac{1}{2}kA^2[/tex]
[tex]47 = \frac{1}{2}k(0.240)^2[/tex]
[tex]k = 1632 J[/tex]
Part b)
At the mean position of SHM whole mechanical energy will convert into kinetic energy
so it is given as
[tex]KE = 47 J[/tex]
Part c)
As per the formula of kinetic energy we know that
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]47 = \frac{1}{2}m(3.45^2)[/tex]
[tex]m = 7.9 kg[/tex]
Part d)
As we know by the equation of the speed of SHM is given as
[tex]v = \sqrt{\frac{k}{m}(A^2 - x^2)}[/tex]
[tex]v = \sqrt{\frac{1632}{7.9}(0.24^2 - 0.16^2)}[/tex]
[tex]v = 2.57 m/s[/tex]
Part e)
As we know by the formula of kinetic energy
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}(7.9)(2.57^2)[/tex]
[tex]KE = 26.1 J[/tex]
Part f)
As per energy conservation we know
KE + PE = Total energy
[tex]26.1 + PE = 47[/tex]
[tex]PE = 20.9 J[/tex]
(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]
(b) The KE at equilibrium point =47j
(c) The mass =7.9kg
(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]
(e)The KE of block 0.160m will be 26.1
(f) The PE will be 20.9j
What will be the asked values of the spring-mass system in the question?(a) for finding spring constant
KE=47 j
A=0.240m
By using the formula
[tex]E=\dfrac{1}{2} kx^2[/tex]
[tex]47=\dfrac{1}{2} k(0.240)^2[/tex]
[tex]k=1632\frac{N}{m^2}[/tex]
(b) At the mean position the whole mechanical energy will be equal to KE so
KE=47j
(c) The mass of the system
[tex]KE =\dfrac{1}{2} mv^2[/tex]
[tex]47=\dfrac{1}{2} m(3.45^2)[/tex]
[tex]m=7.9kg[/tex]
(d)Now the speed of the block
[tex]v=\sqrt{\dfrac{k}{m} (A^2-x^2)}[/tex]
[tex]v=\sqrt{\dfrac{1632}{7.9} (0.24^2-0.16^2)}[/tex]
[tex]v=2.57\frac{m}{s}[/tex]
(e) The KE of the block
[tex]KE=\dfrac{1}{2} mv^2=\dfrac{1}{2} 7.9(2.57)^2=26.1J[/tex]
(f) The PE of the system
[tex]Total Energy = KE+PE[/tex]
[tex]PE= 47-26.1 =20.9J[/tex]
Thus
(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]
(b) The KE at equilibrium point =47j
(c) The mass =7.9kg
(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]
(e)The KE of block 0.160m will be 26.1
(f) The PE will be 20.9j
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5.90. A force is applied to a block to move it up a 30° incline. The incline is frictionless. If F = 65.0 N and M = 5.00 kg , what is the magnitude of the acceleration of the block? Enter your answer in units of m/s^2m/s 2 , without units, to the nearest hundredth.
Answer:
The acceleration of the block is 6.35 m/s².
Explanation:
It is given that,
A force is applied to a block to move it up a 30° incline. The applied force is, F = 65 N
Mass of the block, m = 5 kg
We need to find the acceleration of the block. From the attached figure, it is clear that.
[tex]F_x=ma_x[/tex]
[tex]F\ cos\theta-mg\ sin\theta=ma_x[/tex]
[tex]a_x=\dfrac{F\ cos\theta-mg\ sin\theta}{m}[/tex]
[tex]a_x=\dfrac{65\ N\ cos(30)-5\ kg\times 9.8\ m/s^2\ sin(30)}{5\ kg}[/tex]
[tex]a_x=6.35\ m/s^2[/tex]
So, the acceleration of the block is 6.35 m/s². Hence, this is the required solution.
Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².The magnitude of the acceleration of the block will be 6.35 m/sec².
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
On resolving the given force and accelertaion in the different components and balancing the equation gets.Components in the x-direction.
The given data in the problem,
F is the applied force =65 N
Θ is the angle of inclined plane=30°
m is the mass of the block= 5 kg
We need to find the acceleration of the block in the x-direction
[tex]\rm F_x=ma_x\\\\\rm F_x=Fcos\theta-mgsin\theta\\\\\rm Fcos\theta-mgsin\theta=ma_x\\\\\rm a_x=\frac{Fcos\theta-mgsin\theta}{m}\\\\ \rm a_x=\frac{65\timescos30^0-mgsin30^0}{5} \\\\\rm a_x=6.35 m/sec^2[/tex]
Hence the magnitude of the acceleration of the block will be 6.35 m/sec².
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Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load. Assume that the member is made of steel, which has a modulus of elasticity of E=204.00 N/mm2. Also assume that the member is 3048 mm long and has a cross-sectional area of 1290 mm2
Answer:
The total elongation for the tension member is of 0.25mm
Explanation:
Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:
[tex]\sigma=E*\epsilon[/tex] (1)
Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:
[tex]\delta L=L*\epsilon[/tex] (2)
Here L is the member extension and δL the change total longitudinal elongation.
Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:
[tex]\sigma=P/A[/tex]
[tex]\sigma=22.44N / 1290 mm^2[/tex]
[tex]\sigma=0.0174 N/mm^2[/tex]
Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:
[tex]=\sigma=E*\epsilon[/tex]
[tex]\epsilon=\sigma/E[/tex]
[tex]\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2} [/tex]
[tex]\epsilon=8.53*10^-{5}[/tex]
Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):
[tex]\delta L=3048 mm * 8.53*10^{-5} [/tex]
[tex]\delta L= 0.25 mm [/tex]
A metal ring 4.20 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.240 T/s . (A) What is the magnitude of the electric field induced in the ring? (B) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?
Answer:
09m
Explanation:
yes the current flow in clockwise direction
Explanation:
Given that,
Diameter of the metal ring, d = 4.2 cm
Radius, r = 2.1 cm
Initial magnetic field, B = 1.12 T
The magnetic field is decreasing at the rate of, [tex]\dfrac{dB}{dt}=0.24\ T/s[/tex]
Due to change in magnetic field, an emf is induced in it. And hence, electric field is induced. It is given by :
[tex]\int\limits {E.dl}=\dfrac{d}{dt}(\int\limits{B.dA)}[/tex]
[tex]E.(2\pi r)=\dfrac{dB}{dt}(\pi r^2)[/tex]
[tex]E=\dfrac{dB}{dt}\times \dfrac{r}{2}[/tex]
[tex]E=0.24\times \dfrac{2.1\times 10^{-2}}{2}[/tex]
[tex]E=0.00252\ N/C[/tex]
So, the magnitude of the electric field induced in the ring has a magnitude of 0.00252 N/C.
The direction of electric field will be counter clock wise direction as viewed by someone on the south pole of the magnet.
A motoris st enters a freeway at a speed of 35 mi/h and accelerates uniformly to a speed of 55 mi h. The motorist travels 500 ft while accelerating. I Determine: ) The acceleration of the car 2) The time required to reach 55 mi/h.
Answer:
1) The acceleration of the car = 1.17 m/s²
2) The time required to reach 55 mi/h = 7.59 seconds.
Explanation:
1) Initial speed of motorist, u = 35 mph = 15.56 m/s
Final speed of motorist, v = 55 mph = 24.44 m/s
Distance traveled, s = 500ft = 152.4 m
We have v² = u² + 2as
24.44² = 15.56² + 2 x a x 152.4
a = 1.17 m/s²
The acceleration of the car = 1.17 m/s²
2) We have v = u + at
24.44 = 15.56 + 1.17 x t
t = 7.59 s
The time required to reach 55 mi/h = 7.59 seconds.
A flowerpot falls off a windowsill and passes the win- dow of the story below. Ignore air resistance. It takes the pot 0.380 s to pass from the top to the bottom of this window, which is 1.90 m high. How far is the top of the window below the window- sill from which the flowerpot fell?
Answer:
d = 0.50 m
Explanation:
Let say the speed at the top and bottom of the window is
[tex]v_1 \: and \: v_2[/tex] respectively
now we have
[tex]d = \frac{v_1 + v_2}{2}t[/tex]
[tex]1.90 = \frac{v_1 + v_2}{2} (0.380)[/tex]
[tex]v_1 + v_2 = 10 [/tex]
also we know that
[tex]v_2 - v_1 = 9.8(0.380)[/tex]
[tex]v_2 - v_1 = 3.72[/tex]
now we have from above equations
[tex]v_2 = 6.86 m/s[/tex]
[tex]v_1 = 3.14 m/s[/tex]
now the distance from which it fall down is given as
[tex]v_f^2 - v_i^2 = 2ad[/tex]
[tex]3.14^2 - 0^2 = 2(9.8)d[/tex]
[tex]d = 0.50 m[/tex]
A negative charge -0.550 HC exerts an upward 0.700 N force on an unknown charge 0.220 m directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the -0.550 HC charge? downward upward
Answer:
Part a)
[tex]q_2 = -6.8 \mu C[/tex]
Part b)
[tex]F = 0.700 N[/tex]
direction = downwards
Explanation:
As we know that the negative charge will experience the force due to some other charge below it
the force is given as
[tex]F = 0.700 N[/tex]
now we know that
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now plug in all data
[tex]0.700 = \frac{(9 \times 10^9)(0.550\mu C)q_2}{0.220^2}[/tex]
[tex]0.700 = 1.022\times 10^5 q_2[/tex]
[tex]q_2 = -6.8 \mu C[/tex]
since this is a repulsion force so it must be a negative charge
Part b)
As per Newton's III law it will exert equal and opposite force on it
So here the force on the charge below it will be same in magnitude but opposite in direction
so here we have
[tex]F = 0.700 N[/tex]
direction = downwards
Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B carries a charge of -q. Sphere C carries no net charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Lastly, sphere C is touched to sphere B and separated from it. (a) What is the ratio of the final charge on sphere C to q? What is the ratio of the final total charge on the three spheres to q (b) before they are allowed to touch each other and (c) after they have touched?
Answer:
Part a)
Final charge on C : q = 1.875
Part b)
Ratio for A = 6 : 1.25
Ratio for B = -1 : 1.875
Ratio for C = 0
Explanation:
When two identical metal sphere are connected together then the charge on them will get equally divided on both after connecting them by conducting wire
So here we have
[tex]q_A = + 6q[/tex]
[tex]q_B = -q[/tex]
[tex]q_c = 0[/tex]
Step 1: We connected A and B and then separate them
so we have
[tex]q_A' = q_B' = 2.5q[/tex]
Step 2: We connected A and C and then separate them
so we have
[tex]q_A'' = q_c' = 1.25q[/tex]
Step 3: We connected B and C and then separate them
so we have
[tex]q_c'' = q_b'' = 1.875q[/tex]
The final charge on sphere C is 2.5q. Before touching, the total charge is +5q, which remains the same after all interactions, demonstrating conservation of charge.
Explanation:When identical conductive spheres come into contact, the charges redistribute evenly across both spheres. If Sphere A is initially charged with +6q and Sphere B has a -q charge, touching them together allows their total charge to be shared, resulting in each sphere having (6q - q)/2 = 2.5q. After separation, both spheres A and B would have a charge of 2.5q. By touching Sphere C, which is uncharged, to A and then B in sequence, C gains a fraction of the charge from each, ending up with (2.5q)/2 from A and (2.5q)/2 from B, which totals 2.5q, since touching B does not change the charge obtained from A.
Before contact, the total charge is +5q (+6q from A and -q from B). After all the interactions, the total charge remains the same, +5q, but redistributed: A and B with 2.5q each and C with 2.5q.
In general, the total charge before and after remains constant, demonstrating conservation of charge.
A 900 kg SUV is traveling at a constant speed of 1.44 m/s due north. What is the total force (in N) on the vehicle? (Assume north is the positive direction. Indicate the direction with the sign of your answer.)
Explanation:
It is given that,
Mass of SUV, m = 900 kg
It is moving with a constant speed of 1.44 m/s due south. We need to find the total force on the vehicle. The second law of motion gives the magnitude of force acting on the object. It is given by :
F = m a
Since, [tex]a=\dfrac{dv}{dt}[/tex] i.e. the rate of change of velocity
As SUV is travelling at a constant speed. This gives acceleration of it as zero.
So, F = 0
So, the total force acting on SUV is 0 N. Hence, this is the required solution.
Answer:
The total force acting on the vehicle is 0 N.
Explanation:
Given that,
Mass of SUV = 900 kg
Velocity = 1.44 m/s
SUV is travelling at a constant speed of 1.44 m/s due north.
We need to calculate the force on the vehicle.
Using newton's second law
[tex]F = ma[/tex]....(I)
We know that,
The acceleration is the first derivative of the velocity of the particle.
[tex]a = \dfrac{dv}{dt}[/tex]
Now,put the value of a in equation (I)
[tex]F=m\dfrac{dv}{dt}[/tex]
SUV is traveling at a constant speed it means acceleration is zero.
So, The force will be zero.
Hence, The total force acting on the vehicle is 0 N.
A research-level Van de Graaff generator has a 2.15 m diameter metal sphere with a charge of 5.05 mC on it. What is the potential near its surface in MV? (Assume the potential is equal to zero far away from the surface.)
Answer:
42.3 MV
Explanation:
d = diameter of the metal sphere = 2.15 m
r = radius of the metal sphere
diameter of the metal sphere is given as
d = 2r
2.15 = 2 r
r = 1.075 m
Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C
Potential near the surface is given as
[tex]V = \frac{kQ}{r}[/tex]
[tex]V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}[/tex]
V = 4.23 x 10⁷ volts
V = 42.3 MV
When 1.50 ✕ 10^5 J of heat transfer occurs into a meat pie initially at 20.0°C, its entropy increases by 465 J/K. What is its final temperature (in degrees)?
Answer:
The final temperature is 79.16°C.
Explanation:
Given that,
Heat [tex]Q=1.50\times10^{5}\ J[/tex]
Temperature = 20.0°C
Entropy = 465 J/k
We need to calculate the average temperature
Using relation between entropy and heat
[tex]\Delta S=\dfrac{\Delta Q}{T}[/tex]
[tex]T=\dfrac{\Delta Q}{\Delta S}[/tex]
Where, T = average temperature
[tex]\Delta Q[/tex]= transfer heat
[tex]\Delta S[/tex]= entropy
Put the value into the formula
[tex]T=\dfrac{1.50\times10^{5}}{465}[/tex]
[tex]T=322.58\ K[/tex]
We need to calculate the final temperature
Using formula of average temperature
[tex]T = \dfrac{T_{i}+T_{f}}{2}[/tex]
[tex]T_{f}=2T-T_{i}[/tex]....(I)
Put the value in the equation (I)
[tex]T_{f}=2\times322.58-293[/tex]
[tex]T_{f}=352.16\ K[/tex]
We convert the temperature K to degrees
[tex]T_{f}=352.16-273[/tex]
[tex]T_{f}=79.16^{\circ}\ C[/tex]
Hence, The final temperature is 79.16°C.
A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its normal)?
The light leaves the glass at about 42° relative to its normal
[tex]\texttt{ }[/tex]
Further explanationLet's recall Snell's Law about Refraction as follows:
[tex]\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}[/tex]
where:
n₁ = incident index
θ₁ = incident angle
n₂ = refracted index
θ₂ = refracted angle
[tex]\texttt{ }[/tex]
Given:
incident angle = θ₁ = 30°
index of refraction of air = n₃ = 1.0
index of refraction of glass = n₂ = 1.5
index of refraction of water = n₁ = 1.33
Asked:
refracted angle = θ₃ = ?
Solution:
Surface between Glass - Water:
[tex]n_1 \sin \theta_1 = n_2 \sin \theta_2[/tex]
[tex]\sin \theta_2 = \frac{n_1}{n_2} \sin \theta_1[/tex] → Equation A
[tex]\texttt{ }[/tex]
Surface between Air - Glass:
[tex]n_2 \sin \theta_2 = n_3 \sin \theta_3[/tex]
[tex]\sin \theta_3 = \frac{n_2}{n_3} \sin \theta_2[/tex]
[tex]\sin \theta_3 = \frac{n_2}{n_3} (\frac{n_1}{n_2} \sin \theta_1)[/tex] ← Equation A
[tex]\sin \theta_3 = \frac{n_1}{n_3} \sin \theta_1[/tex]
[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \sin 30^o[/tex]
[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \frac{1}{2}[/tex]
[tex]\sin \theta_3 \approx 0.665[/tex]
[tex]\boxed{\theta_3 \approx 42^o}[/tex]
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Light
The angle at which the light leaves the glass is approximately 19.5°.
Explanation:When a light ray crosses from one medium to another, such as from water to glass, it undergoes refraction, which causes the ray to bend. The angle between the incident ray and the normal to the surface of the glass is called the angle of incidence. To find the angle of incidence, we can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the indices of refraction of the two media.
In this case, the index of refraction of water (n1) is 1.33 and the index of refraction of glass (n2) is 1.5.
The angle of incidence (θ1) is given as 30°. We can use the formula: n1*sin(θ1) = n2*sin(θ2) to solve for θ2, the angle at which the light leaves the glass.
Plugging in the values, we get: 1.33*sin(30°) = 1.5*sin(θ2).
Solving for θ2 gives us θ2 = sin^(-1)((1.33*sin(30°))/1.5).
Evaluating this expression, we find that θ2 is approximately 19.5°.
The earth orbits the sun once per year at the distance of 1.50 x 1011 m. Venus orbits the sun at a distance of 1.08 x 1011 m. These distances are between the centers of the planets and the sun. How long (in earth days) does it take for Venus to make one orbit around the sun
It takes 225 Earth days for Venus to orbit the Sun. Interestingly, Venus spins on its axis so slowly that its day (243 Earth days) is longer than its year. The Sun takes 117 Earth days to return to the same place in Venus' sky.
Explanation:The time it takes for Venus to make one orbit around the sun, also known as its orbital period, is actually 225 Earth days. This is quite different compared to Earth which takes 365.25 days to orbit the Sun. Additionally, Venus spins on its axis very slowly, with its rotational period being 243 Earth days. As a result, a day on Venus - considering its rotation - is longer than its year! Also, this leads to an unusual phenomenon where the sun takes 117 Earth days to return to the same place in the Venusian sky. This suggests that Venus' rotation and orbit display unique characteristics compared to other planets in our solar system, likely due to factors from its formation.
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Venus completes one orbit around the Sun in approximately 225 Earth days. This is shorter than Earth's orbital period due to Venus's closer distance to the Sun.
Orbit of Venus Around the Sun-
The distance between the Earth and the Sun is approximately 1.50 x 1011 meters, while Venus is closer, at 1.08 x 1011 meters. This difference in distance influences the orbital period of each planet. Venus takes about 225 Earth days to complete one full orbit around the Sun.
Venus has a nearly circular orbit and, being closer to the Sun than Earth, receives almost twice as much light and heat. The elliptical orbits mean that different planets have different orbital periods, with those closer to the Sun having shorter years.