At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constant speed. At time t2 = 5.00 s, its acceleration is 4.00 i - 6.00 j m/s2 . What is the radius of the path taken by the particle if t2-t1 is less than one period?
Answers
The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.
At time [tex]t_1[/tex], the acceleration vector has direction [tex]\theta_1[/tex] such that
[tex]\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ[/tex]
which indicates the particle is situated at a point on the lower left half of the circle, while at time [tex]t_2[/tex] the acceleration has direction [tex]\theta_2[/tex] such that
[tex]\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ[/tex]
which indicates the particle lies on the upper left half of the circle.
Notice that [tex]\theta_1-\theta_2=90^\circ[/tex]. That is, the measure of the major arc between the particle's positions at [tex]t_1[/tex] and [tex]t_2[/tex] is 270 degrees, which means that [tex]t_2-t_1[/tex] is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.
Recall that
[tex]\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}[/tex]
where [tex]R[/tex] is the radius of the circle and [tex]T[/tex] is the period. We have
[tex]t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s[/tex]
and the magnitude of the particle's acceleration toward the center of the circle is
[tex]\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}[/tex]
So we find that the path has a radius [tex]R[/tex] of
[tex]7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}[/tex]
The acceleration of a particle moving in a circular path with constant speed is given by the centripetal acceleration. Since the velocity is not given, the radius cannot be determined with this information alone.
Explanation:In circular motion, the acceleration of a particle is given by the centripetal acceleration, which is equal to the square of the speed divided by the radius of the circular path (v^2/r). As the particle moves at a constant speed, we know that the magnitude of acceleration (which gives us the centripetal acceleration) remains the same at both times t1 and t2. Therefore, we can calculate the magnitude of acceleration at both points and set them equal to each other. This will allow us to solve for the radius of the path.
Calculate the magnitude of acceleration at t1 and t2:At t1, a = sqrt((6.00)² + (4.00)²) = 7.21 m/s².
At t2, a = sqrt((4.00)² + (-6.00)²) = 7.21 m/s².Since they're equal, we can set the expressions equal to each other:
(v²/r) = 7.21 m/s².Solving for r, we get:
r = v² / 7.21 m/s². However, since the velocity v is not given, it cannot be determined from this information alone.Learn more about Circular Motion here:
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Related Questions
Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2 where G is the gravitational constant and r is the distance between the bodies. (a) Find dF/dr. dF dr = − 2GmM r3 What is the meaning of dF/dr? dF/dr represents the rate of change of the mass with respect to the distance between the bodies. dF/dr represents the amount of force per distance. dF/dr represents the rate of change of the mass with respect to the force. dF/dr represents the rate of change of the distance between the bodies with respect to the force. dF/dr represents the rate of change of the force with respect to the distance between the bodies. What does the minus sign indicate? The minus sign indicates that as the distance between the bodies increases, the magnitude of the force decreases. The minus sign indicates that the bodies are being forced in the negative direction. The minus sign indicates that the force between the bodies is decreasing. The minus sign indicates that as the distance between the bodies increases, the magnitude of the force increases. The minus sign indicates that as the distance between the bodies decreases, the magnitude of the force remains constant. (b) Suppose it is known that the earth attracts an object with a force that decreases at the rate of 4 N/km when r = 20,000 km. How fast does this force change when r = 10,000 km? N/km
Answers
Answer 1 (a):
According to Newton's Law of Gravitation, the Gravity Force is:
[tex]F=\frac{GMm}{{r}^{2}}[/tex] (1)
This expression can also be written as:
[tex]F=GMm{r}^{-2}[/tex] (2)
If we derive this force [tex]F[/tex] respect to the distance [tex]r[/tex] between the two masses:
[tex]\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr[/tex] (3)
Taking into account [tex]GMm[/tex] are constants:
[tex]\frac{dF}{dr}dFdr=-2GMm{r}^{-3}[/tex] (4)
Or
[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex] (5)
Answer 2 (a): dF/dr represents the rate of change of the force with respect to the distance between the bodies.
In other words, this means how much does the Gravity Force changes with the distance between the two bodies.
More precisely this change is inversely proportional to the distance elevated to the cubic exponent.
As the distance increases, the Force decreases.
Answer 3 (a): The minus sign indicates that the bodies are being forced in the negative direction.This is because Gravity is an attractive force, as well as, a central conservative force.
This means it does not depend on time, and both bodies are mutually attracted to each other.
Answer 4 (b): [tex]X=-32N/km[/tex]
In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit [tex]N/km[/tex]:
[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex] (5)
We have a force that decreases with a rate 1 [tex]\frac{dF_{1}}{dr}dFdr=4N/km[/tex] when [tex]r=20000km[/tex]:
[tex]4N/km=-2\frac{GMm}{{(20000km)}^{3}}[/tex] (6)
Isolating [tex]-2GMm[/tex]:
[tex]-2GMm=(4N/km)({(20000km)}^{3})[/tex] (7)
In addition, we have another force that decreases with a rate 2 [tex]\frac{dF_{2}}{dr}dFdr=X[/tex] when [tex]r=10000km[/tex]:
[tex]XN/km=-2\frac{GMm}{{(10000km)}^{3}}[/tex] (8)
Isolating [tex]-2GMm[/tex]:
[tex]-2GMm=X({(10000km)}^{3})[/tex] (9)
Making (7)=(9):
[tex](4N/km)({(20000km)}^{3})=X({(10000km)}^{3}[/tex] (10)
Then isolating [tex]X[/tex]:
[tex]X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}[/tex]
Solving and taking into account the units, we finally have:
[tex]X=-32N/km[/tex]>>>>This is how fast this force changes when [tex]r=10000 km[/tex]
A 230 g air-track glider is attached to a spring. The glider is pushed in 8.2 cm and released. A student with a stopwatch finds that 13 oscillations take 19.0 s . What is the spring constant? Express your answer using two significant figures.
Answers
Answer:
4.3 N/m
Explanation:
m = 230 g = 0.230 kg, x = 8.2 cm
in 13 oscillations, time taken = 19 s
In 1 oscillation, time taken = 19 / 13 = 1.46 s
By the use of formula of time period , Let k be the spring constant.
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]1.46 = 2\pi \sqrt{\frac{0.230}{k}}[/tex]
0.054 = 0.230 / k
k = 4.26 N/m
k = 4.3 N/m
Three stars, each with the mass of our sun, form an equilateral triangle with sides 1.0×1012m long. (This triangle would just about fit within the orbit of Jupiter.) The triangle has to rotate, because otherwise the stars would crash together in the center. What is the period of rotation?
Answers
Answer:
period of rotation is 9.9843 years
Explanation:
Given data
sides = 1.0×10^12 m
to find out
period of rotation
solution
first we use the gravititional formula i.e
F(g) = 2 F cos 30
here F = G × mass(sun)² / radius²
F = 6.67 × [tex]10^{-11}[/tex] × (1.99× [tex]10^{30}[/tex])² / (1× [tex]10^{30}[/tex])²
so
F(g) = 2 × 6.67 × [tex]10^{-11}[/tex] × (1.99× [tex]10^{30}[/tex])² / (1× [tex]10^{30}[/tex])² × (1.73/2)
F(g) = 4.57 × [tex]10^{26}[/tex] N
and
by the law of sines
r/sin30 = s/sin120
so r will be
r = 1.0×10^12 (0.5/.87)
r = 5.75 × [tex]10^{11}[/tex] m
we know that gravititional force = centripetal force
so here
centripetal force = mv² /r
centripetal force = 1.99× [tex]10^{30}[/tex] v² / 5.75 × [tex]10^{11}[/tex]
so
4.57 × [tex]10^{26}[/tex] = mv² /r
4.57 × [tex]10^{26}[/tex] = 1.99× [tex]10^{30}[/tex] v² / 5.75 × [tex]10^{11}[/tex]
v² = 4.57 × [tex]10^{26}[/tex] × 5.75 × [tex]10^{11} / 1.99× [tex]10^{30}[/tex]
and
v = 11480 m/sec
and we know period of rotation formula
t = d /v
t = 2 [tex]\pi[/tex] r /v
t = 2 × [tex]\pi[/tex] × 5.75 × [tex]10^{11} / 11480
t = 314 × [tex]10^{5} / 31536000 year
t = 9.9843 year
so period of rotation is 9.9843 years
The period of rotation for the three stars can be calculated using Kepler's Third Law and recognizing that the stars form a circular orbital pattern. The relevant distance is from the center of the mass distribution (center of the equilateral triangle) to any of the stars and all three stars have identical masses.
Explanation:This question relates to astrophysics and can be solved using Kepler's Third Law that provides a relationship between the period of revolution (P), semimajor axis (D), and the total mass (M) of a binary system: D³ = (M₁ + M₂)P². However, since we are dealing with three stars of equal mass (each equivalent to the mass of our sun), they form a circular orbital pattern. We use the formula P = √((4π²/G)(r³/3M)), where r is the distance from the center of the mass distribution to any of the stars.
Firstly, we must understand that in an equilateral triangle, the distance from the center to a vertex is r = √3/2 * side. So, the distance from the center to any star is 0.866 * 1.0×10¹²m. The gravitational constant, G, is approximately 6.67 * 10^-11 N(m/kg)².
Substituting these values into our formula, we can calculate the period of rotation of the stars. One should note that all three stars have identical masses which simplifies the calculation.
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An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration due to gravity have the same sign on the way up as on the way down?
Answers
Answer:
a) The velocity is zero when the object is at the highest vertical point (peak). b) Yes, at the peak, the velocity for an instant gets to zero and then the direction of motion and the velocity will be downwards.
C) No, the direction of ‘g’ will be opposite in direction when the object is on the way down when compared to what it was during the upward motion.
Explanation:
a) Since gravity is acting on the object and trying to pull it down to earth, the ball will slow down to a velocity of zero when the ball starts its free fall downward.
b) When the direction of the object changes, its velocity will also change direction.
c) When the ball has a positive velocity, gravity is pulling it in the opposite direction, so it has a negative acceleration trying to slow it down. When the ball's velocity is negative (downward free fall), the acceleration will be positive at the instant when the ball hits the ground. The only way to stop the ball from falling is a positive force upward, which results in a positive acceleration when the ball hits the ground.
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(a) The velocity of the object will be zero at the maximum height
(b) The velocity of the object changes direction.
(c) The acceleration due to gravity will not have the same sign on the way up as on the way down.
(A) The velocity of the object decreases as the object ascends upwards and eventually becomes zero at the maximum height. As the object descends downwards, the velocity increases and becomes maximum before the object hits the ground.
(B) The velocity of the object is always directed in the direction of the motion of the object. As the object moves upward, the direction of the object is upward and as the object moves downwards the direction of the object is downwards. Thus, the velocity of the object changes direction.
(c) The upward motion of the object is opposite direction to acceleration due to gravity and the sign of acceleration due to gravity becomes negative.
As the object moves downwards, it will be moving in the same direction as the acceleration due to gravity. The sign of acceleration due to gravity is positive.
Thus, the acceleration due to gravity will not have the same sign on the way up as on the way down.
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A balloon is 3m in diameter and is filled with 5 kg of N2. Find the specific volume of the gas in the balloon. Express it in both m/kg and in m/kmol. Answers: about 3 and 80, respectively Find the weight of the gas in the balloon, expressed in N. Assume g = 9.8 m/s. Answer: about 50 a. b.
Answers
Answer:
The the specific volume and weight are 79.16m³/k mol and 49 N.
Explanation:
Given that,
Diameter = 3 m
Mass of N₂ = 5 kg
We need to calculate the volume of balloon
Using formula of volume
[tex]V= \dfrac{4}{3}\pi r^3[/tex]
[tex]V= \dfrac{4}{3}\pi \times{\dfrac{d}{2}}^3[/tex]
[tex]V= \dfrac{4}{3}\pi \times(\dfrac{3}{2})^3[/tex]
[tex]V=14.137\ m^3[/tex]
We need to calculate the specific volume in the balloon
[tex] Specific\ volume=\dfrac{Volume}{mass}[/tex]
[tex]Specific\ volume=\dfrac{14.137}{5}[/tex]
[tex]Specific\ volume=2.8274\ m^3/kg[/tex]
Now, Molar mass of N₂ gas is 0.028 kg/mol
Now, Specific volume of N₂ gas in the balloon is
[tex]Specific\ volume =0.028\times2.8274=0.0791672\ m^3/mol[/tex]
[tex]Specific\ volume =79.16m^3/kmol[/tex]
(b). We need to calculate the weight of the gas in the balloon
Weight of the balloon is dependent on the mass of the gas.
The weight of the gas is given by
[tex]W = mg[/tex]
[tex]W = 5\times9.8[/tex]
[tex]W = 49\ N[/tex]
Hence, The the specific volume and weight are 79.16m³/k mol and 49 N.
A dense metal sphere is dropped from a 10-meter tower, and at the exact same time an identical metal sphere is thrown horizontally from the same height. If air friction is ignored, which statement below is true?
A. Both spheres land at the same time.
B. The thrown sphere lands before the dropped sphere.
C. The dropped sphere lands before the thrown sphere.
D. The horizontal speed of the thrown sphere will determine whether it will land first or not.
Answers
Answer:
A. Both spheres land at the same time.
Explanation:
The horizontal motion doesn't affect the vertical motion. Since the two spheres have the same initial vertical velocity and same initial height, they land at the same time.
Answer:
A. Both spheres land at the same time.
Explanation:
If air resistance and friction are ignored, then both spheres land at the same time. Falling objects fall toward the center of the Earth with the same constant acceleration, independent of their mass, when they are defined to be in free-fall. For an object to be in free-fall, it has to be in a situation in which both air resistance and friction are considered negligible. This is true regardless of the direction of the fall.
A force vector has a magnitude of 720 N and a direction of 38 degree north of east. Determine the magnitude and direction of the components of the force that point along the north-south line and along east-west line.
Answers
Answer:
443.3 N
567.4 N
Explanation:
Consider the triangle ABC
AC = hypotenuse = magnitude of force vector = F = 720 N
AB = adjacent = Component of force along east-west line = [tex]F_{x}[/tex]
BC = Opposite = Component of force along north-south line = [tex]F_{y}[/tex]
θ = Angle = 38 deg
In triangle ABC
[tex]Sin38 = \frac{BC}{AC}[/tex]
[tex]Sin38 = \frac{F_{y}}{F}[/tex]
[tex]0.616 = \frac{F_{y}}{720}[/tex]
[tex]F_{y}[/tex] = 443.3 N
Also, In triangle ABC
[tex]Cos38 = \frac{AB}{AC}[/tex]
[tex]Cos38 = \frac{F_{x}}{F}[/tex]
[tex]0.788 = \frac{F_{x}}{720}[/tex]
[tex]F_{x}[/tex] = 567.4 N
A uniform plank is 8-m long and weighs 99 N. It is balanced on a sawhorse at its center. An additional 182 N weight is now placed on the left end of the plank. To keep the plank balanced, it must be moved what distance (in m) to the right? Round your answer to the nearest tenth.
Answers
Answer:
It must be moved by 2.59 m to the right
Explanation:
To keep the plank in uniform state, the moment at sawhorse = 0
Let y be the distance of saw horse from left of plank.
Weight per unit length of plank [tex]=\frac{99}{8}=12.375N/m[/tex]
Taking moment about saw horse
Moment due to left portion,
182 x y + 12.375 x y x 0.5 y = 182 y + 6.1875 y²
Moment due to right portion,
12.375 x (8-y) x 0.5 (8-y) = 396 + 6.1875 y² - 99y
So to keep balance these moments should be same
182 y + 6.1875 y² = 396 + 6.1875 y² - 99y
281 y = 396
y = 1.41 m
Distance moved to right = 0.5 x 8 - 1.41 = 2.59 m
So it must be moved by 2.59 m to the right
The thickness of a $1 bill is 0.11 mm. If you have a stack of $1 bills 450 m tall, how much money do you have?
Answers
Answer:
You will have 4.5 million dollar
Explanation:
The thickness of a $1 bill is 0.11 mm
So we have
1 $ = 0.1 mm
0.1 mm = 1 $
0.0001 m = 1 $
1 m = 10000 $
450 m = 450 x 10000 = 4500000 $
So you will have 4.5 million dollar
Answer:
4090909
Explanation:
Thickness of one bill = 0.11 mm
Total thickness = 450 m
No of $1 bills = total thickness / thickness of one bill
No of $1 bills = 450 / 0.11 × 10^-3
= 4090909
5. A current loop sits in the x ? y plane, centered on the origin. At the origin, the magnetic field due to the loop points in the positive z direction. What direction does the current in the loop run when viewed from the positive z direction?
(a) Counter-clockwise.
(b) Clockwise.
(c) The current is zero.
(d) More information is needed.
Answers
Answer:
option (a)
Explanation:
The direction of current is given by the Maxwell's right hand thumb rule.
If we hold a current carrying conductor in our right hand such that the thumb indicates the direction of magnetic field then the curling of fingers give the direction of current.
Here the direction of current is in counter clockwise direction.
A 92.0kg baseball player slides into second base. The coefficient of friction between the player and the ground is 0.61. If the player comes to rest after 1.2s, what was his initial speed?
Answers
Answer:
The initial speed is 7.174 m/s.
Explanation:
Given that,
Mass = 92.0 kg
Coefficient of friction = 0.61
Time = 1.2 s
We need to calculate the acceleration
Using of friction force
[tex]F = \mu mg[/tex]...(I)
Where, [tex]\mu[/tex] =Coefficient of friction
g = acceleration due to gravity
Using newton's law
[tex]F = ma[/tex]....(II)
m = mass of the baseball
a = acceleration of the baseball
From equation (I) and (II)
[tex]ma=\mu mg[/tex]...(I)
Put the value in the equation
[tex]a=0.61\times9.8[/tex]
[tex]a=5.978\ m/s^2[/tex]
We need to calculate the initial velocity
Using equation of motion
[tex]v = u-at[/tex]
Where, v = final velocity
u = initial velocity
a = acceleration
Here, a is negative because the player comes to rest
t = time
Put the value in the equation
[tex]0=u-5.978\times1.2[/tex]
[tex]u=7.174\ m/s[/tex]
Hence, The initial speed is 7.174 m/s.
The initial speed of the baseball player was approximately 7.176 m/s, calculated using the principles of Physics (Kinematics) and understanding of friction.
Explanation:The subject of the problem involves a concept in Physics known as Kinematics, specifically dealing with friction and motion. The initial speed of the baseball player can be calculated using the equation of motion: final velocity (v_f) = initial velocity (v_i) + acceleration (a) x time (t), where the final velocity in this case is 0 (as the player comes to rest), time is 1.2 seconds, and the acceleration is the frictional force divided by the mass of the player. Frictional force is obtained by multiplying the mass of the player, the gravitational acceleration, and the coefficient of friction. In this case, substituting acceleration = -0.61 * 9.8 m/s² (because it's a decelerating force), the equation becomes 0 = v_i - 0.61 * 9.8 * 1.2 from which initial velocity (v_i) comes out to be approximately 7.176 m/s.
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0. A 3.00-kg block is dropped from rest on a vertical spring whose spring constant is 750 N/m. The block hits the spring, compresses it, and then comes to rest, having fallen through a total vertical distance of 0.45 m. How far is the spring compressed? (Hint: Use conservation of energy.)
Answers
Answer:
the spring compressed is 0.1878 m
Explanation:
Given data
mass = 3 kg
spring constant k = 750 N/m
vertical distance h = 0.45
to find out
How far is the spring compressed
solution
we will apply here law of mass of conservation
i.e
gravitational potential energy loss = gain of eastic potential energy of spring
so we say m×g×h = 1/2× k × e²
so e² = 2×m×g×h / k
so
we put all value here
e² = 2×m×g×h / k
e² = 2×3×9.81×0.45 / 750
e² = 0.0353
e = 0.1878 m
so the spring compressed is 0.1878 m
Given a thermal conductivity constant of a material of 0 4, a surface area of 100 square feet, a temperature on the far side of a material of 165°F and a temperature of 150°F on the near side, and a material thickness of 1.5 inches, what is the conductive heat transfer rate? 2400 F per ft. 4800 F per ft. 3600 F per ft. 6400 F per ft.
Answers
Answer:4800
Explanation:
Given data
k=0.4
surface area(A)=100[tex]ft&^{2}[/tex]
Temprature on near side of material[tex](T_1)[/tex]=150°F
temprature on far side of material[tex](T_2)[/tex]=165°F
thickness(t)=1.5in=0.125[tex]ft^2[/tex]
Conductive heat transfer rate(Q) =[tex]\frac{k\times A\times\left ( T_2-T_1\right )}{y}[/tex]
Q=[tex]\frac{0.4\times 100\left ( 165-150\right )}{1.5}[/tex]
Q=4800°F per feet
A 55-kg box is being pushed a distance of 7.0 m across the floor by a force whose magnitude is 160 N. The force is parallel to the displacement of the box. The coefficient of kinetic friction is 0.25. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.
Answers
Answer:
work done by applied force = 1120J
work done by frictional force = 943.25J
work done by the gravitational force = work done by the normal force = 0J
Explanation:
Given:
Mass of the box = 55kg
Displacement of the box = 7.0m across the floor
Applied force = 160N
Coefficient of kinetic friction ([tex]\mu_k[/tex]) = 0.25
Now the work done (W) is given as:
W = F×d
Where,
F = Applied force on the body
d = Displacement of the body in the direction of the applied force
Forces acting on the box are:
1) Applied force
2) Frictional Force
3) Gravitational force
4) Normal force due to the weight
Now the work done by the respective forces:
1) By the applied force
W = 160 kg × 7 m = 1120 J
2) By the frictional force
[tex]W_f =-\mu mg\times displacement[/tex]
Where
g = acceleration due to the gravity
and the negative sign here depicts that the frictional force is acting the against the applied force and the direction of displacement.
thus,
[tex]W_f =-0.25\times 55\times 9.8\times 7[/tex]
or
[tex]W_f =-943.25J[/tex]
The work done by the normal force and the gravitational force will be zero as there is no displacement in the direction of the application of both the forces
The work done by both the gravitational force and normal force is equal to zero (0).
Given the following data:
Mass = 55 kg.Distance = 7 meters.Magnitude of force = 160 Newton.Coefficient of kinetic friction = 0.25.The forces acting on the box.In this scenario, there are four forces acting on the box and these include:
An applied forceA frictional ForceA gravitational forceA normal force due to weight.For the applied force:
Mathematically, the work done due to an applied force is given by this formula:
[tex]Work\;done = force \times distance\\\\Work\;done = 160 \times 7[/tex]
Work done = 1,120 Nm.
For the frictional force:
Mathematically, the work done due to a frictional force is given by this formula:
[tex]Work\;done = -\mu mg \times distance\\\\Work\;done = -0.25 \times 55 \times 9.8 \times 7[/tex]
Work done = -943.25.
However, the work done by both the gravitational force and normal force is equal to zero (0) because there isn't any displacement or distance covered in the direction of the application of these forces.
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It takes about 8 minutes for light to travel from the Sun to Earth. Pluto is 40 times as far from us as the Sun when Pluto is closest to Earth. How long does it take light to reach Earth from Pluto? (b) Radio waves travel at the speed of light. What problems would you have if you tried to conduct a two-way conversation between Earth and a spacecraft orbiting Pluto?
Answers
Answer:
(a) 5 hours 20 min
(b)I would have a great inconvenience with the delay of the signal
Explanation:
Hello
To find the time it takes for light to reach Earth from Pluto, we use a rule of three
Step 1
define
distance from sun to earth=x
distance from pluto to earth=40x
Step 2
Rule of three
distance time
x ⇒ 8 minutes
40x ⇒ A minutes
[tex]A=\frac{40x*8*min}{x}\\ A=320 min\\\\to\ convert\ to\ hours\\\\A=320\ min*\frac{1 h}{60 mi}\\ A=5.33 hours\\\\A=5\ hours\ and\ 20min\\\\[/tex]
(a) 5 hours 20 min
Step 3
I would have a great inconvenience with the delay of the signal, since the message I send will arrive 5 hours and 20 minutes later, if the message is answered immediately I will listen to it 320 min after that, more than ten and a half hours to give and receive a hello
Consider a 26-MeV proton moving perpendicularly to a 1.35 T field in a cyclotron. Find the radius of curvature, in meters, of the path of the proton while moving through the cyclotron.
Answers
Answer:
0.545 m
Explanation:
K. E = 26 MeV = 26 x 1.6 x 10^-13 J = 4.16 x 10^-12 J
B = 1.35 T
Let r be the radius of curvature
The formula for the kinetic energy of a cyclotron is given by
[tex]K.E. = \frac{B^{2}q^{2}r^{2}}{2m}[/tex]
m = 1.67 x 10^-27 kg, q = 1.6 x 10^-19 c
[tex]4.16\times 10^{-12} = \frac{1.35^{2}\times (1.6\times 10^{-19})^{2}r^{2}}{2\times 1.67\times 10^{-27}}[/tex]
r = 0.545 m
A0.12 kg yo-yo is whirled around by its string in a circle of radius 0.44 m at a constant speed of 6 m/s. What is the required centripetal force in Newtons to keep this yo -yo whirling in a circle?
Answers
Answer:
Centripetal force, F = 9.81 N
Explanation:
It is given that,
Mass of the object (yo-yo), m = 0.12 kg
It is whirled around by its string in a circle
Radius of circle, r = 0.44 m
Speed of object, v = 6 m/s
We need to find the required centripetal force in Newtons to keep this yo -yo whirling in a circle. The centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]F=\dfrac{0.12\ kg\times (6\ m/s)^2}{0.44\ m}[/tex]
F = 9.81 Newton
So, the centripetal force required to keep this yo -yo whirling in a circle is 9.81 N. Hence, this is the required solution.
A particle moves in a 15-cm-radius circle centered at the origin and completes 1.0 rev every 3.0 s. (a) Find the speed of the particle. (b) Find its angular speed o. (c) Write an equation for the x component of the position of the particle as a function of time t, assuming that the particle is on the -x axis at time t 0
Answers
Answer:
a) [tex]V=0.314 m/s[/tex]
b) [tex]\omega=2.09rad/s[/tex]
c) [tex]x(\theta)=0.15\times cos(\frac{2\pi}{3.0} t)[/tex]
Explanation:
Given:
Radius of the circle, r = 15cm = 0.15m
Time period, T =3.0s
a) The velocity (V) of a particle moving in the circular motion is given as:
[tex]V=\frac{2 \pi r}{T}[/tex]
substituting the given values in the above equation we get
[tex]V=\frac{2\times \pi \times 0.15m}{3.0s}[/tex]
or
[tex]V=0.314 m/s[/tex]
b) Angular speed ([tex]\omega[/tex]) is given as:
[tex]\omega=\frac{2\pi}{T}[/tex]
or
[tex]\omega=\frac{2\times \pi}{3.0s}[/tex]
or
[tex]\omega=2.09rad/s[/tex]
c) The position of the particle on the x-position is given as:
[tex]x(\theta)=rcos(\theta)[/tex] (reffer the attached figure)
now the relation between the Θ and the time T is given as:
[tex]\omega = \frac{2\pi}{T}=\frac{\theta}{t}[/tex]
or
[tex]\theta= \frac{2\pi}{T}\times t[/tex]
or
[tex]\theta= \frac{2\pi}{3}\times t[/tex]
substituting the values of r and Θ, we get
[tex]x(\theta)=0.15\times cos(\frac{2\pi}{3.0} t)[/tex]
Final answer:
The particle's speed is 0.1π m/s, its angular speed is 2/3π rad/s, and the x-component of the position function is x(t) = 0.15 * cos((2/3π) * t + π).
Explanation:
(a) Speed of the particle: The speed is the distance traveled by a particle along a circular path per unit time. Given a particle completes 1 revolution (which is 2π times the radius r) every 3 seconds, the speed v is given by: v = (2π * 15 cm) / 3 s. Convert 15 cm to meters by dividing by 100 to get 0.15 meters. So, v = (2π * 0.15 m) / 3 s = 0.1π m/s.
(b) Angular speed of the particle: The angular speed ω is the rate of change of the angular displacement and is measured in radians per second. Since the particle completes one revolution every 3 seconds and one revolution is 2π radians, ω = 2π radians / 3 s = 2/3π rad/s.
(c) Position function for the x-component: Assume the particle starts on the negative x-axis, so the angle θ at t=0 is π. The particle moves in a circle with radius r = 0.15 m. The x-component of the position as a function of time t, using the cosine function is x(t) = r * cos(ω*t + π), where ω = 2/3π rad/s. Plugging values, we get x(t) = 0.15 * cos((2/3π) * t + π).
A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.6 m/s and at an angle of 40.8° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.03 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.73 s, and for the (e) horizontal and (f) vertical components at t = 5.05 s. Assume that the catapult is positioned on a plain horizontal ground.
Answers
To answer this question, let us break down the initial velocity into its horizontal and vertical components.
v_{ix} = vcos(θ)
v_{iy} = vsin(θ)
v_{x} is the horizontal component of the initial velocity, and this value will stay constant over time because we assume only gravity acts on the stone, therefore only the vertical component of the stone's velocity will change over time
v_{iy} is the vertical component of the initial velocity, and this will change over time due to gravity.
v is the magnitude of the initial velocity.
θ is the angle the velocity vector is oriented at with respect to the horizontal.
Given values:
v = 19.6m/s
θ = 40.8°
Plug in these values and solve for v_{ix} and v_{iy}:
v_{ix} = 19.6cos(40.8°) = 14.8m/s
v_{iy} = 19.6sin(40.8°) = 12.8m/s
To find both the horizontal and vertical displacement at any time, we will use this kinematics equation:
D = X + Vt + 0.5At²
When solving for the horizontal displacement, the following values are:
t = elapsed time
D = horizontal displacement
X = initial horizontal displacement
V = initial horizontal velocity
A = horizontal acceleration
There is no initial horizontal displacement, so X = 0m
The initial horizontal velocity V = v_{ix} = 14.8m/s
Assuming we don't care about air resistance, no force has a component acting horizontally on the stone, so A = 0m/s²
Therefore the equation for the stone's horizontal displacement is given by:
D = 14.8t
When solving for the vertical displacement, the following values are:
t = elapsed time
D = vertical displacement
X = initial vertical displacement
V = initial vertical velocity
A = vertical acceleration
There is no initial vertical displacement, so X = 0m
The initial vertical velocity V = v_{iy} = 12.8m/s
Gravity acts downward on the stone, therefore A = -9.81m/s²
Therefore the equation for the stone's vertical displacement is given by:
D = 12.8t - 4.905t²
Now we just plug in various values of t...
a) At t = 1.03s, the horizontal displacement is D = 14.8(1.03) = 15.2m
b) At t = 1.03s, the vertical displacement is D = 12.8(1.03)-4.905(1.03)² = 7.98m
c) At t = 1.73s, the horizontal displacement is D = 14.8(1.73) = 25.6m
d) At t = 1.73s, the vertical displacement is D = 12.8(1.73)-4.905(1.73)² = 7.46m
Before you write down the following results, read the following explanation.
e) At t = 5.05s, the horizontal displacement is D = 14.8(5.05) = 74.7m
f) At t = 5.05s, the vertical displacement is D = 12.8(5.05)-4.905(5.05)² = -60.4m
Notice that we have a negative value for the vertical displacement. This isn't possible within the context of the problem, so the vertical displacement at t = 5.05s is actually 0m
The problem hasn't stated whether the ground is frictionless or not. I'm going to assume it is frictionless and therefore the stone will keep moving across the ground after landing, so the horizontal displacement at t = 5.05s is 74.7m
A student is skateboarding down a ramp that is 8.51 m long and inclined at 23.7° with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 4.97 m/s. Neglect friction and find the speed at the bottom of the ramp.
Answers
To find the speed at the bottom of the ramp, we can use the principles of physics. By equating the potential energy at the top to the kinetic energy at the bottom, we can solve for the velocity at the bottom of the ramp. Using the given values and trigonometry, we can calculate the height of the ramp and then determine the velocity.
Explanation:To find the speed of the skateboarder at the bottom of the ramp, we can use the principles of physics. Since we are neglecting friction, we can assume that the mechanical energy of the skateboarder is conserved. The mechanical energy consists of both kinetic energy and potential energy. At the top of the ramp, the skateboarder has only potential energy, which is given by the formula: potential energy = mass * gravity * height. At the bottom of the ramp, the skateboarder has only kinetic energy, which is given by the formula: kinetic energy = 1/2 * mass * velocity^2. By equating the potential energy at the top to the kinetic energy at the bottom, we can solve for the velocity at the bottom of the ramp.
In this case, the mass of the skateboarder is not provided, but it cancels out when equating the potential and kinetic energies. So we can assume any mass value as long as it is consistent throughout the calculation. Let's assume the mass of the skateboarder is 1 kg for simplicity. The height of the ramp is not provided, but we can calculate it using trigonometry. The height is the vertical component of the ramp's length, which is given by: height = length * sin(angle). Plug in the values and calculate the height.
Next, we calculate the potential energy at the top of the ramp using the calculated height. The potential energy is then equated to the kinetic energy at the bottom of the ramp and solve for the velocity. Plug in the values and calculate the velocity at the bottom of the ramp.
What is the area outside the thermodynamic system called? O A) Phase O B) Universe ° C) Isolated 。D) Surroundings 0 E) External Conditions Save
Answers
Answer:
Correct option is 'd' Surroundings
Explanation:
The space surrounding a thermodynamic system is known as Surroundings
At one instant, a 19.0-kg sled is moving over a horizontal surface of snow at 4.50 m/s. After 7.80 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving.
Answers
Answer:
Frictional force, F = -10.96 N
Explanation:
It is given that,
Mass of the sled, m = 19 kg
Initial velocity of the sled, u = 4.5 m/s
After 7.80 s has elapsed, the sled stops. We need to find the magnitude of the average friction force acting on the sled while it was moving. The change in momentum is equal to the impulse imparted i.e.
[tex]F.\Delta t=m(v-u)[/tex]
[tex]F=\dfrac{m(v-u)}{\Delta t}[/tex]
[tex]F=\dfrac{19\ kg(0-4.5\ m/s)}{7.8\ s}[/tex]
F = -10.96 N
So, the average frictional force acting on the sled is 10.96 N. Hence, this is the required solution.
At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda decreases at a constant rate of 0.25 in/sec.
Answers
Answer:
The soda is being sucket out at a rate of 3.14 cubic inches/second.
Explanation:
R= 2in
S= π*R²= 12.56 inch²
rate= 0.25 in/sec
rate of soda sucked out= rate* S
rate of soda sucked out= 3.14 inch³/sec
Final answer:
Soda is being pulled out of a cylindrical glass at the rate of -3.14 cubic inches per second.
Explanation:
The question asks at what rate soda is being absorbed from a cylindrical glass. Given the cylindrical glass's dimensions and the rate at which the depth of soda decreases, we need to find the volume rate of change.
He formula for the volume of a cylinder is V = πr2h, where r is the radius and h is the height of the cylinder. To find the rate at which volume changes, we differentiate with respect to time to obtain dV/dt = πr2(dh/dt). Plugging in the values, we have r = 2 inches and dh/dt = -0.25 inches/second (negative because the depth is decreasing).
Using the given values, the rate at which soda is being pulled out of the glass is
dV/dt = π(2 in)2(-0.25 in/sec) = π(4 in2)(-0.25 in/sec) = -3.14 in3/sec.
Therefore, the rate at which the soda is being pulled out of the glass is -3.14 in3/sec.
A spring-loaded toy gun is used to shoot a ball of mass 1.5 kg straight up in the air, as shown. The spring has spring constant 777 N/m. If the spring is compressed a distance of 20.0 centimeters from its equilibrium position and thern released, the ball reaches a maximum height (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis. (a) How much energy is stored in the spring when it is compressed 20.0 cm? (b) Find the maximum height of the ball. (C) Find the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position)
Answers
The energy stored in the spring when it is compressed 20.0 cm is 15.54J, the maximum height of the ball is 1.056 meters. while the muzzle velocity of the ball is 4.55 m/s.
a) Potential Energy in the Spring
Potential energy = 1/2 * spring constant * (compression distance)^2
Potential energy = 1/2 * 777 N/m * (0.2 m)^2 = 15.54 J
b) Maximum Height of the Ball
Maximum height = potential energy / weight of the ball
Maximum height = 15.54 J / (1.5 kg * 9.81 m/s^2) = 1.056 m
Therefore, the maximum height reached by the ball is 1.056 meters.
(c) Muzzle Velocity of the Ball
muzzle velocity = sqrt(2 * potential energy / mass)
muzzle velocity = sqrt(2 * 15.54 J / 1.5 kg) = 4.55 m/s
Hence, the muzzle velocity of the ball is 4.55 m/s.
A) The energy stored in the spring when compressed is 15.54 J. B) The maximum height reached by the ball is 1.05 m. C) The muzzle velocity of the ball is 4.56 m/s.
A) The energy stored in a spring, also known as elastic potential energy, can be calculated using the formula:
Espring = ½ * k * x2
where k is the spring constant, and x is the compression distance. By plugging in the given values (k = 777 N/m and x = 0.20 m), we get:
Espring = ½ * 777 N/m * (0.20 m)2 ≈ 15.54 J
B) At the maximum height, all the elastic potential energy is converted into gravitational potential energy (Egravitational), given by:
Egravitational = m * g * h
Where m is the mass of the ball, g is the acceleration due to gravity (9.81 m/s2), and h is the height. Since Espring = Egravitational, we can solve for h:
h = Espring / (m * g) = 15.54 J / (1.5 kg * 9.81 m/s2) ≈ 1.05 m
C) The muzzle velocity of the ball can be found using the conservation of energy principle. The elastic potential energy stored in the spring is converted entirely into the kinetic energy of the ball at the spring's equilibrium position:
Kinetic energy, Ekinetic = ½ * m * v2
Since Espring = Ekinetic, solving for v gives:
v = √(2 * Espring / m) = √(2 * 15.54 J / 1.5 kg) ≈ 4.56 m/s
The peak electric field in an intense IR laser beam is about 25,000 V/m. How much energy does this laser beam deliver in 1 minute if it makes a spot with a diameter of 2mm?
Answers
Answer:
156.3 J
Explanation:
E = Amplitude of the electric field = 25000 V/m
t = time interval = 1 minute = 60 sec
d = diameter of the spot = 2 mm = 2 x 10⁻³ m
Area of the spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Energy delivered to the spot is given as
U = (0.5)ε₀ E² c A t
Inserting the values
U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)
U = 156.3 J
To find the energy delivered by the laser beam in 1 minute, we need to calculate the power of the laser beam and then multiply it by the duration.
E = Amplitude of the electric field = 25000 V/m
t = time interval = 1 minute = 60 sec
d = diameter of the spot = 2 mm = 2 x 10⁻³ m
Area of the spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Energy delivered to the spot is given as
U = (0.5)ε₀ E² c A t
Inserting the values
U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)
U = 156.3 J
X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{\circ}θ=30 ∘ relative to the incident X-rays, what is the energy of the recoiling electron?
Answers
Answer:
37.91594 keV
Explanation:
[tex]E_i[/tex] = Incident energy = 400 keV
θ = 30°
h = Planck's constant = 4.135×10⁻¹⁵ eV s = 6.626×10⁻³⁴ J s
Incident photon wavelength
[tex]\lambda_i=\frac{hc}{E_i}\\\Rightarrow \lambda_i=\frac{4.135\times 10^{-15}\times 3\times 10^8}{400\times 10^3}\\\Rightarrow \lambda_i=3.101\times 10^{-12}\ m[/tex]
Difference in wavelength
[tex]\Delta \lambda=\frac{h}{m_ec}(1-cos\theta)\\\Rightarrow \Delta \lambda=\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8}(1-cos30)\\\Rightarrow \Delta \lambda=3.248\times 10^{-13}\ m[/tex]
[tex]\lambda_f=\lambda_i+\Delta \lambda\\\Rightarrow \lambda_f=3.101\times 10^{-12}+3.248\times 10^{-13}\\\Rightarrow \lambda_f=3.426\times 10^{-12}[/tex]
Final photon wavelength
[tex]\lambda_f=\frac{hc}{\lambda_f}\\\Rightarrow E_f=\frac{4.135\times 10^{-15}\times 3\times 10^8}{3.426\times 10^{-12}}\\\Rightarrow E_f=362084.06\ eV = 362.08406\ keV[/tex]
Energy of the recoiling electron
[tex]\Delta E=E_i-E_f\\\Rightarrow \Delta E=400-362.08406=37.91594\ keV[/tex]
Energy of the recoiling electron is 37.91594 keV
Answer:
The energy of recoiling electron=192.44 keV
Explanation:
Energy of x-ray[tex]E_o=400 keV[/tex]
Web know that compton shift is
[tex]\Delta \lambda =\dfrac{h}{m_eC(1-cos\theta)}[/tex]
[tex]m_e[/tex] is the mass of electron and C is the velocity of sound.
Given that θ=30°
Now by putting the values
[tex]\Delta \lambda =\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8(1-cos30)}[/tex]
[tex]\Delta \lambda =2.8\times 10^{-3}nm[/tex]
[tex]\Delta \lambda_o=\dfrac{hc}{E_o}[/tex]
By putting the values
[tex]\Delta \lambda_o=\dfrac{6.63\times 10^{-34}3\times 10^8}{400\times 1.602\times 10^{-16}}[/tex]
[tex]\Delta \lambda_o=3.31times 10^{-3}nm[/tex]
[tex]\lambda =\Delta \lambda +\lambda _o[/tex]
[tex]\lambda=2.8\times 10^{-3}+3.31\times 10^{-3}nm[/tex]
[tex]\lambda=6.11`\times 10^{-3}nm[/tex]
Energy [tex]E=\dfrac{hC}{\lambda }[/tex]
So [tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{6.11\times 10^{-12}}[/tex]
E=207.55 keV
The energy of recoiling electron=400-207.55 keV
The energy of recoiling electron=192.44 keV
A particle starts from the origin at t = 0 with an initial velocity of 5.3 m/s along the positive x axis.If the acceleration is (-2.6 i^ + 4.7 j^)m/s2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x coordinate.
Answers
Answer:
Velocity at the point of maximum x cordinate is 9.578m/s
Position vector of the particle when it reaches point of maximum x ordinate is [tex]\overrightarrow{r}=5.402\widehat{i}+9.76\widehat{j}[/tex]
Explanation:
We shall resolve the motion of the particle along x and y direction separately
The particle will reach it's maximum x coordinate when it's velocity along x axis shall become 0
We have acceleration along x-axis = [tex]-2.6m/s^{2}[/tex]
acceleration along y-axis = [tex]4.7m/s^{2}[/tex]
Thus using the first equation of motion along x axis we get
[tex]v_{x}=u_{x}+a_{x}t\\\\[/tex]
Applying values we get
[tex]0=5.3-2.6t\\\\\therefore t=\frac{5.3}{2.6}sec\\\\t=2.038sec[/tex]
Now to obtain it's position we shall use third equation of motion
[tex]v_{x}^{2}=u_{x}^{2}+2as_{x}\\\\0=(5.3)^{2}+2(-2.6)s_{x}\\\\\therefore s_{x}=\frac{-28.09}{-5.2}m\\\\s_{x}=5.402m[/tex]
Now it's location along y- axis can be obtained using 2nd equation of motion along the y axis
[tex]s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}[/tex]
Applying values as follows we get
[tex]u_{y}=0\\a_{y}=4.7m/s^{2}\\t=2.038s[/tex]
[tex]s_{y}=0\times 2.038+\frac{1}{2}\times 4.7m/s^{2}\times2.038^{2}\\\\s_{y}=9.76m[/tex]
thus the position vector of the particle when it reaches it's maximum x co-ordinate is
[tex]\overrightarrow{r}=5.402\widehat{i}+9.76\widehat{j}[/tex]
Now velocity of the particle at the position of maximum x co-ordinate shall be zero along x-axis and along the y-axis it can be found along the first equation of motion along y axis
[tex]v_{y}=u_{y}+a_{y}t\\\\v_{y}=0+4.7\times 2.038\\\\v_{y}=9.578m/s[/tex]
Discuss Joule-Thompson effect with relevant examples and formulae.
Answers
Answer:
[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]
Explanation:
Joule -Thompson effect
Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.
Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.
Now lets take Steady flow process
Let
[tex]P_1,T_1 [/tex] Pressure and temperature at inlet and
[tex]P_2,T_2 [/tex] Pressure and temperature at exit
We know that Joule -Thompson coefficient given as
[tex]\mu _j=\left(\frac{\partial T}{\partial p}\right)_h[/tex]
Now from T-ds equation
dh=Tds=vdp
So
[tex]Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp[/tex]
⇒[tex]dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]
So Joule -Thompson coefficient
[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]
This is Joule -Thompson coefficient for all gas (real or ideal gas)
We know that for Ideal gas Pv=mRT
[tex]\dfrac{\partial v}{\partial T}=\dfrac{v}{T}[/tex]
So by putting the values in
[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]
[tex]\mu _j=0[/tex] For ideal gas.
If the speed of light in a substance is 2.26 x 10^8 m/s, what is the index of refraction of that substance?
Answers
The index of refraction can be calculated using the speed of light in vacuum divided by the speed of light in a specific medium. Given the speed of light in a medium is 2.26 x 10^8 m/s, the index of refraction would be calculated as roughly 1.33.
Explanation:To calculate the index of refraction for a certain medium, you would use the speed of light in vacuum (c) divided by the speed of light in that medium (v). This equation is denoted as n = c/v.
The speed of light in vacuum is approximately 3.00×10^8 m/s. Given that the speed of light in a certain substance is 2.26 x 10^8 m/s, we can substitute these values into the equation to find the material's index of refraction.
So, n = (3.00×10^8 m/s) / (2.26 x 10^8 m/s). Therefore, the index of refraction of that substance is approximately 1.33. This property of a substance tells us how much the path of light is bent, or refracted, when entering the material.
Learn more about Index of Refraction here:https://brainly.com/question/30761100
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A communication satellite is orbiting at an altitude of 3 earth radii above the ground. what is the speed?
Answers
Answer:
4557.88 m/s
Explanation:
M = mass of earth = 5.98 x 10²⁴ kg
R = radius of earth = 6.4 x 10⁶ m
r = orbital radius of satellite = 3R = 3 x 6.4 x 10⁶ = 19.2 x 10⁶ m
[tex]v[/tex] = speed of the satellite
speed of the satellite is given as
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
[tex]v=\sqrt{\frac{(6.67\times 10^{-11})(5.98\times 10^{24})}{19.2\times 10^{6}}}[/tex]
[tex]v[/tex] = 4557.88 m/s