Answer:
29274.93096 m/s
[tex]2.73966\times 10^{33}\ J[/tex]
[tex]-5.39323\times 10^{33}\ J[/tex]
[tex]2.56249\times 10^{33}\ J[/tex]
[tex]-5.21594\times 10^{33}[/tex]
Explanation:
[tex]r_p[/tex] = Distance at perihelion = [tex]1.471\times 10^{11}\ m[/tex]
[tex]r_a[/tex] = Distance at aphelion = [tex]1.521\times 10^{11}\ m[/tex]
[tex]v_p[/tex] = Velocity at perihelion = [tex]3.027\times 10^{4}\ m/s[/tex]
[tex]v_a[/tex] = Velocity at aphelion
m = Mass of the Earth = 5.98 × 10²⁴ kg
M = Mass of Sun = [tex]1.9889\times 10^{30}\ kg[/tex]
Here, the angular momentum is conserved
[tex]L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s[/tex]
Earth's orbital speed at aphelion is 29274.93096 m/s
Kinetic energy is given by
[tex]K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J[/tex]
Kinetic energy at perihelion is [tex]2.73966\times 10^{33}\ J[/tex]
Potential energy is given by
[tex]P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}[/tex]
Potential energy at perihelion is [tex]-5.39323\times 10^{33}\ J[/tex]
[tex]K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J[/tex]
Kinetic energy at aphelion is [tex]2.56249\times 10^{33}\ J[/tex]
Potential energy is given by
[tex]P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}[/tex]
Potential energy at aphelion is [tex]-5.21594\times 10^{33}\ J[/tex]
Answer:
:) *_* :3 ^-^ {.}{.}
Explanation:
The uniform crate has a mass of 150 kg. The coefficient of static friction between the crate and the floor is μs = 0.2. The coefficient of static friction between the man's shoes and the floor is μ′s = 0.4. Assume the man only exerts a horizontal force on the crate. Determine whether the 85-kg man can move the crate.
Answer:
The man will be able to move the crate.
Explanation:
[tex]\mu_s[/tex] = Coefficient of static friction between the crate and the floor = 0.2
[tex]\mu'_s[/tex] = Coefficient of static friction between the man's shoes and the floor = 0.4
[tex]m_c[/tex] = Mass of crate = 150 kg
[tex]m_p[/tex] = Mass of man = 85 kg
g = Acceleration due to gravity = 9.81 m/s²
Horizontal force in order to move the crate is given by
[tex]F_h=\mu_sm_cg\\\Rightarrow F_h=0.2\times 150\times 9.81\\\Rightarrow F_h=294.3\ N[/tex]
Maximum force that the man can apply
[tex]F_m=\mu'_sm_pg\\\Rightarrow F_m=0.4\times 85\times 9.81\\\Rightarrow F_m=333.54\ N[/tex]
Here it can be seen that [tex]F_m>F_h[/tex].
So, the man will be able to move the crate.
Final answer:
To determine whether the 85-kg man can move the 150-kg crate, we need to compare the force of static friction between the crate and the floor with the force exerted by the man. If the force of static friction is greater than the force exerted by the man, the crate will not move. If it is less than or equal to the force exerted by the man, the crate will move.
Explanation:
The force of static friction can be calculated using the equation:
fs = μsN
where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.
In this case, the normal force can be calculated as:
N = mass × gravity
where the mass is the total mass of the crate and the man, and gravity is the acceleration due to gravity.
If the force of static friction is greater than the force exerted by the man, the crate will not move. If the force of static friction is less than or equal to the force exerted by the man, the crate will move.
In this case, we have:
Mass of crate = 150 kg
Mass of man = 85 kg
Coefficient of static friction between crate and floor (μs) = 0.2
Coefficient of static friction between man's shoes and floor (μ's) = 0.4
Force exerted by man = mass of man × acceleration due to gravity
The normal force can be calculated as the sum of the weight of the crate and the weight of the man:
Normal force (N) = (mass of crate + mass of man) × acceleration due to gravity
The force of static friction between the crate and floor can be calculated as:
fs = μs × N
If this force is greater than the force exerted by the man, the crate will not move. If it is less than or equal to the force exerted by the man, the crate will move.
A wave pulse travels down a slinky. The mass of the slinky is m = 0.89 kg and is initially stretched to a length L = 6.2 m. The wave pulse has an amplitude of A = 0.27 m and takes t = 0.408 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.44 Hz.1) What is the speed of the wave pulse? 17.035 m/s
2) What is the tension in the slinky? 33.92 N
3) What is the average speed of a piece for the slinky as a complete wave pulse passes? .4704 m/s
4) What is the wavelength of the wave pulse? 34.765 m
1) The wave speed is 15.2 m/s
2) The tension in the slinky is 33.2 N
3) The average speed of a piece of the slinky during one pulse is 0.475 m/s
4) The wavelength is 34.5 m
Explanation:
1)
The motion of a wave pulse along the slinky is a uniform motion, therefore its speed is given by the equation for uniform motion:
[tex]v=\frac{L}{t}[/tex]
where
L is the length covered
t is the time elapsed
For the wave in this problem, we have:
L = 6.2 m is the length of the slinky
t = 0.408 s is the time taken for a pulse to travel across the length os the slinky
Substituting, we find the wave speed
[tex]v=\frac{6.2}{0.408}=15.2 m/s[/tex]
2)
The speed of a wave on a slinky can be found with the same expression for the wave speed along a string:
[tex]v=\sqrt{\frac{T}{m/L}}[/tex]
where
T is the tension in the slinky
m is the mass of the slinky
v is the wave speed
L is the length
In this problem, we have:
m = 0.89 kg is the mass of the slinky
L = 6.2 m is the length
Therefore, we can re-arrange the equation to find the tension in the slinky, T:
[tex]T=v^2 (\frac{m}{L})=(15.2)^2 (\frac{0.89}{6.2})=33.2 N[/tex]
3)
The average speed of a piece of the slinky as a complete wave pulse passes is the total displacement done by a piece of slinky during one period, which is 4 times the amplitude, divided by the time taken for one complete oscillation, the period:
[tex]v_{avg} = \frac{4A}{T}[/tex]
where
A is the amplitude
T is the period
Here we have:
A = 0.27 m is the amplitude of the wave
The period is the reciprocal of the frequency, therefore
[tex]T=\frac{1}{f}[/tex]
where f = 0.44 Hz is the frequency of this wave. Substituting and solving, we find
[tex]v_{avg} = \frac{4A}{1/f}=4Af=4(0.27)(0.44)=0.475 m/s[/tex]
4)
The wavelength of the wave pulse can be found by using the wave equation:
[tex]v=f\lambda[/tex]
where
v is the wave speed
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the pulse in this problem, we have
v = 15.2 m/s
f = 0.44 Hz
Substituting, we find the wavelength:
[tex]\lambda=\frac{v}{f}=\frac{15.2}{0.44}=34.5 m[/tex]
Learn more about waves:
brainly.com/question/5354733
brainly.com/question/9077368
#LearnwithBrainly
The speed of the wave pulse is 17.035 m/s, the tension in the slinky is 33.92 N, the average speed of a piece is 17.035 m/s, and the wavelength of the wave pulse is 34.765 m.
Explanation:1) The speed of the wave pulse can be calculated using the formula v = λf, where v is the speed, λ is the wavelength, and f is the frequency. In this case, the frequency is given as 0.44 Hz and the wavelength can be calculated as λ = v/f = 17.035/0.44 = 38.716 m.
The speed of the wave pulse is therefore 17.035 m/s.
2) The tension in the slinky can be determined using the formula T = 2μv², where T is the tension, μ is the mass per unit length, and v is the speed of the wave pulse. The mass per unit length can be calculated as μ = m/L = 0.89/6.2 = 0.143 kg/m.
The tension is then T = 2 * 0.143 * (17.035)² = 33.92 N.
3) The average speed of a piece of the slinky can be calculated as v_avg = λ/T, where λ is the wavelength and T is the period of the wave pulse. The period can be calculated as T = 1/f = 1/0.44 = 2.2727 s.
The average speed is then v_avg = 38.716/2.2727 = 17.035 m/s.
4) The wavelength of the wave pulse is given as λ = v/f = 17.035/0.44 = 34.765 m.
Learn more about Wave pulse here:https://brainly.com/question/14885673
#SPJ3
10%) Problem 7: Water flows through a water hose at a rate of Q1 = 620 cm3/s, the diameter of the hose is d1 = 1.99 cm. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of v2 = 10.4 m/s. show answer No Attempt 17% Part (a) Enter an expression for the cross-sectional area of the hose, A1, in terms of its diameter, d1.
Answer:
a) [tex]A_1 = \frac{\pi d_1^2}{4}[/tex]
Explanation:
a) the cross-sectional area of the hose would be the square of radius times pi. And since the sectional radius is half of its diameter d. We can express the cross-sectional area A1 in term of diameter d1
[tex]A_1 = \pi r_1^2 = \pi (d_1/2)^2 = \frac{\pi d_1^2}{4}[/tex]
A satellite is put into an elliptical orbit around the Earth. When the satellite is at its perigee, its nearest point to the Earth, its height above the ground is hp = 229.0 km, and it is moving with a speed of up = 8.050 km/s. The gravitational constant G equals 6.67 x 10-11 mº.kg-1.5-2 and the mass of Earth equals 5.972 x 1024 kg. When the satellite reaches its apogee, at its farthest point from the Earth, what is its height ha above the ground? For this problem, choose gravitational potential energy of the satellite to be 0 at an infinite distance from Earth. km
Answer:
The height above the ground is 1279.51 km
Solution:
As per the question:
Height above the ground, [tex]h_{p} = 229.0\ km[/tex]
Speed of the satellite, [tex]v_{p} = 8.050\ km/s = 8050\ m/s[/tex]
Gravitational constant, [tex]G = 6.67\times 10^{- 11}\ m.kg[/tex]
Mass of the Earth, m = [tex]5.972\times 10^{24}\ kg[/tex]
Now,
The distance of the earth from the perigee is given by:
[tex]R_{p} = h_{p} + R_{E}[/tex]
where
[tex]R_{E} = 6371\ km[/tex]
[tex]R_{p} = 229.0 + 6371 = 6600\ km[/tex]
Now,
To calculate the distance of the earth from the earth from the apogee:
[tex]R_{A} = \frac{R_{p}}{\frac{2Gm}{v_{p}^{2}R_{p} - 1}}[/tex]
[tex]R_{A} = \frac{6600\times 1000}{\frac{2\times 6.67\times 10^{- 11}\times 5.972\times 10^{24}}{8050^{2}6600\times 1000 } - 1}}[/tex]
[tex]R_{A} = 7650.51\ km[/tex]
Height above the ground, [tex]h_{A} = R_{A} - R_{E} = 7650.51 - 6371 = 1279.51\ km[/tex]
A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 40 kJ/min. Determine the rate of heat transfer to the kitchen air in kilojoules per minute to three significant digits. Pay attention to the units asked for in the answer !!!
Answer:
73.3 kJ / min
Explanation:
COP or coefficient of performance of a refrigerator is defined as ratio of heat extracted from the refrigerator to electrical imput to the refrigerator
If Q₁ be the heat extracted out and Q₂ be the heat given out to the surrounding
Imput energy = Q₂ - Q₁
so COP = Q₁ / Q₂ - Q₁
Given
COP = 1.2
Q₁ = 40kJ
Substituting the values
1.2 = 40 / (Q₂ - 40)
1.2 (Q₂ - 40) = 40
1.2 Q₂ = 2.2 X 40
Q₂ = 73.3 kJ / min
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1800 kg and was approaching at 5.00 m/s due south. The second car has a mass of 700 kg and was approaching at 21.0 m/s due west. (a) Calculate the final velocity of the cars. (b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.)
Answer:
The final velocity of the cars is 6.894 m/s
The kinetic energy lost is 117,441 J
Explanation:
Using the conservation of the linear momentum :
[tex]P_i = P_f[/tex]
Where [tex]P_i[/tex] is the inicial linear momentum and [tex]P_f[/tex] is the final linear momemtum.
The linear momentum is calculated by:
[tex]P = MV[/tex]
where M is the mass and V is the velocity.
First we identify the directions of the velocity of both cars:
the first car is moving in the axis y for the south direction, we will take this direction like positive.the second car is moving in the axis x to the west, we will take this direction like positive.The conservation of the linear momentum is made on direction x, so:
[tex]P_i = P_f\\M_{1}V_{1ix} +M_{2}V_{2ix} = V_{sx}(M_1+M_2)[/tex]
where [tex]M_1[/tex] is the mass of the car 1, [tex]V_{1ix}[/tex] is the car 1's inicial velocity in the axis x, [tex]M_2[/tex] is the mass of the car 2, [tex]V_{2ix}[/tex] is the car 2's inicial velocity in the axis x and [tex]V_{sx}[/tex] is the velocity of the car 1 and car 2 after the collition.
The car 1 just move in the axis y so, it dont have horizontal velocity ([tex]V_{1ix}[/tex] = 0)
now the equation is:
[tex]M_2V_{2ix} = V_{sx}(M_1 +M_2)[/tex]
Replacing the values, we get:
(700 kg)(21 m/s) = [tex]V_{sx}[/tex]( 1800 kg + 700 kg)
solving for [tex]V_{sx}[/tex]:
[tex]V_{sx} = \frac{700(21)}{1800+700}[/tex]
[tex]V_{sx}[/tex] = 5.88 m/s
Now we do the conservation of the linear momentum on direction y:
[tex]P_i = P_f\\M_{1}V_{1iy} +M_{2}V_{2iy} = V_{sy}(M_1+M_2)[/tex]
where [tex]M_1[/tex] is the mass of the car 1, [tex]V_{1iy}[/tex] is the car 1's inicial velocity in the axis y, [tex]M_2[/tex] is the mass of the car 2, [tex]V_{2iy}[/tex] is the car 2's inicial velocity in the axis y and [tex]V_{sy}[/tex] is the velocity of the car 1 and car 2 together after the collition.
The car 2 just move in the axis x so, it don't have horizontal velocity ([tex]V_{2iy}[/tex] = 0)
now the equation is:
[tex]M_1V_{1iy} = V_{sy}(M_1 +M_2)[/tex]
Replacing the values, we get:
[tex]1800(5 m/s) = V_{sy}(1800 +700)[/tex]
solving for [tex]V_{sx}[/tex]:
[tex]V_{sy} = \frac{1800(5)}{1800+700}[/tex]
[tex]V_{sy}[/tex] = 3.6 m/s
Now, we have the two components of the velocity and using pythagorean theorem we find the answer as:
[tex]V_f = \sqrt{5.88^2+3.6^2}[/tex]
[tex]V_f =[/tex] 6.894 m/s
Finally we have to find the kinetic energy lost, so the kinetic energy is calculated by:
K = [tex]\frac{1}{2}MV^2[/tex]
so, ΔK = [tex]K_f -K_i[/tex]
where [tex]K_f[/tex] is the final kinetic energy and [tex]K_i[/tex] is the inicial kinetic energy.
then:
[tex]K_i = \frac{1}{2}(1800)(5)^2+\frac{1}{2}(700)(21)^2= 176,850 J[/tex]
[tex]K_f = \frac{1}{2}(1800+700)(6.894)^2 = 59,409 J[/tex]
so:
ΔK = 59409 - 176850 = -117441 J
Final answer:
The final velocity of the cars is -9.48 m/s due south. The amount of kinetic energy lost in the collision is 49275 J.
Explanation:
To find the final velocity of the cars, we can use the conservation of momentum principle. Since the cars stick together after the collision, their combined mass is 1800 kg + 700 kg = 2500 kg. We can calculate the momentum of each car before the collision and add them together. Car A's momentum is 1800 kg * (-5.00 m/s) = -9000 kg·m/s, and Car B's momentum is 700 kg * (-21.0 m/s) = -14700 kg·m/s. The combined momentum of the cars is -9000 kg·m/s + (-14700 kg·m/s) = -23700 kg·m/s. Since momentum is conserved, the combined momentum of the cars after the collision is also -23700 kg·m/s. Dividing this momentum by the combined mass of the cars gives us the final velocity: -23700 kg·m/s / 2500 kg = -9.48 m/s. The magnitude of the final velocity is 9.48 m/s, and the direction is south since Car A originally had a southward velocity.
To calculate the amount of kinetic energy lost in the collision, we need to find the initial and final kinetic energies. The initial kinetic energy is given by the formula KE = 0.5 * mass * velocity^2. For Car A, the initial kinetic energy is 0.5 * 1800 kg * (5.00 m/s)^2 = 11250 J. For Car B, the initial kinetic energy is 0.5 * 700 kg * (21.0 m/s)^2 = 51450 J. The total initial kinetic energy is 11250 J + 51450 J = 62700 J.
The final kinetic energy is given by the formula KE = 0.5 * mass * velocity^2. For the cars combined, the final kinetic energy is 0.5 * 2500 kg * (9.48 m/s)^2 = 111975 J.
The kinetic energy lost in the collision is the difference between the initial kinetic energy and the final kinetic energy, which is 62700 J - 111975 J = 49275 J.
Consider a rod of length L rotated about one of its ends instead of about its center of mass. If the mass of the rod is 5 kg, and the length is 2 meters, calculate the magnitude of the moment of inertia (I).
Answer:
[tex]I_{edge} = 6.67 kg.m^2[/tex]
Explanation:
given,
mass of rod = 5 Kg
Length of rod = 2 m
R = 1 m
moment of inertial from one edge of the rod = ?
moment of inertia of rod through center of mass
[tex]I_{CM}= \dfrac{1}{12}M(L)^2[/tex]
using parallel axis theorem
[tex]I_{edge} = I_{CM} + MR^2[/tex]
[tex]I_{edge} =\dfrac{1}{12}ML^2+ M(\dfrac{L}{2})^2[/tex]
[tex]I_{edge} =\dfrac{1}{12}ML^2+ \dfrac{ML^2}{4}[/tex]
[tex]I_{edge} =\dfrac{1}{3}ML^2[/tex]
now, inserting all the given values
[tex]I_{edge} =\dfrac{1}{3}\times 5 \times 2^2[/tex]
[tex]I_{edge} = 6.67 kg.m^2[/tex]
A food department is kept at 2128C by a refrigerator in an environment at 308C. The total heat gain to the food department is estimated to be 3300 kJ/h and the heat rejection in the condenser is 4800 kJ/h. Determine the power input to the compressor, in kW and the COP of the refrigerator.
Answer:
2.2
Explanation:
[tex]Q_u[/tex] = Heat rejection in the condenser = 3300 kJ/h
[tex]Q_L[/tex] = Heat gain to the food department = 4800 kJ/h
Power output is given by
[tex]W=Q_u-Q_L\\\Rightarrow W=4800-3300\\\Rightarrow W=1500\ kJ/h[/tex]
COP of a refrigerator is given by
[tex]COP=\frac{Desired\ effect}{Work}\\\Rightarrow COP=\frac{Q_L}{W}\\\Rightarrow COP=\frac{3300}{1500}\\\Rightarrow COP=2.2[/tex]
The COP of the refrigerator is 2.2
A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a distance of 0.38 cm. Find the speed of the bullet as it emerges from the block if its initial speed is 450 m/s
Answer:395.6 m/s
Explanation:
Given
mass of bullet [tex]m=5 gm[/tex]
mass of wood block [tex]M=1 kg[/tex]
Length of string [tex]L=2 m[/tex]
Center of mass rises to an height of [tex]0.38 cm[/tex]
initial velocity of bullet [tex]u=450 m/s[/tex]
let [tex]v_1[/tex] and [tex]v_2[/tex] be the velocity of bullet and block after collision
Conserving momentum
[tex]mu=mv_1+Mv_2[/tex] -------------1
Now after the collision block rises to an height of 0.38 cm
Conserving Energy for block
kinetic energy of block at bottom=Gain in Potential Energy
[tex]\frac{Mv_2^2}{2}=Mgh_{cm}[/tex]
[tex]v_2=\sqrt{2gh_{cm}} [/tex]
[tex]v_2=\sqrt{2\times 9.8\times 0.38}[/tex]
[tex]v_2=0.272 m/s[/tex]
substitute the value of [tex]v_2[/tex] in equation 1
[tex]5\times 450=5\times v_1+1000\times 0.272[/tex]
[tex]v_1=395.6 m/s[/tex]
Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission is in the infrared region of the spectrum. The other has a temperature of 12,000 K, so its frequency of maximum emission is in the ultraviolet region of the spectrum.
Compared to the object at 3000 K, the object at 12,000 K:
A. emits more light in the infrared region of the spectrum.
B. emits more light in the visible region of the spectrum.
C. emits more light in the ultraviolet region of the spectrum
D. emits more light in only two of the three listed spectral regions.
E. emits more light in all three spectral regions.
The object at 12,000 K emits more light in the ultraviolet region of the spectrum compared to the object at 3000 K.
Explanation:When comparing the two objects, we can see that the object at 12,000 K emits more light in the ultraviolet region of the spectrum compared to the object at 3000 K. This is because as the temperature of an object increases, the frequency of maximum emission also increases. The object at 12,000 K has a higher frequency of maximum emission, which falls in the ultraviolet region of the spectrum.
A lidless shoebox is made of five rectangular pieces of cardboard forming its base and four sides. Its dimensions are: base length L = 48 cm, base width W = 25 cm, and side height H = 11 cm. The base lies in quadrant I of the xy-plane, with its length along the x-axis and its width along the y-axis. The box’s sides extend in the positive z-direction. The cardboard is thin and has an area mass density of s = 0.075 g/cm2.(a) What is the x-coordinate, in centimeters, of the center of mass of the shoebox's base?
The x-coordinate of the center of mass of the lidless shoebox's base, which is a rectangle, will be at half its length. Therefore, the x-coordinate is 24 cm.
Explanation:The center of mass is the coordinate that averages the distribution of mass in a physical object. To calculate the center of mass, one must consider the dimensions of the object. For a rectangular object like the shoebox's base with length, L = 48 cm, and width, W = 25 cm, the x-coordinate of the center of mass will be half the length. Therefore, the x-coordinate of the center of mass of the shoebox's base is L/2 = 48 cm / 2 = 24 cm. This is because in this rectangular coordinate system, the center of mass is located at the middle along each dimension (L, W).
Learn more about Center of Mass here:https://brainly.com/question/33607198
#SPJ12
The x-coordinate of the center of mass of the shoebox's base is half its length which is 24 cm.
Explanation:
The center of mass of an object is basically the average position of all its parts. Given that the shoebox's base lies on the xy-plane, the x-coordinate for its center of mass will simply be half the length of the base. Hence, using the length L = 48 cm, the x-coordinate for the center of mass is: L/2 = 48/2 = 24 cm. Therefore, the center of mass of the shoebox's base along the x-axis lies at 24 cm.
Learn more about Center of Mass here:https://brainly.com/question/33607198
#SPJ11
The radius of Earth is 6370 km in the Earth reference frame. The cosmic ray is moving at 0.880Co relative to Earth.
a. In the reference frame of a cosmic ray how wide does Earth seem along the flight direction?
b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?
Express your answer with the appropriate units.
Answer:
6052114.67492 m
[tex]12.742\times 10^{6}\ m[/tex]
Explanation:
v = Velocity of cosmic ray = 0.88c
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
d = Width of Earth = Diameter of Earth = [tex]12.742\times 10^{6}\ m[/tex]
When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original
This happens due to length contraction
Length contraction is given by
[tex]d_e=d\sqrt{1-\frac{v^2}{c^2}}\\\Rightarrow d_e=12.742\times 10^{6}\sqrt{1-\frac{0.88^2c^2}{c^2}}\\\Rightarrow d_e=6052114.67492\ m[/tex]
The Earth's width is 6052114.67492 m
Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.
Hence, the width is [tex]12.742\times 10^{6}\ m[/tex]
Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500 km (or even higher) above the surface. Io has a mass of 8.93×1022 kg and a radius of 1821 km . For this calculation, ignore any variation in gravity over the 500-km range of the debris.
Final answer:
Io remains volcanically active due to tidal heating caused by Jupiter's gravitational forces.
Explanation:
Io, the innermost of Jupiter's Galilean moons, is known for its high level of volcanism, making it the most volcanically active body in the solar system. Despite its small size, Io remains volcanically active due to the effect of gravity, specifically tidal heating caused by the massive gravity of Jupiter. Jupiter's strong gravitational forces pull Io into an elongated shape, resulting in tidal heating that generates enough internal heat to drive volcanic activity on the moon.
The equation of motion for a simple harmonic oscillator (SHO) is:
m(d2x/ dt2) = −kx
where m is the mass and k the spring constant. A generic solution of the above differential equation can be written in the form:
x(t) = A cos(ωt + φ0).
Where ω = p k/m and A and φ0 are arbitrary constants to be determined by the initial conditions of the motion. You are given the following information: a block of mass m = 10 kg is attached to a spring with spring constant k = 2.5N/m; at time t = 0 the amplitude of the motion is 0, and the velocity is -5 m/s.
(a) Using the generic solution (1) above find the form x(t) for the position as a function of time for the block (you need to determine the constants A, ω and φ0). Do not forget to write clearly the appropriate dimension for each quantity, A, φ0, ω.
(b) Plot the function x(t) that you found in part (a) from t = 0 until t = 20s.
(You can use a graphical calculator or WolframAlpha or Desmos to make your plot)
Answer:
Explanation:
ω = [tex]\sqrt{\frac{k}{m} }[/tex]
k = 2.5 N/m
m = 10 kg
[tex]\omega = \sqrt{\frac{2.5}{10} }[/tex]
ω = .5 rad /s
x(t) = A cos(ωt + φ₀)
When t = 0 , x(t) = 0
0 = A cos(ωx 0 + φ₀)
cos φ₀ = 0
φ₀ = π /2
x(t) = A cos(ωt +π /2 )
Putting the value of ω
x(t) = A cos(.5 t +π /2 )
Differentiating on both sides
dx(t)/dt = - .5 A sin(.5 t +π /2 )
v(t) = - .5 A sin(.5 t +π /2 )
Given t =0 , v(t) = -5 m/s
-5 = - .5 A sin(.5 x0 +π /2 )
-5 = - .5 A sinπ /2
A = 10 m
x(t) = 10 cos( .5 t +π /2 )
b )
when t = π ( 3.14 s )
x(t) = - 10 m
when t = 2π ( 6.28s )
x(t) = 0
when t = 3π ( 9.42 s )
x(t) = 10 m
and so on
Final answer:
The position as a function of time for the block in the given scenario is x(t) = 0 for any value of t.
Explanation:
To find the form x(t) for the position as a function of time for the block in the given scenario, we first need to determine the constants A, ω, and φ0. We are given that at time t = 0, the amplitude of the motion is 0 and the velocity is -5 m/s.
Using the equation x(t) = A cos(ωt + φ0), we can determine the values of A, ω, and φ0. Since the amplitude is 0, we have A = 0. The velocity is given by v(t) = - Umax sin(ωt + φ0), where Umax = √(k/m)A. Plugging in the given velocity of -5 m/s, we can solve for Umax and find its value to be 5 m/s.
Using the equation ω = √(k/m), we can substitute in the given values to find ω. Therefore, ω = √(2.5 N/m /10 kg) = 0.5 rad/s. Finally, we can substitute the values of A = 0, ω = 0.5 rad/s, and φ0 = 0 into the equation x(t) = A cos(ωt + φ0) to find the form of x(t) as x(t) = 0 cos(0.5t + 0) = 0 for any value of t.
A conducting sphere of radius R carries positive charge q. Calculate the amount of work that would be required to move a small positive test charge q0 slowly from r=5R to r=3R. Assume that the presence of q0 has no effect on how the charge q is distributed over the sphere.
Express your answer in terms of the electric constant ϵ0 and some or all of the variables q, q0, and R.
The work done to move a positive test charge from a point at a radial distance of 5R from the center of a charged sphere to a point at a radial distance of 3R is given by the difference in the electric potentials at these points times the charge of the test charge.
Explanation:The work done, W, in moving a small positive test charge, q0, in an electric field produced by another charged object is given by the expression W = q0 x (Vf - Vi), where Vf and Vi are> the final and initial electric potentials, respectively. The electric potential, V, at a point located a distance r from the center of a conducting sphere carrying a charge q is V = 1/4πε0 x q/r. So, the work done to move the test charge from r = 5R to r = 3R is W = q0 x {[1/4πε0 x q/(3R)] - [1/4πε0 x q/(5R)]}, which simplifies to W = q0q/4πε0R x (5/15 - 3/25).
Learn more about Electric Field and Work here:https://brainly.com/question/12578759
#SPJ12
The work required to move a small positive test charge from a radius of 5R to 3R on a conducting sphere of radius R carrying positive charge q is calculated to be W = q*q0/(30πϵ0R) using Coulomb's law and the work-energy theorem.
Explanation:The work needed to move a charge in an electric field is given by the integral of force times distance. The force on a charge in an electric field is given by Coulomb's law: F = k*q*q0/r^2, where k is Coulomb's constant = 1/4π*ϵ0. The work done in moving a charge from r1 to r2 is given by the integral from r1 to r2 of dr, which results in k*q*q0*(1/r1 - 1/r2).
In this case, with r1=5R and r2=3R, the work is W = k*q*q0*(1/5R - 1/3R), which simplifies to W = 2k*q*q0/(15R), or W = q*q0/(30πϵ0R) when substituting k = 1/4π*ϵ0.
Learn more about Work Done in Electric Field here:https://brainly.com/question/32913781
#SPJ11
To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity as measured in W/m2 changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I isβ=10log(II0)dB,where I0 is a reference intensity. For sound waves, I0 is taken to be 10−12W/m2. Note that log refers to the logarithm to the base 10.Part AWhat is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity (i.e., I=10I0)?Part BWhat is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity (i.e. I=100I0)?Express the sound intensity numerically to the nearest integer.
Most of the information to solve this problem is provided in the statement, therefore we will apply the concepts related to the intensity of the sound and its method of representation across the logarithmic scale.
By definition as we saw the level of sound intensity in decibels is represented by
[tex]\beta = 10log(\frac{I}{I_0})dB[/tex]
Where, I = Intensity for which decibels is to be calculated
[tex]I_0[/tex]= Reference intensity (at this case is [tex]10^{-12}W/m^2[/tex]
PART A ) Intensity is 10 times the reference intensity.
Here [tex]I = 10I_0[/tex], replacing
[tex]\beta = 10log(\frac{10I_0}{I_0})dB[/tex]
[tex]\beta = 10log(10)dB[/tex]
[tex]\beta = 10dB[/tex]
Therefore the sound intensity in decibels of a sound wave 10 times stronger than reference intensity is 10dB
PART B) Intensity is 100 times the reference intensity
Here [tex]I = 100I_0[/tex], replacing
[tex]\beta = 10log(\frac{100I_0}{I_0})dB[/tex]
[tex]\beta = 10log(100)dB[/tex]
[tex]\beta = 20dB[/tex]
Therefore the sound intensity in decibels of a sound wave 10 times stronger than reference intensity is 20dB
For a sound intensity 10 times the reference level, the decibel level is 10 dB, and for 100 times the reference level, it is 20 dB.
To understand the decibel scale for measuring sound intensity, we need to recognize that it is a logarithmic scale. The formula for the sound intensity level β in decibels is given by:
β = 10 log(I/I₀) dB,
where I₀ is the reference intensity, taken as 10⁻¹² W/m², the threshold of hearing.
Part A
To find the sound intensity level β for a sound wave whose intensity is 10 times the reference intensity (i.e., I = 10 I₀):
Substitute I = 10 I₀ into the formula:β = 10 log(10 I₀ / I₀).Since I₀ cancels out, we get:β = 10 log(10).We know log10(10) = 1, so:β = 10 x 1 = 10 dB.Part B
To find the sound intensity level β for a sound wave whose intensity is 100 times the reference intensity (i.e., I = 100 I₀):
Substitute I = 100 I₀ into the formula:β = 10 log(100 I₀ / I₀).Since I₀ cancels out, we get:β = 10 log(100).We know log10(100) = 2, so:β = 10 x 2 = 20 dB.A solid cylinder and a thin-walled hollow cylinder have the same radii. They are rolling horizontally toward the bottom of an incline. Assuming there is no friction, what is the maximum height each one reaches (in terms of the variables involved)?
Answer:
height solid / height hollow = 3/4
Explanation:
the height reached is directly proportional to the original KE of the object.
Total KE solid
=[tex]1/2mv^2+1/2\times1/2m r^2(v^2 / r^2 )=3/4mv^2[/tex]
Total KE hollow = [tex]1/2mv^2+1/2mr^2(v^2 / r^2 )=mv^2[/tex]
height solid / height hollow = 3/4
A wire carries a current of 4.1 A. How many electrons per second are passing any cross sectional area of the wire? Enter your answer in the format *.**E** (for example, the fundamental unit of charge, which is 1.60 cross 10^(-19) C, would be entered by typing 1.60E-19). electrons per second
To solve this problem it is necessary to apply the concepts related to Current and Load.
The current in terms of the charge of an electron can be expressed as
[tex]i = \frac{q}{t}[/tex]
Where,
q = Charge
t = time
At the same time the Charge is the amount of electrons multiplied by the amount of these, that is
q = ne
Replacing in the first equation we have to
[tex]i = \frac{q}{t}[/tex]
[tex]i = \frac{ne}{t}[/tex]
Clearing n,
[tex]n = \frac{it}{e}[/tex]
Here the time is one second then
[tex]n = \frac{i}{e}[/tex]
[tex]n = \frac{4.1}{1.6*10^{-9}}[/tex]
[tex]n = 2.56*10^{19}electrons[/tex]
Therefore the number of electrons per second are passing any cross sectional area of the wire are [tex]2.56*10^{19}electrons[/tex]
Jack stands at rest on a skateboard. The mass of Jack and the skateboard together is 75 kg. Ryan throws a 3.0 kg ball horizontally to the right at 4.0 m/s to Jack, who catches it. What is the initial momentum of the system in kg m/s?
Answer:
The initial momentum of the system is 12 kg m/s
Explanation:
Hi there!
The momentum of the system is calculated as the sum of the momenta of the objects that compose the system, in this case, Jack and the ball.
The momentum of each object is calculated as follows:
p = m · v
Where:
p = momentum
m = mass
v = velocity
The initial momentum of the system is the momentum before Jack catches the ball and is given by the sum of the momentum of the ball (pb) plus the momentum of Jack (pj) (let´s consider the right direction as positive):
Initial momentum = pb + pj
initial momentum = mb · vb + mj · vj
initial momentum = 3.0 kg · 4.0 m/s + 75 kg · 0 m/s
initial momentum = 12 kg m/s
The initial momentum of the system is 12 kg m/s
Final answer:
The initial momentum of the system, consisting of Jack, the skateboard, and the ball, is 0 kg m/s.
Explanation:
Initial momentum of the system:
To calculate the initial momentum of the system consisting of Jack, the skateboard, and the ball, we can use the principle of conservation of momentum. The initial momentum of the system is equal to the sum of the individual momenta of the objects involved.
Initial momentum = Total mass × Initial velocity = (75 kg + 3.0 kg) × 0 m/s = 0 kg m/s
The ratio of carbon-14 to nitrogen in an artifact is 1:1 . Given that the half-life of carbon-14 is 5730 years, how old is the artifact?
Answer: 5,728.5 years
Explanation:
By measuring the current ratio of daughter to parent, that is (Dt/Pt) one can deduce the age of any sample. The assumption is that there is no daughter atom present at time,t=0 and that the daughter atoms are due to the parent decay(where none has been lost).
Lamda= ln 2÷ half life,t(1/2)
Lamda= ln2= 0.693/5730 years
Lamda=1.21 × 10^-4
Using the formula below;
Age,t= 1/lamda× (ln {1+ Dt/Pt})------------------------------------------------------------------------------------(1)
Slotting our values into equation (1) above.
Age,t= 1/lamda(ln[1+1/1])
Age,t= 1/1.2×10^-4(ln 1+1)
Age,t= 1/1.21×10^-4(ln 2)
Age,t= ln 2/ 1.21×10^-4
Age, t= 5,728.5 years.
Answer:
11,460 years
Explanation:
half-life= 5,730 years
you are looking for the 1 from the ration 1:3
multiply the half-life by two in order to get the 1
5,730 x 2 = 11,640
=11,460 years
The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assuming the alveolus acts like a spherical bubble, what is the surface tension of the fluid and membrane around the outside of the alveolus? How does this compare to the surface tension of water?
Answer:
The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.
Solution:
As per the question:
Radius of an alveolus, R = [tex]200\mu m = 200\times 10^{- 6}\ m[/tex]
Gauge Pressure inside, [tex]P_{in} = 25\ mmHg[/tex]
Blood Pressure outside, [tex]P_{o} = 10\ mmHg[/tex]
Now,
Change in pressure, [tex]\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa[/tex]
Since the alveolus is considered to be a spherical shell
The surface tension can be calculated as:
[tex]\Delta P = \frac{4\pi T}{R}[/tex]
[tex]T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m[/tex]
And we know that the surface tension of water is 72.8 mN/m
Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.
When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amplitude of 5.70 10-7 m and a maximum speed of 2.66 10-3 m/s. (a) What is the frequency (in Hz) of the eardrum's vibration
Answer:
Frequency will be 743.09 Hz
Explanation:
We have given maximum speed [tex]v_m=2.66\times 10^{-3}m/sec[/tex]
Amplitude [tex]A=5.7\times 10^{-7}m[/tex]
We have to find the frequency of the vibration
We know that angular frequency is given by
[tex]\omega =\frac{v_m}{A}=\frac{2.66\times 10^{-3}}{5.7\times 10^{-7}}=4666.666rad/sec[/tex]
[tex]2\pi f=4666.66[/tex]
[tex]f=743.09Hz[/tex]
Two wheels having the same radius and mass rotate at the same angular velocity. One wheel is made with spokes so nearly all the mass is at the rim. The other is a solid disk. How do their rotational kinetic energies compare?
a. The wheel with spokes has higher KE, but not twice as high.
b. The solid wheel has higher KE, but not twice as high.
c. The wheel with spokes has about twice the KE.
d. The solid wheel has about twice the KE.
e. They are nearly the same.
Answer:
C. The wheel with spokes has about twice the KE.
Explanation:
Given that
Mass , radius and the angular speed for both the wheels are same.
radius = r
Mass = m
Angular speed = ω
The angular kinetic energy KE given as
[tex]KE=\dfrac{1}{2}I\omega ^2[/tex]
I=Moment of inertia for wheels
Wheel made of spokes
I₁ = m r²
Wheel like a disk
I₂ = 0.5 m r²
Now by comparing kinetic energy
[tex]\dfrac{KE_1}{KE_2}=\dfrac{I_1}{I_2}[/tex]
[tex]\dfrac{KE_1}{KE_2}=\dfrac{mr^2}{0.5mr^2}[/tex]
[tex]\dfrac{KE_1}{KE_2}=2[/tex]
KE₁= 2 KE₂
Therefore answer is C.
Answer:
The wheel with spokes has almost twice the KE.
A mirror faces a cliff, located some distance away. Mounted onthe cliff is a second mirror, directly opposite the first mirrorand facing toward it. A gun is fired very close to the firstmirror. The speed of sound is 342 m/s.How many times does the flash of the gunshot travel the round-tripdistance between the mirrors before the echo of the gunshot isheard?
To develop this problem, we simply have to make a relationship between speeds. This is because the echo of the sound is traveling a distance equal to that which the light travels.
We must calculate how much is that relationship of round trips between the two, as well
[tex]N = \frac{c}{s}[/tex]
Where
c = Speed of light
s = Speed of sound
By definition we know that
[tex]c = 3*10^8m/s[/tex]
[tex]s = 342,2m/s[/tex](Normal conditions)
Then,
[tex]N = \frac{3*10^8m/s}{342,2m/s}[/tex]
[tex]N = 876680.3[/tex]
Therefore the flash of the gunshot travel the round-trip distance between the mirror around to 876680.3 times before the echo of the gunshot is heard.
To determine how many times the flash of the gunshot travels the round-trip distance between the mirrors before the echo is heard, we need to consider the time it takes for sound to travel from the gun to the first mirror, and then from the first mirror to the second mirror and back.
Explanation:To determine how many times the flash of the gunshot travels the round-trip distance between the mirrors before the echo is heard, we need to consider the time it takes for sound to travel from the gun to the first mirror, and then from the first mirror to the second mirror and back.
The total round-trip distance between the mirrors is twice the distance between the mirrors. Let's say this distance is 'd'.
The time it takes for sound to travel from the gun to the first mirror and back is given by 2d/v, where 'v' is the speed of sound.
So, the number of times the flash of the gunshot travels the round-trip distance before the echo is heard is given by the time taken for the sound divided by the time taken by the flash to travel the same distance.
Therefore, the answer to the question is the speed of sound divided by the speed of light (ignoring any delays due to the reflection in the mirrors).
Learn more about Echoes here:https://brainly.com/question/10399979
#SPJ3
Two identical cars are driving in opposite directions at the same speed. Their kinetic energies have ____.
O the same magnitude, but opposite signs
O the same magnitude and sign
O different magnitudes, but the same sign
O different magnitudes and opposite signs
Answer:
Their kinetic energies have the same magnitude and sign.
Explanation:
Hi there!
Kinetic energy is not a vector, then it has no direction and therefore it does not matter the sense of movement of the car relative to a system of reference. Mathematically it would be also impossible to obtain a negative kinetic energy. The equation of kinetic energy (KE) is the following:
KE = 1/2 · m · v²
where m is the mass of the car (always positive) and v is its speed (not velocity, remember that the speed is the magnitude of the velocity vector, that´s why the kinetic energy is not a vector. I agree that the "v" in the formula is confusing).
So, even if we use a negative speed (that would be wrong), the kinetic energy will be positive because the speed is squared.
Then, if the cars have the same mass and speed, they will have the same kinetic energy, magnitude and sign (positive).
3) A 0.060-kg tennis ball, initially moving at a speed of 12 m/s, is struck by a racket causing it to rebound in the opposite direction at a speed of 18 m/s. What is the change in momentum of the ball?
Answer:
Change in momentum will be 1.8 kgm/sec
Explanation:
We have given mass of the ball m = 0.060 kg
Initial speed = 12 m /sec
And final speed = 18 m/sec
We have to find the change in momentum
Change in momentum is given by [tex]\Delta P=m(v_f-v_i)[/tex]
So [tex]\Delta P=0.060\times (18-(-12))=0.060\times 30=1.8kgm/sec[/tex] ( negative sign is due to finally opposite direction of ball )
So change in momentum will be 1.8 kgm/sec
The change of momentum of the tennis ball after being hit by the racket, reversing its direction, is calculated to be -1.8 kg*m/s.
Explanation:To determine the change in momentum of a tennis ball following a bounce, one must calculate the initial and final momentum and find the difference between the two. Momentum (p) is the product of mass (m) and velocity (v), hence p = mv. The tennis ball in question has a mass of 0.06 kg. The initial momentum is 0.06 kg * 12 m/s = 0.72 kg*m/s. After it is struck by the racket, it rebounds in the opposite direction at a speed of 18 m/s giving it a momentum of 0.06 kg * -18 m/s = -1.08 kg*m/s (negative because direction changed). The change in momentum is the final momentum minus the initial momentum, hence -1.08 kg*m/s - 0.72 kg*m/s = -1.8 kg*m/s.
Learn more about Change in Momentum here:https://brainly.com/question/2193212
#SPJ3
A ball is dropped from the top of a tall building. At the same instant, a second ball is thrown upward from the ground level. When the two balls pass one another, one on the way up, the other on the way down, compare the magnitudes of their acceleration:
a. The acceleration changes during the motion, so you cannot predict the exact value when the two balls pass each other.
b. The acceleration of both balls is the same.
c. The acceleration of the ball thrown upward is greater.
d. The accelerations are in opposite directions.
e. The acceleration of the dropped ball is greater.
Answer:B
Explanation:
For the ball dropped From a tall ball building the direction of acceleration is downward and its magnitude is [tex]9.8 m/s^2[/tex]
For the ball thrown upward towards the dropped ball the direction and magnitude of acceleration is same i.e. downwards as gravity always acting downward with constant magnitude.
Thus option B is correct
In this case, the acceleration of both balls is the same (Option B).
What is acceleration?Acceleration is a physical property to measure the movement of an object in a given period of time.
Acceleration can be estimated as a change in velocity over the change in time (.e.g., meters per second squared).The conventionally used unit of measurement of the acceleration is meter per second squared.In conclusion, in this case, the acceleration of both balls is the same (Option B).
Learn more in:
https://brainly.com/question/605631
A rubber ball filled with air has a diameter of 24.2 cm and a mass of 0.459 kg. What force is required to hold the ball in equilibrium immediately below the surface of water in a swimming pool? (Assume that the volume of the ball does not change. Indicate the direction with the sign of your answer.)
To solve this problem, it is necessary to apply the concepts related to Newton's second Law as well as to the expression of mass as a function of Volume and Density.
From Newton's second law we know that
F= ma
Where,
m = mass
a = acceleration
At the same time we know that the density is given by,
[tex]\rho = \frac{m}{V} \rightarrow m = \rho V[/tex]
Our values are given as,
[tex]g = 9.8m/s^2[/tex]
[tex]m =0.459 kg[/tex]
D=0.242 m
Therefore the Force by Weight is
[tex]F_w = mg[/tex]
[tex]F_w = 0.459kg * 9.8m/s^2 = 4.498N[/tex]
Now the buoyant force acting on the ball is
[tex]F_B=\rho V g[/tex]
The value of the Volume of a Sphere can be calculated as,
[tex]V = \frac{4}{3} \pi r^3[/tex]
[tex]V = \frac{4}{3} \pi (0.242/2)^3[/tex]
[tex]V = 0.007420m^3[/tex]
[tex]\rho_w = 1000kg/m^3 \rightarrow[/tex] Normal conditions
Then,
[tex]F_B=0.007420*(1000)*(9.8) = 72.716 N[/tex]
Therefore the Force net is,
[tex]F_{net} = F_B -F_w[/tex]
[tex]F_{net} = 72.716N - 4.498N =68.218 N[/tex]
Therefore the required Force is 68.218N
A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration α. The flywheel is assumed to be at rest at time t=0 in Parts A and B of this problem.
Part A
Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α.
Express your answer in terms of ω1 and α.
Part B
Find the angle θ1 through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity ω1.
Express your answer in terms of some or all of the following:
A. ω1
B. α
C. t1.
Part C
Assume that the motor has accelerated the wheel up to an angular velocity ω1 with angular acceleration α in time t1. At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of −5α. Find t2, the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity).
Express your answer in terms of some or all of the following:
A. ω1
B. α,
C. t1.
Answer:
a)[tex]t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec[/tex]
b)[tex]\theta_1=\frac{w_1^2}{2\alpha}rad[/tex]
c)[tex]t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec[/tex]
Explanation:
1) Basic concepts
Angular displacement is defined as the angle changed by an object. The units are rad/s.
Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:
[tex]w=\frac{\Delat \theta}{\Delta t}[/tex]
Angular acceleration is the rate of change of the angular velocity respect to the time
[tex]\alpha=\frac{dw}{dt}[/tex]
2) Part a
We can define some notation
[tex]w_o=0\frac{rad}{s}[/tex],represent the initial angular velocity of the wheel
[tex]w_1=?\frac{rad}{s}[/tex], represent the final angular velocity of the wheel
[tex]\alpha[/tex], represent the angular acceleration of the flywheel
[tex]t_1[/tex] time taken in order to reach the final angular velocity
So we can apply this formula from kinematics:
[tex]w_1=w_o +\alpha t_1[/tex]
And solving for t1 we got:
[tex]t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec[/tex]
3) Part b
We can use other formula from kinematics in order to find the angular displacement, on this case the following:
[tex]\Delta \theta=wt+\frac{1}{2}\alpha t^2[/tex]
Replacing the values for our case we got:
[tex]\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2[/tex]
And we can replace [tex]t_1[/tex]from the result for part a, like this:
[tex]\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2[/tex]
Since [tex]\theta_o=0[/tex] and [tex]w_o=0[/tex] then we have:
[tex]\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}[/tex]
And simplifying:
[tex]\theta_1=\frac{w_1^2}{2\alpha}rad[/tex]
4) Part c
For this case we can assume that the angular acceleration in order to stop applied on the wheel is [tex]\alpha_1 =-5\alpha \frac{rad}{s}[/tex]
We have an initial angular velocity [tex]w_1[/tex], and since at the end stops we have that [tex]w_2 =0[/tex]
Assuming that [tex]t_2[/tex] represent the time in order to stop the wheel, we cna use the following formula
[tex]w_2 =w_1 +\alpha_1 t_2[/tex]
Since [tex]w_2=0[/tex] if we solve for [tex]t_2[/tex] we got
[tex]t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}[/tex]
And from part a) we can see that [tex]w_1=\alpha t_1[/tex], and replacing into the last equation we got:
[tex]t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec[/tex]
A copper ring and a wooden ring of the same dimensions are placed in magnetic fields so that there is the same change in magnetic flux through them. Compare the induced EMFs and currents in the rings.
Answer:
emf induces in both iron and copper is same
Explanation:
The induced emf in both the rings would be same.
The induced emf in any material is given by
[tex]\epsilon= -N\frac{d\phi}{dt}[/tex]
N= number of turns in the coil
dΦ= change in magnetic flux
dt= change in time
clearly, induced emf is independent of the material of the object,it only depends upon rate of change of flux.
therefore, emf induces in both iron and copper is same.
The induced emf and currents in the copper ring and the wooden ring are compared based on their conductivity.
Explanation:When a magnetic field changes, it induces an electric field and hence an emf (electromotive force) in a conducting loop. The magnitude of the induced emf depends on the rate of change of the magnetic field. In this case, both the copper ring and the wooden ring experience the same change in magnetic flux, which means they have the same induced electric fields. However, the copper ring, being a good conductor, has a much higher induced emf compared to the wooden ring, which has a lower conductivity.
Wooden Ring:
Conductivity: Wood is a poor conductor of electricity. Its conductivity is significantly lower compared to metals like copper.
Induced Emf and Currents: Due to its poor conductivity, a wooden ring will exhibit a much weaker induced emf and induce much smaller currents when subjected to a changing magnetic field.
In summary, the difference in conductivity between copper and wood directly influences the induced emf and currents in the respective rings. The high conductivity of copper facilitates efficient response to changes in magnetic fields, while the low conductivity of wood results in a much weaker and less efficient induction process.
Learn more about induced emf and currents here:https://brainly.com/question/34238669
#SPJ3