To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.
Kepler's third law tells us that
[tex]T^2 = \frac{4\pi^2}{GM}a^3[/tex]
Where
T= Period
G= Gravitational constant
M = Mass of the sun
a= The semimajor axis of the comet's orbit
The period in years would be given by
[tex]T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s[/tex]
PART A) Replacing the values to find a, we have
[tex]a^3= \frac{T^2 GM}{4\pi^2}[/tex]
[tex]a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}[/tex]
[tex]a^3 = 6.46632*10^{39}[/tex]
[tex]a = 1.86303*10^{13}m[/tex]
Therefore the semimajor axis is [tex]1.86303*10^{13}m[/tex]
PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by
[tex]R = a(1-e)[/tex]
[tex]R = 1.86303*10^{13}(1-0.997)[/tex]
[tex]R= 5.58*10^{10}m[/tex]
A softball is hit over a third baseman's head with some speed v0 at an angle θ above the horizontal. Immediately after the ball is hit, the third baseman turns around and begins to run at a constant velocity V=7.00m/s. He catches the ball t=2.00s later at the same height at which it left the bat. The third baseman was originally standing L=18.0m from the location at which the ball was hit.
A) Find v0. Use g=9.81m/s2 for the magnitude of the acceleration due to gravity.
Express the initial speed numerically in units of meters per second to three significant figures.
B) Find the angle θ in degrees.
Express your answer numerically in degrees to three significant figures.
Answer:
A) v0 = 18.8 m/s
B) θ = 31.5°
Explanation:
On the x-axis:
[tex]X_{ball} = X_{player}[/tex]
[tex]v0*cos\theta*t=L+V*t[/tex]
where
t = 2s; L = 18m; V = 7m/s
[tex]v0*cos\theta*2=32[/tex]
[tex]v0=16/cos\theta[/tex]
On the y-axis for the ball:
[tex]\Delta Y=v0*sin\theta*t-1/2*g*t^2[/tex]
[tex]0 = v0*sin\theta*2-19.62[/tex]
Replacing v0:
[tex]0 = tan\theta*2*16-19.62[/tex]
[tex]\theta=atan(19.62/32)=31.5\°[/tex]
Now, the speed v0 was:
v0 = 18.8m/s
The initial speed of the ball is 9.00 m/s and the angle is 0 degrees above the horizontal.
Explanation:Part A:
To find the initial speed, we can use the horizontal distance traveled by the ball and the time it takes to reach the third baseman. The horizontal distance is the initial distance between the third baseman and the location where the ball was hit, which is 18.0m. The time is given as 2.00s. So, using the formula: distance = speed × time, we can rearrange it to find the initial speed: speed = distance / time. Plugging in the values, we get: speed = 18.0m / 2.00s = 9.00m/s.
Part B:
To find the angle, we can use the vertical distance traveled by the ball. The vertical distance is the same as the height at which the ball left the bat. Since the third baseman catches the ball at the same height, the vertical distance is zero. Using the formula: vertical distance = initial velocity × time + 0.5 × acceleration due to gravity × time squared, we can rearrange it to find the angle: angle = arctan( (2 × vertical distance × gravity) / (initial speed squared) ). Plugging in the values, we get: angle = arctan( (2 × 0 × 9.81) / (9.00^2) ) = 0 degrees.
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The density of the liquid flowing through a horizontal pipe is 1200 kg/m3. The speed of the fluid at point A is 7.5 m/s while at point B it is 11 m/s. What is the difference in pressure, PB – PA, between points B and A?A. +5.0 × 104 PaB. –1.9 × 103 PaC. –2.5 × 104 PaD. –3.9 × 104 PaE. +3.8 × 103 Pa
Answer:
The difference in pressure between points B and A is [tex]-3.9\times10^{4}\ Pa[/tex]
(D) is correct option.
Explanation:
Given that,
Density of the liquid = 1200 kg/m³
Speed of fluid at point A= 7.5 m/s
Speed of fluid at point B = 11 m/s
We need to calculate the difference in pressure between points B and A
Using formula of change in pressure
[tex]\Delta P=\dfrac{1}{2}D(v_{2}^2-v_{1}^2)[/tex]
Where, [tex]v_{1}[/tex] = Speed of fluid at point A
[tex]v_{2}[/tex] = Speed of fluid at point B
D = Density of the liquid
Put the value into the formula
[tex]\Delta P=\dfrac{1}{2}\times1200\times(11^2-7.5^2)[/tex]
[tex]\Delta P=-38850\ Pa[/tex]
[tex]\Delta P=-3.9\times10^{4}\ Pa[/tex]
Hence, The difference in pressure between points B and A is [tex]-3.9\times10^{4}\ Pa[/tex]
The difference in pressure, PB – PA, between points B and A can be calculated using Bernoulli's equation. Plugging in the given values, we find that the pressure difference is approximately -39,000 Pa.
Explanation:
The difference in pressure between points B and A can be calculated using Bernoulli's equation. Bernoulli's equation states that the pressure difference between two points in a fluid flow system is equal to the difference in kinetic energy and potential energy between the two points. In this case, we can calculate the pressure difference using the equation:
PB - PA = (1/2) * ρ * (VB^2 - VA^2)
PB - PA = (1/2) * 1200 kg/m3 * (11 m/s)^2 - (7.5 m/s)^2
Simplifying the equation, we find that the difference in pressure, PB - PA, is approximately -39,000 Pa. Therefore, the correct answer is D. -3.9 × 104 Pa.
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A simple harmonic oscillator consists of a block of mass 4.10 kg attached to a spring of spring constant 240 N/m. When t = 1.70 s, the position and velocity of the block are x = 0.165 m and v = 3.780 m/s.
(a) What is the amplitude of the oscillations?
What were the (b) position and (c) velocity of the block at t = 0 s?
Answer:
(a) the amplitude of the oscillations is 0.52 m
(b) position of the block at t = 0 s is - 0.06 m
(c) the velocity of the block at t = 0 s is 3.96 m/s
Explanation:
given information:
m = 4.10 kg
k = 240 N/m
t = 1.70 s
x = 0.165 m
v = 3.780 m/s
(a) the amplitude of the oscillations
x(t) = A cos (ωt+φ)
v(t) = dx(t)/dt
= - ω A sin (ωt+φ)
ω = [tex]\sqrt{\frac{k}{m} }[/tex]
= [tex]\sqrt{\frac{240}{4.10} }[/tex]
= 7.65 rad/s
v(t)/x(t) = - ω A sin (ωt+φ)/A cos (ωt+φ)
v(t)/x(t) = - ω sin (ωt+φ)/cos (ωt+φ)
v(t)/x(t) = - ω tan (ωt+φ)
(ωt+φ) = [tex]tan^{-1}[/tex] (-v/ωx)
= [tex]tan^{-1}[/tex] (-3.780/(7.65)(0.165))
= - 1.25
(ωt+φ) = - 1.25
φ = - 1.25 - ωt
= - 1.25 - (7.65 x 1.70)
= - 14.26
x(t) = A cos (ωt+φ)
A = x(t) / cos (ωt+φ)
= 0.165/cos(-1.25)
= 0.52 m
(b) position, at t=0
x(t) = A cos (ωt+φ)
x(0) = A cos (ω(0)+φ)
x(0) = A cos (φ)
= 0.52 cos (-14.26)
= - 0.06 m
(c) the velocity, at t=0
v(t) = - ω A sin (ωt+φ)
v(0) = - ω A sin (ω(0)+φ)
= - ω A sin (φ)
= - (7.65)(0.52) sin (-14.26)
= 3.96 m/s
A proton moves through a uniform magnetic field given by vector B = (10 i hat- 24.0 j hat + 30 k hat) mT. At time t1, the proton has a velocity given by vector v = v x i hat + vy j hat + (2000 m/s) k hat and the magnetic force on the proton is vector Fb = (3.82x10-17 N) i hat + (1.59x10-17 N) j hat. (a) At that instant, what is vx? m/s (b) What is vy?
a. The value of vₓ = 33.77 km/s
b. The value of Vy = 81.05 km/s
The magnetic force on the proton
The magnetic force, F on the proton moving in the uniform magnetic field is given by
F = qv × B where
F = force on proton = (3.82 × 10⁻¹⁷ N)l + (1.59 × 10⁻¹⁷ N)j + 0k, q = charge on proton = 1.602 × 10⁻¹⁹ C, v = velocity of proton = Vxl + Vyj + (2000 m/s)k and B = magnetic field = (10i - 24.0j + 30k) mT = (10i - 24.0j + 30k) × 10⁻³ TRe-writing the force in matrix form, we have
[tex]\left[\begin{array}{ccc}F_{x} &F_{y} &F_{z} \\\end{array}\right] = q\left[\begin{array}{ccc}i&j&k\\v_{x} &v_{y}&v_{z}\\B_{x}&B_{y}&B_{z}\end{array}\right][/tex]
Taking the determinant, we have
[tex]F_{x}i + F_{y}j + F_{z}k = q[(v_{y}B_{z} - v_{z}B_{y})]i + q[(v_{x}B_{z} - v_{z}B_{x})]j + q[(v_{x}B_{y} - v_{y}B_{x})]k[/tex]
Equating the components of the force, we have
[tex]F_{x} = q[(v_{y}B_{z} - v_{z}B_{y})] (1)\\F_{y} = q[(v_{x}B_{z} - v_{z}B_{x})] (2)\\F_{z} = q[(v_{x}B_{y} - v_{y}B_{x})] (3)[/tex]
[tex]F_{x}/q = [(v_{y}B_{z} - v_{z}B_{y})] (4)\\F_{y}/q = [(v_{x}B_{z} - v_{z}B_{x})] (5)\\F_{z}/q = [(v_{x}B_{y} - v_{y}B_{x})] (6)[/tex]
Since F = (3.82 × 10⁻¹⁷ N)l + (1.59 × 10⁻¹⁷ N)j + 0k, equation (5) and (6) become
[tex]1.59 X 10^{-17} /1.602 X 10^{-19} = [(v_{x}B_{z} - v_{z}B_{x})] (5)\\993 = [(v_{x}B_{z} - v_{z}B_{x})] (7)\\Also\\0/q = [(v_{x}B_{y} - v_{y}B_{x})] (6)\\0 = (v_{x}B_{y} - v_{y}B_{x}) \\v_{x}B_{y} = v_{y}B_{x}\\v_{y} = \frac{v_{x}B_{y}}{B_{x}} (8)[/tex]
a. The value of VxThe value of vₓ = 33.77 km/s
Since [tex]B_{x}[/tex] = 10 × 10 ⁻³ T, [tex]B_{z}[/tex] = 30 × 10 ⁻³ T and [tex]v_{z}[/tex] = 2000 m/s, substituting the values of the variables into equation (7), we have
[tex]v_{x}B_{z} - v_{z}B_{x} = 993 (7)[/tex]
vₓ(30 × 10 ⁻³ T) - 2000 m/s × 10 × 10 ⁻³ T = 993 N
vₓ(30 × 10 ⁻³ T) - 20 m/sT = 993 N
vₓ(30 × 10 ⁻³ T) = 993 N + 20 m/sT
vₓ(30 × 10 ⁻³ T) = 1013 N
vₓ = 1013 N/(30 × 10 ⁻³ T)
vₓ = 33.77 × 10³ m/s
vₓ = 33.77 km/s
So, the value of vₓ = 33.77 km/s
b. The value of VyThe value of Vy = 81.05 km/s
Since [tex]B_{x}[/tex] = 10 × 10 ⁻³ T, [tex]B_{y}[/tex] = 24.0 × 10 ⁻³ T and [tex]v_{x}[/tex] = 33.77 × 10³ m/s, substituting the values of the variables into equation (8), we have
[tex]v_{y} = \frac{v_{x}B_{y}}{B_{x}} (8)[/tex]
Vy = 33.77 × 10³ m/s × 24.0 × 10 ⁻³ T/10 × 10 ⁻³ T
Vy = 810.48 m/sT/10 × 10 ⁻³ T
Vy = 81.048 × 10³ m/s
Vy ≅ 81.05 km/s
So, the value of Vy = 81.05 km/s
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A small artery has a length of 1.10 × 10 − 3 m and a radius of 2.50 × 10 − 5 m. If the pressure drop across the artery is 1.45 kPa,what is the flow rate through the artery? Assume that the temperature is 37 °C and the viscosity of whole blood is 2.084 × 10 − 3 Pa·s.
Answer:
The flow rate is [tex]9.7\times 10^{- 11}\ m^{3}/s[/tex]
Solution:
As per the question:
Pressure drop, = 1.45 kPa = 1450 Pa
Radius of the artery, R = [tex]2.50\times 10^{- 5}\ m[/tex]
length of the artery, L = [tex]1.10\times 10^{- 3}\ m[/tex]
Temperature, T = [tex]37^{\circ}C[/tex]
Viscosity, [tex]\eta = 2.084\times 10^{- 3}\ Pa.s[/tex]
Now,
The flow rate is given by:
[tex]Q = \frac{\pi R^{4}P}{8\eta L}[/tex]
[tex]Q = \frac{\pi (2.50\times 10^{- 5})^{4}\times 1450}{8\times 2.084\times 10^{- 3}\times 1.10\times 10^{- 3}} = 9.7\times 10^{- 11}\ m^{3}/s[/tex]
The flow rate through the artery can be calculated using Poiseuille's law, which takes into account the pressure difference, radius, length, and viscosity of the fluid. Using the given values, the flow rate is approximately 0.00686 m^3/s.
Explanation:The flow rate through an artery can be calculated using Poiseuille's law, which states that the flow rate is directly proportional to the pressure difference and the fourth power of the radius, and inversely proportional to the length and viscosity of the fluid.
Using the given values, we can calculate the flow rate as follows:
Flow rate = (pressure difference * pi * radius^4) / (8 * viscosity * length)
Substituting the given values into the equation gives:
Flow rate = (1.45 * 10^3 * 3.14 * (2.5 * 10^-5)^4) / (8 * 2.084 * 10^-3 * 1.10 * 10^-3) = 0.00686 m^3/s
Therefore, the flow rate through the artery is approximately 0.00686 m^3/s.
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A weather balloon is designed to expand to a maximum radius of 16.2 m when in flight at its working altitude, where the air pressure is 0.0282 atm and the temperature is â65âC. If the balloon is filled at 0.873 atm and 21âC, what is its radius at lift-off?
Answer:
5.78971 m
Explanation:
[tex]P_1[/tex] = Initial pressure = 0.873 atm
[tex]P_2[/tex] = Final pressure = 0.0282 atm
[tex]V_1[/tex] = Initial volume
[tex]V_2[/tex] = Final volume
[tex]r_1[/tex] = Initial radius = 16.2 m
[tex]r_2[/tex] = Final radius
Volume is given by
[tex]\frac{4}{3}\pi r^3[/tex]
From the ideal gas law we have the relation
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{0.873\times \frac{4}{3}\pi r_1^3}{294.15}=\frac{0.0282\frac{4}{3}\pi r_2^3}{208.15}\\\Rightarrow \frac{0.873r_1^3}{294.15}=\frac{0.0282\times 16.2^3}{208.15}\\\Rightarrow r_1=\frac{0.0282\times 16.2^3\times 294.15}{208.15\times 0.873}\\\Rightarrow r_1=5.78971\ m[/tex]
The radius of balloon at lift off is 5.78971 m
To find the radius of the weather balloon at lift-off, the ideal gas law can be used. Using the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off, the radius at lift-off can be calculated to be approximately 4.99 m.
Explanation:To find the radius of the weather balloon at lift-off, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
In this case, we know that the number of moles is constant, as the balloon is filled with the same amount of helium at lift-off and in flight. Therefore, we can write the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off.
Plugging in the given values, we have (0.873 atm)(V1) = (0.0282 atm)(16.2 m)^3. Solving for V1, we find that the volume at lift-off is approximately 110.9 m^3. The radius can then be calculated using the formula for the volume of a sphere: V = (4/3) * π * r^3, where r is the radius.
Therefore, the radius at lift-off is approximately 4.99 m.
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A single slit of width d = 0.08 mm is illuminated by light of two wavelengths, l = 446 nm and l = 662 nm. The diffraction pattern appears on a screen 1.05 m away. (a) Calculate the angles at which the third dark fringe appears for each wavelength. q446 = rad q662 = rad (b) Calculate the width of the central bright fringe for each wavelength. d446 = m d662 = m
Final answer:
The question is about using the single slit diffraction formula to calculate the angles at which the third dark fringe appears for two different wavelengths and the width of the central bright fringe for each wavelength, involving physics and high school level understanding.
Explanation:
The student's question relates to the concept of single slit diffraction, which is a phenomenon in physics where light spreads out after passing through a narrow opening, resulting in a characteristic pattern. Specifically, the angles at which dark fringes appear and the width of the central bright fringe are calculated for two different wavelengths using the formulas for single slit diffraction.
For part (a), to calculate the angles for the third dark fringe for each wavelength, we can use the formula dsin(θ) = mλ, where d is the slit width, θ is the angle of the dark fringe, m is the order of the dark fringe, and λ is the wavelength of the light. The third dark fringe corresponds to m = 3 for this calculation. For part (b), the width of the central bright fringe is calculated by considering the angles to the first dark fringes on either side of the central peak and then using trigonometry to determine the width on the screen.
A light string is wrapped around the outer rim of a solid uniform cylinder of diameter 75.0 cm that can rotate without friction about an axle through its center. A 3.00 kg stone is tied to the free end of the string. When the system is released from rest, you determine that the stone reaches a speed of 3.40 m/s after having fallen 2.40 m. What is the mass of the cylinder?
Since a light string is wrapped around the outer rim of a solid uniform cylinder, the mass of the cylinder is equal to 18.4 kilograms.
Given the following data:
Diameter = 75.0 cmMass of stone = 3 kgInitial velocity = 0 m/s (since the system was released from rest)Final velocity = 3.40 m/sDistance = 2.40 meters.To calculate the mass of the cylinder:
First of all, we would determine the acceleration of the stone by using the the third equation of motion;
[tex]V^2 = U^2 + 2aS[/tex]
Where:
V is the final speed. U is the initial speed. a is the acceleration. S is the distance covered.Substituting the given parameters into the formula, we have;
[tex]3.4^2 = 0^2 + 2a(2.40)\\\\11.56 = 0 + 4.8a\\\\11.56 = 4.8a\\\\a = \frac{11.56}{4.8}[/tex]
Acceleration, a = 2.41 [tex]m/s^2[/tex]
Next, we would solve for the torque by using the formula:
[tex]T = m(g - a)\\\\T = 3(9.8 - 2.41)\\\\T = 3 \times 7.39[/tex]
Torque, T = 22.17 Newton.
In rotational motion, torque is given by the formula:
[tex]T = I\alpha \\\\Tr = \frac{Mr^2}{2} \times \frac{a}{r} \\\\Tr = \frac{Mra}{2}\\\\2Tr = Mra\\\\M = \frac{2T}{a}\\\\M = \frac{2\times 22.17}{2.41}\\\\M = \frac{44.34}{2.41}[/tex]
Mass = 18.4 kilograms
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The system described is based on rotational dynamics and energy conservation principles in physics. The initial potential energy of the stone is converted into kinetic energy, and the stone's tensional force is equated with the cylinder's net torque. By systematically resolving these equations, we find the mass of the cylinder to be about 66.16 kg.
Explanation:The subject of your question pertains to rotational dynamics in Physics. In this particular system, there are two main things to consider. First, we need to acknowledge the law of conservation of energy. The entire mechanical energy of the system would remain constant because there is neither any frictional force present nor any external force doing work. Second, we have to consider the moment of inertia for the cylinder.
For the stone, the initial potential energy (mgh) is converted into kinetic energy (0.5*m*v^2). So, 3*9.81*2.4 = 0.5*3*(3.4)^2, which solves to be accurate.
The tension in the string is equal to the force exerted by the stone, so T = m*g = 3*9.81 = 29.43 N.
As the string unwinds, the cylinder spins faster so, we equate the tensional force with the net torque, τ_net = I*α. Here, α = tangential acceleration (which equals g) divided by radius, hence, α = g/radius.
By simplifying these equations, we find that the mass of the cylinder is approximately 66.16 kg.
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As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150-T magnetic field near the center of the solenoid. You have enough wire for 3100 circular turns. This solenoid must be 57.0 cm long and 7.00 cm in diameter.What current will you need to produce the necessary field?
Answer:
I = 21.94 A
Explanation:
Given that
B= 0.15 T
Number of turns N= 3100
Length L = 57 cm
Diameter ,d= 7 cm
We know that magnetic field in the solenoid given as
[tex]B=\dfrac{\mu_0IN}{L}[/tex]
[tex]I=\dfrac{BL}{\mu_0N}[/tex]
Now by putting the values
[tex]I=\dfrac{0.15\times 0.57}{4\pi \times 10^{-7}\times 3100}[/tex]
I = 21.94 A
Therefore current need to produce 0.15 T magnetic filed is 21.94 A.
An arteriole has a radius of 25 μm and it is 1000 μm long. The viscosity of blood is 3 x 10-3 Pa s and its density is 1.055 g cm-3 . Assume the arteriole is a right circular cylinder. A. Assuming laminar flow, what is the resistance of this arteriole?
Answer: [tex]1.955(10)^{13} \frac{Pa.s}{m^{3}}[/tex]
Explanation:
This can be solved by the Poiseuille’s law for a laminar flow:
[tex]R=\frac{8 \eta L}{\pi r^{4}}[/tex]
Where:
[tex]R[/tex] is the resistance of the arteriole
[tex]\eta=3(10)^{-3} Pa.s[/tex] is the viscosity of blood
[tex]L=1000 \mu m=1000(10)^{-6}m[/tex] is the length of the arteriole
[tex]r=25 \mu m=25(10)^{-6}m[/tex] is the radius of the arteriole
[tex]R=\frac{8 (3(10)^{-3} Pa.s)(1000(10)^{-6}m)}{\pi (25(10)^{-6}m)^{4}}[/tex]
[tex]R=1.955(10)^{13} \frac{Pa.s}{m^{3}}[/tex]
The resistance of the arteriole is calculated using the Hagen-Poiseuille equation, resulting in an approximate value of 1.96 × 10¹⁵ Pa·s·m⁻⁴. The formula determines resistance by varying viscosity, length, and radius values, which is essential for understanding blood flow dynamics in physiological settings.
We can solve this problem using the formula for fluid resistance in a cylindrical tube (Hagen-Poiseuille equation):
[tex]R = (8 \times \eta \times L) / (\pi \times r_4)[/tex]
Where:
R is the resistance.η is the viscosity of the fluid (3 × 10⁻³ Pa·s).L is the length of the tube (1000 × 10⁻⁶ m).r is the radius of the tube (25 × 10⁻⁶ m).Substituting the values:
[tex]R = (8 \times 3 \times 10^{-3} Pa\cdot s \times 1000 \times 10^{-6} m) / (\pi \times (25 \times 10^{-6} m))[/tex]
Calculating the values inside the formula:
[tex]R = (24 \times 10^{-3} Pa\cdot s\cdot m) / (\pi \times 3.90625 \times 10^{-18} m^{4})[/tex][tex]R = (24 \times 10^{-3} Pa\cdot s\cdot m) / (1.227 \times 10^{-17} m^{4})[/tex][tex]R \approx 1.96 \times 1015 Pa\cdot s/m^{-4}[/tex]So, the resistance of the arteriole is approximately 1.96 × 10¹⁵ Pa·s·m⁻⁴.
The angular momentum of a flywheel having a rotational inertia of 0.142 kg·m2 about its central axis decreases from 7.20 to 0.140 kg·m2/s in 2.10 s. (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? (b) Assuming a constant angular acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) What is the magnitude of the average power done on the flywheel?
Answer:
3.3619 Nm
54.27472 rad
182.46618 J
86.88 W
Explanation:
[tex]L_i[/tex] = Initial angular momentum = 7.2 kgm²/s
[tex]L_f[/tex] = Final angular momentum = 0.14 kgm²/s
I = Moment of inertia = 0.142 kgm²
t = Time taken
Average torque is given by
[tex]\tau_{av}=\frac{L_f-L_i}{\Delta t}\\\Rightarrow \tau_{av}=\frac{0.14-7.2}{2.1}\\\Rightarrow \tau_{av}=-3.3619\ Nm[/tex]
Magnitude of the average torque acting on the flywheel is 3.3619 Nm
Angular speed is given by
[tex]\omega_i=\frac{L_i}{I}[/tex]
Angular acceleration is given by
[tex]\alpha=\frac{\tau}{I}[/tex]
From the equation of rotational motion
[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=\frac{L_i}{I}\times t+\frac{1}{2}\times \frac{\tau}{I}\times t^2\\\Rightarrow \theta=\frac{7.2}{0.142}\times 2.1+\frac{1}{2}\times \frac{-3.3619}{0.142}\times 2.1^2\\\Rightarrow \theta=54.27472\ rad[/tex]
The angle the flywheel turns is 54.27472 rad
Work done is given by
[tex]W=\tau\theta\\\Rightarrow W=-3.3619\times 54.27472\\\Rightarrow W=-182.46618\ J[/tex]
Work done on the wheel is 182.46618 J
Power is given by
[tex]P=\frac{W}{t}\\\Rightarrow P=\frac{-182.46618}{2.1}\\\Rightarrow P=-86.88\ W[/tex]
The magnitude of the average power done on the flywheel is 86.88 W
Fish are hung on a spring scale to determine their mass. (a) What is the force constant of the spring in such a scale if the spring stretches 8.00 cm for a 10.0 kg load? (b) What is the mass of a fish that stretches the spring 5.50 cm? (c) How far apart are the half-kilogram marks on the scale?
a) The spring constant is 1225 N/m
b) The mass of the fish is 6.88 kg
c) The marks are 0.4 cm apart
Explanation:
a)
When the spring is at equilibrium, the weight of the load applied to the spring is equal (in magnitude) to the restoring force of the spring, so we can write
[tex]mg = kx[/tex]
where
m is the mass of the load
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
k is the spring constant
x is the stretching of the spring
For the load in this problem we have
m = 10.0 kg
x = 8.00 cm = 0.08 m
Substituting, we find the spring constant
[tex]k=\frac{mg}{x}=\frac{(10)(9.8)}{0.08}=1225 N/m[/tex]
b)
As before, at equilibrium, the weight of the fish must balance the restoring force in the spring, so we have
[tex]mg=kx[/tex]
where this time we have:
m = mass of the fish
[tex]g=9.8 m/s^2[/tex]
k = 1225 N/m is the spring constant
x = 5.50 cm = 0.055 m is the stretching of the spring
Substituting,
[tex]m=\frac{kx}{g}=\frac{(1225)(0.055)}{9.8}=6.88 kg[/tex]
c)
To solve this part, we just need to find the change in stretching of the spring when a load of half-kilogram is hanging on the spring. Using again the same equation,
[tex]mg=kx[/tex]
where this time we have:
m = 0.5 kg
[tex]g=9.8 m/s^2[/tex]
k = 1225 N/m
x = ? is the distance between the half-kilogram marks on the scale
Substituting,
[tex]x=\frac{mg}{k}=\frac{(0.5)(9.8)}{1225}=0.004 m = 0.4 cm[/tex]
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If the particle is in the ground state, what is the probability that it is in a window Δx=0.0002L wide with its midpoint at x=0.700L? You should be able to answer this part without evaluating any integrals! Since Δx is so small, you can assume that ψ(x) remains constant over that interval, so the integral is approximately
To solve this problem it is necessary to apply the concepts related to the wave function of a particle and the probability of finding the particle in the ground state.
The wave function is given as
[tex]\phi = \sqrt{\frac{2}{L}} sin(\frac{\pi x}{L})[/tex]
Therefore the probability of finding the particle must be
[tex]P = |\phi(x)^2| \Delta X[/tex]
[tex]P = |\sqrt{\frac{2}{L}} sin(\frac{\pi x}{L})|^2 \Delta x[/tex]
[tex]P = (\frac{2}{L})(sin(\frac{\pi x}{L}))^2\Delta x[/tex]
[tex]P = (\frac{2}{L})(sin(\frac{\pi (0.7L)}{L}))^2 (0.0002L)[/tex]
[tex]P = 2*(sin(0.7\pi))^2(0.0002)[/tex]
[tex]P = 2.618*10^{-4}[/tex]
Therefore the probability is [tex]2.618*10^{-4}[/tex]
The probability that the particle is in a window [tex]\(\Delta x = 0.0002L\)[/tex]wide with its midpoint at [tex]\(x = 0.700L\)[/tex] is given by [tex]\(P = |\psi(0.700L)|^2 \Delta x\).[/tex]
In quantum mechanics, the probability of finding a particle in a certain position is given by the square of the wave function's magnitude at that position, multiplied by the width of the position interval. For a particle in the ground state, the wave function \(\psi(x)\) is given by the solution to the Schrödinger equation for the particular potential well the particle is in. For a simple one-dimensional infinite potential well, the ground state wave function is a half-sine wave.
Given that the particle is in the ground state and the window[tex]\(\Delta x\)[/tex] is very small, we can make the approximation that the wave function \(\psi(x)\) does not change significantly over this interval.
Since we are assuming [tex]\(\psi(x)\[/tex]is constant over [tex]\(\Delta x\)[/tex], the integral simplifies to the product of [tex]\(|\psi(0.700L)|^2\)[/tex] and the width of the interval [tex]\(\Delta x\):[/tex]
[tex]\[ P = \int_{0.700L - \frac{\Delta x}{2}}^{0.700L + \frac{\Delta x}{2}} |\psi(x)|^2 dx \approx |\psi(0.700L)|^2 \Delta x \][/tex]
This is a standard result for the probability of finding a particle in a small interval around a point in quantum mechanics when the wave function is approximately constant over that interval. The actual value of [tex]\(|\psi(0.700L)|^2\)[/tex] would depend on the specific form of the wave function for the ground state of the system in question.
However, the integral does not need to be evaluated explicitly due to the assumption of a constant wave function over the small interval [tex]\(\Delta x\).[/tex]
Tom and Mary are riding a merry-go-round. Tom is on a horse about half way between the center and the outer rim, and Mary is on a horse at the outer rim.
a) Which child has the larger angular speed? Explain
b) Which child has the larger linear speed? Explain
c) Which child has the larger radial acceleration? Explain
Answer:
Explanation:
Given
Tom is on half way between the center and the outer rim
Mary is on the extreme outer rim
Suppose [tex]\omega [/tex] is the angular velocity of ride with radius of outer rim be R
[tex]\alpha =[/tex]angular acceleration of Ride
At any instant both Tom and Mary experience the same angular speed
(b)Linear velocity at any instant
[tex]v_{Tom}=\omega \times R[/tex]
[tex]v_{Mary}=\omeag \times \frac{R}{2}[/tex]
Thus Tom has higher linear speed
(c) For radial Acceleration
[tex]a_r=\omega ^2\times r[/tex]
[tex]a_{Tom}=\omega ^2\times R[/tex]
[tex]a_{Mary}=\omega ^2\times \frac{R}{2}[/tex]
Tom has higher radial Acceleration as it is directly Proportional of radius
A guitar string is plucked and set into vibration. The vibrating string disturbs the surrounding air, resulting in a sound wave. Which of the following is correct? Group of answer choices If the temperature of air changes, the speed of the Wave in the string Doesn't change, but the speed of the Sound wave in air changes . Both the Wave in the string and the Sound wave in air are transverse waves . Both the Wave in the string and the Sound wave in air are longitudinal waves . The Wave in the string is longitudinal , the Sound wave in air is transverse .
Answer:
a) True. The speed of the wave in the string does not depend on the temperature, but the speed of the wave the air depends on the Temperatue.
Explanation:
Let's analyze the vibration of the string, when the string is touched a transverse wave is produced whose speed is determined by the tension and density of the string
V = √T/μ
This wave advances and bounces at one of the ends forming a standing wave that induces a vibration in the surrounding air that therefore also produces a longitudinal wave whose velocity is a function of time
v = 331 √( 1+ T/273)
With this information we can review the statements given
a) True. The speed of the wave in the string does not depend on the temperature, but the speed of the wave the air depends on the Temperatue.
b) False. The wave in the string is transverse, but the wave in the air is longitudinal
c) False. The wave in the string is transverse
d) False. It's the other way around
Answer:
A
Explanation:
Just got it right on edge
Burns produced by steam at 100°C are much more severe than those produced by the same mass of 100°C water. Calculate the quantity of heat in (Cal or kcal) that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C. Specific heat of water = 1.00 kcal/(kg · °C); heat of vaporization = 539 kcal/kg; specific heat of human flesh = 0.83 kcal/(kg · °C).
To calculate the quantity of heat that must be removed from 6.1 g of 100°C steam, we need to consider both the change in temperature and the phase change from steam to liquid. The specific heat of water is used to calculate the heat required to lower the temperature, while the heat of vaporization is used to calculate the heat required to condense the steam. Adding these two heat values together gives us the total amount of heat that must be removed from the steam, which is approximately 3.61164 kcal.
Explanation:When steam at 100°C condenses and its temperature is lowered to 46°C, heat must be removed from the steam. To calculate the amount of heat, we can use the specific heat of steam and the latent heat of vaporization. First, we calculate the heat required to lower the temperature of the steam from 100°C to 46°C using the specific heat of water. We then calculate the heat required to condense the steam using the latent heat of vaporization. Finally, we add these two heat values together to obtain the total amount of heat that must be removed from the steam.
Given:
Mass of steam = 6.1 gTemperature change = 100°C - 46°C = 54°CSpecific heat of water = 1.00 kcal/(kg · °C)Heat of vaporization = 539 kcal/kg
Calculations:
Calculation:
Q1 = 0.32874 kcalQ2 = 3.2829 kcalQ = Q1 + Q2 = 0.32874 kcal + 3.2829 kcal = 3.61164 kcal
Therefore, the quantity of heat that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C is approximately 3.61164 kcal.
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To condense and cool 6.1 g of 100°C steam to 46°C, 3.2879 kcal must be removed for condensation, and 0.3304 kcal for cooling, for a total of 3.6183 kcal.
Explanation:Calculating the Quantity of Heat for Condensation and Cooling
To calculate the quantity of heat that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C, we need to consider two processes: condensation and cooling. For condensation, we use the heat of vaporization, and for cooling, we use the specific heat of water.
Calculate the heat released during condensation of steam into water at 100°C:Total heat removed is the sum of the heat from both steps: 3.2879 kcal + 0.3304 kcal = 3.6183 kcal.
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During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h).
What is the average force exerted on the 0.057-kg tennis ball by Venus Williams’ racquet, assuming that the ball’s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)?
a)660 N
b)1660 N
c)25 N
d)4000 N
Final answer:
The average force exerted on the tennis ball by Venus Williams' racquet during her serve is calculated using the formula F = Δp/Δt, resulting in approximately 660 N, with the correct answer being option a).
Explanation:
The question relates to determining the average force exerted by Venus Williams' racquet on a tennis ball during her serve at the 2007 French Open. To find the average force, we can use the formula derived from Newton's second law of motion, F = ma. However, instead of finding the acceleration first, we use the formula F = Δp/Δt (force is the change in momentum over the change in time), because we know the velocity after impact and the time of contact.
The initial momentum (pi) is 0, as the initial velocity is negligible, so the final momentum (pf) is mv, where m is the mass of the tennis ball (0.057 kg) and v is the final velocity (58 m/s). As the time of contact with the racquet is 5.0 ms (or 0.005 seconds), we calculate the force as follows:
F = (m × v)/Δt = (0.057 kg × 58 m/s)/0.005 s = (3.306 kg·m/s)/0.005 s = 661.2 N
Therefore, the correct answer is approximately 660 N, which matches option a).
If the surface temperature of that person's skin is 30∘C (that's a little lower than healthy internal body temperature because your skin is usually a little colder than your insides), what is the total power that person will radiate? For now, assume the person is a perfect blackbody (so ϵ=1).
Answer:
[tex]E=477.92\ W.m^{-2}[/tex]
Explanation:
Given that:
Absolute temperature of the body, [tex]T=273+30=303\ K[/tex]
emissivity of the body, [tex]\epsilon=1[/tex]Using Stefan Boltzmann Law of thermal radiation:
[tex]E=\epsilon. \sigma.T^4[/tex]
where:
[tex]\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}[/tex] (Stefan Boltzmann constant)
Now putting the respective values:
[tex]E=1\times 5.67\times 10^{-8}\times 303^4[/tex]
[tex]E=477.92\ W.m^{-2}[/tex]
A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her height. The observer measures her relativistic momentum to have a magnitude of 2.1 × 1010 kg·m/s. What does the observer measure for her height?
To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.
By definition we know that the relative velocity of an object with reference to the Light, is defined by
[tex]V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}[/tex]
Where,
V = Speed from relative point
c = Speed of light
On the other hand we have that the linear momentum is defined as
P = mv
Replacing the relative velocity equation here we have to
[tex]P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}[/tex]
[tex]P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}[/tex]
[tex]P^2 = \frac{P^2V^2}{c^2}+m^2V^2[/tex]
[tex]P^2 = V^2 (\frac{P^2}{c^2}+m^2)[/tex]
[tex]V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}[/tex]
[tex]V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}[/tex]
[tex]V = 2.81784*10^8m/s[/tex]
Therefore the height with respect the observer is
[tex]l = l_0*\sqrt{1-\frac{V^2}{c^2}}[/tex]
[tex]l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}[/tex]
[tex]l = 0.56m[/tex]
Therefore the height which the observerd measure for her is 0.56m
A lightning flash releases about 1010J of electrical energy. Part A If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37∘C, what are the final state and temperature of the water? The specific heat of water is 4180 J/kg⋅ ∘C, heat of vaporization at the boiling temperature for water is 2.256×106J/kg, the specific heat of steam is 1970 J/kg⋅ ∘C.
Answer:
water is in the vapor state,
Explanation:
We must use calorimetry equations to find the final water temperatures. We assume that all energy is transformed into heat
E = Q₁ + [tex]Q_{L}[/tex]
Where Q1 is the heat required to bring water from the current temperature to the boiling point
Q₁ = m [tex]c_{e}[/tex] ( [tex]T_{f}[/tex] -T₀)
Q₁ = 50 4180 (100 - 37)
Q₁ = 1.317 10⁷ J
Let's calculate the energy so that all the water changes state
[tex]Q_{L}[/tex] = m L
[tex]Q_{L}[/tex] = 50 2,256 106
[tex]Q_{L}[/tex] = 1,128 10⁸ J
Let's look for the energy needed to convert all the water into steam is
Qt = Q₁ + [tex]Q_{L}[/tex]
Qt = 1.317 107 + 11.28 107
Qt = 12,597 10⁷ J
Let's calculate how much energy is left to heat the water vapor
ΔE = E - Qt
ΔE = 10¹⁰ - 12,597 10⁷
ΔE = 1000 107 - 12,597 107
ΔE = 987.4 10⁷ J
With this energy we heat the steam, clear the final temperature
Q = ΔE = m [tex]c_{e}[/tex] ( [tex]T_{f}[/tex]-To)
( [tex]T_{f}[/tex]-T₀) = ΔE / m [tex]c_{e}[/tex]
[tex]T_{f}[/tex] = T₀ + ΔE / m [tex]c_{e}[/tex]
[tex]T_{f}[/tex] = 100 + 987.4 10⁷ / (50 1970)
[tex]T_{f}[/tex] = 100 + 1,002 10⁵
[tex]T_{f}[/tex] = 1,003 10⁵ ° C
This result indicates that the water is in the vapor state, in realizing at this temperature the water will be dissociated into its hydrogen and oxygen components
Calculating the final state and temperature of 50 kg of water at 37°C with 10¹°J of energy involves determining the energy needed to heat the water to 100°C and the energy to vaporize it. Using the given specific heat of water, heat of vaporization, and specific heat of steam, one can find whether all the water turns to steam and whether any energy is left to further heat the steam.
Explanation:If a lightning flash releases about 1010J of electrical energy, and all this energy is added to 50 kg of water at 37°C, we can determine the final state and temperature of the water. Given that the specific heat of water is 4180 J/kg·°C, the heat of vaporization at the boiling temperature for water is 2.256×106J/kg, and the specific heat of steam is 1970 J/kg·°C, these values can be used to calculate the amount of water that can be heated, brought to a boil, and then completely turned to steam. We can break this problem down into different stages, each requiring a certain amount of energy.
Firstly, we would need to calculate how much energy is required to bring the water from 37°C to 100°C (its boiling point):
Energy to heat the water to boiling point (Q1) = mass (m) × specific heat of water (c) × change in temperature (ΔT) = 50 kg × 4180 J/kg·°C × (100°C - 37°C).Afterward, we would need to calculate the energy required to vaporize the water:
Energy to vaporize the water (Q2) = mass (m) × heat of vaporization (Lv) = 50 kg × 2.256×106J/kg.The sum of Q1 and Q2 should be equal to or less than the energy released by the lightning flash for all the water to be vaporized. If there is any excess energy after vaporizing the water, it would further heat the steam. If all the energy is not used in heating and vaporizing the water, the final state would still include some liquid water. A detailed calculation using the provided information and considering the stages involved will give the final temperature and state of the water.
Ball made of wood and ball made of aluminum, of the same radius are placed into reservoir with water. Aluminum ball after sinking rests at the bottom, wooden ball floats. Which ball experiences greater buoyancy force? (14.4)
Answer:Aluminium
Explanation:
Given
Aluminium ball sinks after placing inside the water while wooden block floats.
Buoyancy on a body is given by
[tex]volume\ inside\ the\ water \times density\ of\ fluid\times g[/tex]
here [tex]\rho _{Al}>\rho _{water}[/tex]
thus aluminium ball sinks
but for
[tex]\rho _{wood}<\rho _{water}[/tex]
only a part of wood is under the water i.e. not whole volume is not under water therefore net buoyancy is less in case of wood
A 1.8-m uniform rod is being used to balance two buckets of paint. One pucket is full, but the other is mostly empty, and has only 0.25 the inertia of the full bucket. How far away from the full bucket is the center of mass of the system located?
Answer: 0.36 m
Explanation:
By definition, the x coordinate of the center of mass of the system obeys the following equation:
Xm = (m1x1 + m2x2 + …….+ mnxn) / m1+m2 +……+ mn
Neglecting the mass of the rod, and choosing our origin to be coincident with the location of the full bucket, we can write the following expression for the X coordinate of the center of mass (Assuming that both masses are aligned over the x-axis, so y-coordinates are zero):
Xm = 0.25 mb . 1.8 m / (1+0.25) mb
Simplifying mb, we get:
Xm= 0.36 m (to the right of the full bucket).
Final answer:
To find the center of mass of the system, use the principle of moments to balance the torques produced by the full and empty buckets. The distance from the fulcrum to the center of mass can be calculated using the equation for torque.
Explanation:
To find the center of mass of the system, we can use the principle of moments. The center of mass of the system will be located at a distance from the full bucket that balances the torques produced by the two buckets.
Let's denote the distance from the fulcrum to the center of mass of the system as x. The torques produced by the full and empty buckets can be calculated using the formula, Torque = Force * Distance.
Since the empty bucket has 0.25 the inertia of the full bucket, the torque produced by the empty bucket will be 0.25 times the torque produced by the full bucket. Setting up the equation using the principle of moments, we can solve for x.
A baseball has a mass of 0.15 kg and radius 3.7 cm. In a baseball game, a pitcher throws the ball with a substantial spin so that it moves with an angular speed of 45 rad/s and a linear speed of 42 m/s. Assuming the baseball to be a uniform solid sphere, determine the rotational and translational kinetic energies of the ball in joules.
Answer:
Explanation:
Given
mass of baseball [tex]m=0.15 kg[/tex]
radius of ball [tex]r=3.7 cm[/tex]
angular speed of ball [tex]\omega =45 rad/s[/tex]
linear speed of ball [tex]v=42 m/s[/tex]
Transnational Kinetic Energy is given by
[tex]K.E.=\frac{mv^2}{2}[/tex]
[tex]K.E.=\frac{1}{2}\times 0.15\times 42^2[/tex]
[tex]k.E.=\frac{1}{2}\times 0.15\times 1764[/tex]
[tex] k.E.=132.3 J[/tex]
Considering the ball as solid sphere its moment of inertia is given by
[tex]I=\frac{2}{5}mr^2=\frac{2}{5}\times 0.15\times (0.037)^2[/tex]
[tex]I=8.21\times 10^{-5} kg-m^2[/tex]
Rotational Kinetic Energy
[tex]=\frac{1}{2}\times I\times \omega ^2[/tex]
[tex] =\frac{1}{2}\times 8.21\times 10^{-5}\times 45^2[/tex]
[tex] =\frac{1}{2}\times 8.21\times 10^{-5}\times 2025[/tex]
[tex] =0.0831 J[/tex]
A 10 m beam of mass 280 kg extends over a ledge. The beam is not attached, but simply rests on the surface. A 50 kg student intends to position the beam so that he can walk to the end of it. How far from the edge of the ledge can the beam extend?
To determine how far the beam can extend over the ledge, we need to consider the principle of equilibrium. The beam will be in equilibrium when the clockwise moments equal the counterclockwise moments. Calculating the moments, we find that the beam can extend 7.0 meters from the edge of the ledge.
Explanation:To determine how far the beam can extend over the ledge, we need to consider the principle of equilibrium. The beam will be in equilibrium when the clockwise moments equal the counterclockwise moments. The moment is a force multiplied by the distance from the pivot point. In this case, the pivot point is the edge of the ledge and the force is the weight of the beam and the student.
Let's calculate the moments: The moment of the beam is 280 kg * 9.8 m/s² * x, where x is the distance from the edge of the ledge. The moment of the student is 50 kg * 9.8 m/s² * (10 m - x), where 10 m is the length of the beam.
Setting the clockwise moments equal to the counterclockwise moments, we have 280 kg * 9.8 m/s² * x = 50 kg * 9.8 m/s² * (10 m - x). Solving for x gives us x = 7.0 m. Therefore, the beam can extend 7.0 meters from the edge of the ledge.
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Final answer:
To ensure the beam does not tip, calculate the torque around the pivot at the ledge. The beam's weight produces a torque at its center of mass, while the student's weight produces a torque at the beam's end. Solving for equilibrium, the beam can extend 5.6 meters.
Explanation:
The problem presented involves a horizontal beam that a student wishes to extend over a ledge without it falling over due to its own weight and the weight of the student. To determine how far the beam can extend over the ledge, one must consider the torque that is acting on the beam around the pivot point, which is at the edge of the ledge. The beam must remain in rotational equilibrium, meaning the total torque around the pivot must be zero.
Let's assume the beam extends a distance x over the edge. The weight of the beam acts at its center of mass, which is x/2 from the pivot (edge of the ledge) if x is the length beyond the ledge. The moment (or torque) due to the beam's weight is (280 kg × 9.8 m/s² × x/2). The moment due to the student's weight at the end of the beam is (50 kg × 9.8 m/s² × x). For rotational equilibrium, these moments must be equal.
Setting the torques equal gives us: 280 kg × 9.8 m/s² × x/2 = 50 kg × 9.8 m/s² × x. Solving for x gives x = (280/100) × 2, so x must be 5.6 meters. Thus the beam can safely extend 5.6 meters over the ledge without tipping over.
Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e., protons with one orbiting electron) are fed into a chamber where there is a strong magnetic field. Electrons in this chamber are trapped in tight orbits, which greatly increases the chance that they will collide with a hydrogen atom and ionizes it. One such source uses a magnetic field of 90 mT, and the electrons' kinetic energy is 2.0 eV.If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?
Answer:
Explanation:
Electron's kinetic energy = 2 eV
= 2 x 1.6 x 10⁻¹⁹ J
1/2 m v² = 3.2 x 10⁻¹⁹
1/2 x 9.1 x 10⁻³¹ x v² = 3.2 x 10⁻¹⁹
v² = .703 x 10¹²
v = .8385 x 10⁶ m/s
Electrons revolve in a circular orbit when forced to travel in a magnetic field whose radius can be expressed as follows
r = mv / Bq
where m , v and q are mass , velocity and charge of electron .
here given magnetic field B = 90 mT
= 90 x 10⁻³ T
Putting these values in the expression above
r = mv / Bq
= [tex]\frac{9\times10^{-31}\times.8385\times10^6}{90\times10^{-3}\times1.6\times10^{-19}}[/tex]
= .052 mm.
Hospital waiting room times are normally distributed with a mean of 38.12 minutes and a standard deviation of 8.63 minutes. What is the shortest wait time that would still be in the worst 20% of wait times?
Answer:
[tex]x= 45.46[/tex]
Explanation:
given,
mean of hospital waiting (μ) = 38.12 min
standard deviation (σ) = 8.63 min
worst waiting time = 20%
the are will be same as the = 100 - 20%
= 80%
we have to determine the z- value according to 80% or 0.80
using z-table
z-value = 0.85
now, using formula
[tex]Z = \dfrac{x-\mu}{\sigma}[/tex]
[tex]0.85 = \dfrac{x-38.12}{8.63}[/tex]
[tex]x-38.12 = 0.85\times {8.63}[/tex]
[tex]x= 45.46[/tex]
waiting times follow a normal distribution with
Mean, \mu=38.12Mean,μ=38.12
Standard\ deviation,\sigma=8.63Standard deviation,σ=8.63
Longer waiting times are worse than shorter waiting times. Hence the worst 20% of wait times are wait times on the right tail of the distribution. The inferred level of confidence is 0.80.
The z value corresponding to the right tail probability of 0.2 is
Z=0.85Z=0.85
But
Z = \frac{x-\mu}{\sigma}Z=
σ
x−μ
x =Z*\sigma +\mux=Z∗σ+μ
=0.85 * 8.63 +38.12 =45.4555=0.85∗8.63+38.12=45.4555
answer:
the shortest wait time that would still be in the worst 20% of wait times is 45.4555 minutes
A Martian leaves Mars in a spaceship that is heading to Venus. On the way, the spaceship passes earth with a speed v = 0.80c relative to it. Assume that the three planets do not move relative to each other during the trip. The distance between Mars and Venus is 1.20 × 1011 m, as measured by a person on earth. What does the Martian measure for the distance between Mars and Venus?
To find the relative distance from one point to another it is necessary to apply the Relativity equations.
Under the concept of relativity the distance measured from a spatial object is given by the equation
[tex]l = l_0 \sqrt{1-\frac{v^2}{c^2}}[/tex]
Where
[tex]l_0[/tex]= Relative length
v = Velocity of the spaceship
c = Speed of light
Replacing with our values we have that
[tex]l = l_0 \sqrt{1-\frac{v^2}{c^2}}[/tex]
[tex]l = 1.2*10^{11} \sqrt{1-\frac{0.8c^2}{c^2}}[/tex]
[tex]l = 1.2*10^{11} \sqrt{1-0.8^2}[/tex]
[tex]l = 7.2*10^{10}m[/tex]
Therefore the distance between Mars and Venus measured by the Martin is [tex]7.2*10^{10}m[/tex]
Which of the following statements are true?
A. Heat is converted completely into work during isothermal expansion.
B. Isothermal expansion is reversible under ideal conditions.
C. During the process of isothermal expansion, the gas does more work than during an isobaric expansion (at constant pressure) between the same initial and final volumes.
WHAT I KNOW:Curve C= Isothermal process and A=Isobaric processThe first law of thermodynamics is a form of conservation of energy where the change of internal energy of a closed system will be equal to the energy added to the system minus work done by the system on its surroundings. deltaU=Q-W. Isothermal process is when the temperature doesnt change and the system is in contact with a heat reservoir. delta U=0 b/c because temperature is constant so the first law can be rewritten as Q=W, because Q=W that means #1 is true, right?.I think that the process could be reversible if heat is added or lost? which would make #2 true..?Work can be calulated by equation W=PdeltaV or by the area under the PV diagram curve. so is #3 false?
Answer:
Statements A & B are true.
Explanation:
Heat is converted completely into work during isothermal expansion.
Work done in an isothermal process is given as:
[tex]W=n.R.T.ln(\frac{V_f}{V_i} )[/tex]
where the subscripts denote final and initial conditions.
In case of an ideal gas the change in internal energy is proportional to the change in temperature. Here the temperature is constant in the process, therefore ΔU=0.
From the first law of thermodynamics:
[tex]dW=dQ+dU[/tex]
for isothermal process it becomes
[tex]dW=dQ[/tex]
Isothermal expansion is reversible under ideal conditions.
Ideally there is no friction involved as all the heat is converted into work, so the process is reversible.
During the process of isothermal expansion, the gas does more work than during an isobaric expansion (at constant pressure) between the same initial and final volumes.
As we know that the work done is given by the area under the P-V curve so in case of the work done between two specific states of volume by an isothermal process we have only one path and by isobaric process we have 2 paths among which one is having the higer work done than the isothermal process nd the other one is having the lesser area under the curve than that of under isothermal curve.
If there were a great migration of people toward the Earth's equator, the length of the day would
a. decrease because of conservation of angular momentum.
b. remain unaffected.
c. increase because of conservation of angular momentum.
d. decrease because of conservation of energy.
e. increase because of conservation of energy.
Answer:
C. increase because of conservation of angular momentum.
Explanation:
We know that if there is no any external torque on the system then the angular momentum of the system will be conserved.
Angular momentum L
L = I ω
I=Mass moment of inertia , ω=Angular speed
If angular momentum is conserved
I₁ω₁ = I₂ω₂
If people migrates towards Earth's equator then mass moment of inertia will increases.
I₂ > I₁
Then we can say that ω₁ > ω₂ ( angular momentum is conserve)
We know that time period T given as
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]T_1=\dfrac{2\pi}{\omega_1}[/tex]
[tex]T_2=\dfrac{2\pi}{\omega_2}[/tex]
If ω₂ decrease then T₂ will increase .It means that length of the day will increase.
Therefore answer is C.
If a large number of people migrated towards the Earth's equator, the day length would remain largely unaffected. This is due to the law of conservation of angular momentum, considering the insignificant change to the Earth's total mass.
Explanation:The scenario mentioned in the question refers to a law in physics called the conservation of angular momentum. The length of a day on Earth is predominantly determined by how fast our planet spins on its axis. If there were a large migration of people towards the Earth's equator, hypothetically, the day length would remain largely unaffected. The mass of the people in this context is insignificant when compared to the total mass of the Earth. As such, their migration would not have a noticeable impact on Earth's rotational speed due to conservation of angular momentum. Therefore, the correct answer is that the length of the day would remain unaffected (option b).
Learn more about Conservation of Angular Momentum here:https://brainly.com/question/32554788
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2. A package with a heat sink and fan has a thermal resistance to the ambient of 8 °C/W. The thermal resistance from the die to the package is 2 °C/W. If the package is in a chassis that will never exceed 50 °C and the maximum acceptable die temperature is 110 °C, how much power can the chip dissipate?
To solve this problem it is necessary to apply the concepts related to the Power depending on the temperature and the heat transferred.
By definition the power can be expressed as
[tex]P = \frac{\Delta T}{\Delta Q}[/tex]
Where,
[tex]\Delta T = T_m - T_a =[/tex] Change at the temperature, i.e, the maximum acceptable die temperature ([tex]T_m[/tex]) with the allowable temperature in chassis ([tex]T_A[/tex])
[tex]\Delta Q = Q_A-Q_D =[/tex] Change in the thermal resistance to ambient ([tex]Q_A[/tex]) and the Thermal resistance from die to package ([tex]Q_D[/tex])
Our values are given as,
[tex]T_m=110\°C[/tex]
[tex]T_a= 50\°C[/tex]
[tex]Q_A= 8\°C/W[/tex]
[tex]Q_D= 2\°C/W[/tex]
Replacing we have,
[tex]P = \frac{110-50}{8-2}[/tex]
[tex]P = 10W[/tex]
The power that can dissipate the chip is 10W