Answer:
A. The pressure denoted as Pa and Pb at the surfaces of A and B in the tube is
PA= Pgas
PB= Patmos
B. The second sketch
C. The gas pressure is
Pgas= Patmos+ rho.g(h2-h1)
= 1atm + rho.g (h2-h1)
Explanation:
The gas pressure inside the box is [tex]\( 1.018 \times 10^5 \, \text{Pa} \)[/tex] to three significant figures.
To find the gas pressure [tex]\( p_{\text{gas}} \)[/tex] inside the box using the information provided, we'll follow the hydrostatic equilibrium principles as outlined.
The density of mercury [tex](\( \rho \)) = \( 1.36 \times 10^4 \, \text{kg/m}^3 \)[/tex]
The height of the mercury column in the left arm [tex](\( h_1 \)) = 10.0 cm = 0.10 m[/tex]
The height of the mercury column in the right arm [tex](\( h_2 \)) = 6.00 cm = 0.06 m[/tex]
Atmospheric pressure [tex](\( p_{\text{atm}} \)) = 1.00 atm = \( 1.013 \times 10^5 \, \text{Pa} \)[/tex]
Acceleration due to gravity [tex](\( g \)) = \( 9.81 \, \text{m/s}^2 \)[/tex]
1. Convert the heights to meters:
[tex]\[ h_1 = 0.10 \, \text{m} \] \[ h_2 = 0.06 \, \text{m} \][/tex]
2. Draw a schematic:
The left arm of the U-tube is connected to the box containing the gas. The right arm is open to the atmosphere. Mercury levels in the U-tube indicate that the pressure on the left side (due to the gas) is higher than the pressure on the right side (atmospheric pressure).
3. Pressure at the interface between mercury and gas (left side):
[tex]\[ p_{\text{left}} = p_{\text{gas}} \][/tex]
4. Pressure at the interface on the open side (right side):
[tex]\[ p_{\text{right}} = p_{\text{atm}} \][/tex]
5. Difference in mercury column height:
[tex]\[ \Delta h = h_1 - h_2 = 0.10 \, \text{m} - 0.06 \, \text{m} = 0.04 \, \text{m} \][/tex]
6. Hydrostatic pressure difference:
The difference in height of the mercury columns corresponds to the pressure difference due to the gas in the box and the atmospheric pressure:
[tex]\[ \Delta p = \rho g \Delta h \] \[ \Delta p = (1.36 \times 10^4 \, \text{kg/m}^3) \times (9.81 \, \text{m/s}^2) \times (0.04 \, \text{m}) \] \[ \Delta p = 1.36 \times 10^4 \times 9.81 \times 0.04 \] \[ \Delta p = 532.224 \, \text{Pa} \][/tex]
7. Calculate the gas pressure: Since the pressure in the gas must account for the atmospheric pressure plus the pressure due to the height difference of the mercury:
[tex]\[ p_{\text{gas}} = p_{\text{atm}} + \Delta p \] \[ p_{\text{gas}} = 1.013 \times 10^5 \, \text{Pa} + 532.224 \, \text{Pa} \] \[ p_{\text{gas}} = 1.01832 \times 10^5 \, \text{Pa} \][/tex]
[tex]\[p_{\text{gas}} = 1.018 \times 10^5 \, \text{Pa}\][/tex]
Complete question:- To practice Tactics Box 13.1 Hydrostatics.
In problems about liquids in hydrostatic equilibrium, you often need to find the pressure at some point in the liquid. This Tactics Box outlines a set of rules for thinking about such hydrostatic problems.
Answer:
Assume patmos=1.00atmpatmos=1.00atm. What is the gas pressure?
Express your answer in pascals to three significant figures.
Pgas=_____ Pa
EDIT (Needed Information):
Draw a picture. Show open surfaces, pistons, boundaries, and other features that affect pressure. Include height and area measurements and fluid densities. Identify the points at which you need to find the pressure. These objects make up the system; the environment is everything else.
Determine the pressure p0p0 at surfaces.
Surface open to the air: p0=patmosp0=patmos, usually 1 atmatm.
Surface in contact with a gas: p0=pgasp0=pgas.
Closed surface: p0=F/Ap0=F/A, where FFF is the force the surface, such as a piston, exerts on the fluid and AAA is the area of the surface.
Use horizontal lines. The pressure in a connected fluid (of one kind) is the same at any point along a horizontal line.
Allow for gauge pressure. Pressure gauges read pg=p−1atmpg=p−1atm.
Use the hydrostatic pressure equation: p=p0+ρghp=p0+ρgh, where ρρrho is the density of the fluid, ggg is the acceleration due to gravity, and hhh is the height of the fluid.
Use these rules to work out the following problem: A U-shaped tube is connected to a box at one end and open to the air at the other end. The box is full of gas at pressure pgaspgasp_gas, and the tube is filled with mercury of density 1.36×104 kg/m3kg/m3 . When the liquid in the tube reaches static equilibrium, the mercury column is h1h1h_1 = 10.0 cmcm high in the left arm and h2h2h_2 = 6.00 cmcm high in the right arm, as shown in the figure.(Figure 1) What is the gas pressure pgaspgasp_gas inside the box?
A simple pendulum is made my attaching a rod of negligible mass to a 2.0 kg pendulum bob at the end. It is observed that on Earth, the period of small-angle oscillations is 1.0 second. It is also observed that on Planet X this same pendulum has a period of 1.8 seconds. How much does the pendulum bob weigh on Planet X
Final answer:
The weight of the pendulum bob on Planet X remains 2.0 kg, the same as on Earth, because the mass of the pendulum bob does not affect the period of a simple pendulum. The difference in periods indicates a change in gravitational strength, not mass.
Explanation:
The question revolves around understanding how the period of oscillation of a simple pendulum changes with gravity on different planets. Since the mass of the pendulum bob does not affect the period of a simple pendulum, which depends only on the length of the pendulum and the gravitational acceleration (g), the weight of the pendulum bob on Planet X would still be 2.0 kg, the same as on Earth. However, the difference in periods between Earth and Planet X indicates a difference in gravitational acceleration, implying that g on Planet X is weaker than on Earth. The formula for the period (T) of a simple pendulum is T = 2π√(l/g), where l is the length of the pendulum and g is the gravitational acceleration. Since the mass of the pendulum bob does not factor into this equation, the weight of the pendulum bob on Planet X remains the same, but its apparent weight will change according to the planet's gravitational pull.
A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\text{m}})x - (4.64 ~\frac{\text{rad}}{\text{sec}})t +(1.33 ~\text{rad}) \big)y=(5.26 m)⋅sin((1.65 m rad )x−(4.64 sec rad )t+(1.33 rad)) How much time will it take for a peak on this traveling wave to propagate a distance of 5.00 meters along the length of the string?
Answer:
t = 1.77 s
Explanation:
The equation of a traveling wave is
y = A sin [2π (x /λ -t /T)]
where A is the oscillation amplitude, λ the wavelength and T the period
the speed of the wave is constant and is given by
v = λ f
Where the frequency and period are related
f = 1 / T
we substitute
v = λ / T
let's develop the initial equation
y = A sin [(2π / λ) x - (2π / T) t +Ф]
where Ф is a phase constant given by the initial conditions
the equation given in the problem is
y = 5.26 sin (1.65 x - 4.64 t + 1.33)
if we compare the terms of the two equations
2π /λ = 1.65
λ = 2π / 1.65
λ = 3.81 m
2π / T = 4.64
T = 2π / 4.64
T = 1.35 s
we seek the speed of the wave
v = 3.81 / 1.35
v = 2.82 m / s
Since this speed is constant, we use the uniformly moving ratios
v = d / t
t = d / v
t = 5 / 2.82
t = 1.77 s
A pilot flies in a straight path for 1 hour 30 minutes. She then makes a course correction, heading 45 degrees to the right of her original course, and flies 2 hours 15 minutes in the new direction. If she maintains a constant speed of 345 mi/h, how far is she from her starting position? Give your answer to the nearest mile.
Answer:
1199 miles
Explanation:
1 hour 30 minutes = 1 + 30/60 = 1.5 hours
2 hours 15 minutes = 2 + 15/60 = 2.25 hours
The distance she flew in the 1st segment is:
1.5*345 = 517.5 miles
The distance she flew in the 2nd segment is:
2.25 * 345 = 776.25 miles
Since the 2nd segment is 45 degree with respect to the 1st segment, this means that she has flown
776.25 * cos(45) = 549 miles in-line with the 1st segment and
776.25* sin(45) = 549 miles perpendicular to the 1st segment:
So the distance from the end to her starting position is
[tex]\sqrt{(517.5 + 549)^2 + 549^2} = 1199 miles[/tex]
A rectangular neoprene sheet has width W = 1.00 m and length L = 4.00 m. The two shorter edges are affixed to rigid steel bars that are used to stretch the sheet taut and horizontal. The force applied to either end of the sheet is F = 81.0 N. The sheet has a total mass M = 4.00 kg. The left edge of the sheet is wiggled vertically in a uniform sinusoidal motion with amplitude A = 10.0 cm and frequency f = 1.00 Hz. This sends waves spanning the width of the sheet rippling from left to right. The right side of the sheet moves upward and downward freely as these waves complete their traversal.
(A) Derive an expression for the velocity with which the waves move along the sheet. Express your answer in terms of the variables F, L, and M.
(B) What is the value of this speed for the specified choices of these parameters? Express your answer with the appropriate units.
Answer:
(a) [tex]\sqrt{FL/M}[/tex]
(b) [tex]9 ms^{-1}[/tex]
Explanation:
(a)
The speed of the wave depends on the type of the material. Here we have neoprene sheet so the speed of the wave will depend on the linear density of neoprene sheet. Linear Density is defined as Mass per unit length of the material (as materials of same type can have different thickness). The symbol used for the Linear Density is .
μ = Mass of the sheet / Length of the sheet
The speed of the wave in such material can be be found by using the equation:
[tex]\frac{1}{v^{2} } = {\frac{Linear Density}{Force} }[/tex] (where v is speed)
The equation can be rearranged:
v = [tex]\sqrt{Force/Linear Density}[/tex]
so,
v = sqrt(F/μ)
v = sqrt(F/ (M/L))
v = [tex]\sqrt{FL/M}[/tex] (answer)
(b)
Putting the values
F = 81 N
M = 4kg
L = 4m
v = [tex]\sqrt{FL/M}[/tex]
v = 9m/s
A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the ends. (a) What is the current in amperes in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the material of which the wire is made.
Answer:
(a) Current is 2831.93 A
(b) [tex]8.40A/m^2[/tex]
(c) [tex]\rho =15.52\times 10^{-9}ohm-m[/tex]
Explanation:
Length of wire l = 3.22 m
Diameter of wire d = 7.32 mm = 0.00732 m
Cross sectional area of wire
[tex]A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2[/tex]
Resistance [tex]R=11.9mohm=11.9\times 10^{-3}ohm[/tex]
Potential difference V = 33.7 volt
(A) current is equal to
[tex]i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A[/tex]
(B) Current density is equal to
[tex]J=\frac{i}{A}[/tex]
[tex]J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2[/tex]
(c) Resistance is equal to
[tex]R=\frac{\rho l}{A}[/tex]
[tex]11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}[/tex]
[tex]\rho =15.52\times 10^{-9}ohm-m[/tex]
One consequence of turbulence is mixing. Different layers of fluid flow cross over one another very easily, and get blended together. This is another kind of "transport", which lets atoms which might have started out all in one place get uniformly mixed around. Would you expect turbulent mixing to happen most easily in:
A. waterB. motor oilC. airD. honey
Answer:
Turbulence mixing will happen mostly in air
Explanation:
When the temperature is at 30∘C, the A-36 steel pipe fits snugly between the two fuel tanks. When fuel flows through the pipe, the temperature at ends A and B rise to 130∘C and 80∘C, respectively. If the temperarture drop along the pipe is linear, determine the average normal stress developed in the pipe. Assume the walls of each tank acts as a spring, each having a stiffness of k=900 MN/m.
In a system where an A-36 steel pipe is snugly fit between two fuel tanks, the rise in fuel temperature will cause thermal stresses in the pipe due to the restraint from the fuel tanks. This is related to the thermal properties of the material and the rate of change due to temperature rising, from which the average normal stress can be calculated.
Explanation:Determining the average normal stress developed in an A-36 steel pipe when fuel flows through it at varying temperatures requires knowledge of thermal expansion and associated stress in materials. This is related to thermal properties and the rate of change due to temperature rise.
When temperature rises along the pipe such as mentioned, from room temperature to 130∘C and 80∘C, the steel pipe expands. However, the pipe is restrained by the fuel tanks acting as springs, leading to development of stress within the pipe. The average normal stress can be calculated by dividing the force exerted by the expansion (or contraction) by the cross-sectional area of the pipe:
F/A = σ
Where, F is the force and A is the cross-sectional area of the pipe. The force can be obtained from Hooke's law for springs (F = k Δx), where k is the stiffness of the tank walls acting as springs, and Δx is the change in length due to thermal expansion.
The average normal stress is a measure of the extent to which the pipe is going through physical changes due to thermal variations.
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Find the corresponding initial value problem of the following spring-mass systems, that if solved, would give the position u of the mass at any time t. Use 9.8 m/s 2 for the acceleration due to gravity. DO NOT SOLVE THE IVP. A mass of 60 kg stretches a spring 19.6 m. The mass is acted on by an external force of 6 sin(4t) N and moves in a medium that imparts a viscous force of 8 N when the speed of the mass is 16 cm/s. The mass is pushed up 5 cm and set in motion an initial downward velocity of 2 cm/s. Write the IVP so that u would be in meters if solved.
Answer:
Explanation:
The picture attached shows the full explanation
A student builds and calibrates an accelerometer and uses it to determine the speed of her car around a certain unbanked highway curve. The accelerometer is a plumb bob with a protractor that she attaches to the roof of her car. A friend riding in the car with the student observes that the plumb bob hangs at an angle of 15.0° from the vertical when the car has a speed of 21.5 m/s. (a) What is the centripetal acceleration of the car rounding the curve? m/s2 (b) What is the radius of the curve? m (c) What is the speed of the car if the plumb bob deflection is 7.00° while rounding the same curve? m/s
(a) [tex]\(a_c \approx 39.8 \, \text{m/s}^2\)[/tex]
(b) [tex]\(r \approx 11.6 \, \text{m}\)[/tex]
(c) [tex]\(v_2 \approx 28.4 \, \text{m/s}\)[/tex]
(a) calculation of the centripetal acceleration [tex](\(a_c\))[/tex],
[tex]\[ a_c = \frac{g}{\sin(\theta_1)} \][/tex]
Given [tex]\(g \approx 9.8 \, \text{m/s}^2\) and \(\theta_1 = 15.0°\)[/tex], we find:
[tex]\[ a_c = \frac{9.8 \, \text{m/s}^2}{\sin(15.0°)} \][/tex]
Calculating this gives [tex]\(a_c \approx 39.8 \, \text{m/s}^2\).[/tex]
(b) The radius ((r)) of the curve can be found using the formula:
[tex]\[ r = \frac{v^2}{a_c} \][/tex]
Substituting [tex]\(v = 21.5 \, \text{m/s}\) and \(a_c \approx 39.8 \, \text{m/s}^2\),[/tex]we get:
[tex]\[ r = \frac{(21.5 \, \text{m/s})^2}{39.8 \, \text{m/s}^2} \][/tex]
This calculation yields [tex]\(r \approx 11.6 \, \text{m}\).[/tex]
(c) For a new deflection angle [tex]\(\theta_2 = 7.00°\)[/tex], we find the new centripetal acceleration [tex](\(a_{c2}\))[/tex] using:
[tex]\[ a_{c2} = \frac{g}{\sin(\theta_2)} \][/tex]
Substituting [tex]\(g \approx 9.8 \, \text{m/s}^2\) and \(\theta_2 = 7.00°\)[/tex], we get:
[tex]\[ a_{c2} = \frac{9.8 \, \text{m/s}^2}{\sin(7.00°)} \][/tex]
Calculating this gives [tex]\(a_{c2} \approx 84.7 \, \text{m/s}^2\).[/tex]
Then, using [tex]\(a_{c2} = \frac{v_2^2}{r}\)[/tex], we find the new speed [tex](\(v_2\)):[/tex]
[tex]\[ v_2 = \sqrt{a_{c2} \cdot r} \][/tex]
Substituting [tex]\(a_{c2} \approx 84.7 \, \text{m/s}^2\) and \(r \approx 11.6 \, \text{m}\), we get \(v_2 \approx 28.4 \, \text{m/s}\).[/tex]
You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it. Imagine that you slowly pull down with a spring scale so the block is now below the equilibrium position where it was hanging at rest. The scale reading when you let go of the block is 3.0 N.
Part A
Where was the block when you let go? Assume y0 is the equilibrium position of the block and that "down" is the positive direction.
Part B
Determine the work you did stretching the spring.
Express your answer to two significant figures and include the appropriate units.
Part C
What was the energy of the spring-Earth system when you let go (assume that zero potential energy corresponds to the equilibrium position of the block)?
Express your answer to two significant figures and include the appropriate units.
Part D
How far will the block rise after you release it?
Express your answer to two significant figures and include the appropriate units.
Answer:
a) x = 0.144 m
b) W = 0.15 J
c) E = 0.14 J
d) The block will rise 0.07m after it is released
Explanation:
a) The elastic force equals the gravitational force
F = kx = mg
x = 0.07, m = 0.1 kg, g = 9.81 m/s²
0.07k = 0.1 * 9.8
k = (0.1*9.8)/0.07
k = 14 N/m
When the force, F = 3N
F = kx
3 = 14x
x = 3/14
x = 0.214 m
The position of the block = 0.214 - 0.07 = 0.144m
B) Determine the work you did stretching the spring.
Energy stored in the spring when x = 0.07
E = 0.5 kx²
E = 0.5 * 14 * 0.07²
E = 0.0343 J
Energy stored in the spring when x = 0.214
E = 0.5 kx²
E = 0.5 * 14 * 0.214²
E = 0.32 J
Potential energy lost due to gravity = mgh
PE = 0.1 * 9.81 * 0.144
PE = 0.141 J
So to calculate the work done:
0.0343 + W = 0.32 - 0.141
W = 0.15 J
c) Energy in the spring
E = 0.32 - 0.0343 - 0.15
E = 0.1357 = 0.14 J
d)
[tex]1/2 *k *0.214^{2} = 1/2 kx^{2} + mg(0.214+x)\\0.32 = 7x^{2} + 0.1*9.8(0.214+x)\\[/tex]
Solving for x, x = 0.07 m
The block will rise 0.07m after it is released
A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352 kgm2 and a radius of 12.5 cm. A mass of 423 grams is attached to the free end of the string. With the string vertical and taut, the mass is gently released so it can descend under the influence of gravity. As the mass descends, the string unwinds and causes the pulley to rotate, but does not slip on the pulley. What is the speed (in m/s) of the mass after it has fallen through 1.25 m
Answer:
the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s
Explanation:
Given that :
Mass attached to the free end of the string, m = 423 g = 0.423 kg
Moment of inertia of pulley, I = 0.0352 kg m²
Radius of the pulley, r = 12.5 cm = 0.125 meters
Depth of fallen mass, h = 1.25 m
Acceleration due to gravity, g = 9.8 m/s²
Change in potential energy = mgh
= 0.423 × 9.8 × 1.25
=5.18175 J
From the question, we understand that the change in potential energy is used to raise and increase the kinetic energy of hanging mass and the rotational kinetic energy of pulley.
As such;
[tex]5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2[/tex]
where;
[tex]\omega[/tex] is the angular velocity of the pulley
v is the velocity of the mass after falling 1.25 m
where:
[tex]v = r \omega[/tex]
[tex]\omega = \frac{v}{r}[/tex]
replacing [tex]\omega = \frac{v}{r}[/tex] into above equation; we have:
[tex]5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I( \frac{v}{r})^2[/tex]
[tex]5.18175= (\frac{1}{2} m + \frac{1}{2}* (\frac{I}{r^2})v^2 \\ \\ v^2 = \frac{5.18175}{0.5*0.423+0.5*\frac{0.0352}{0.1252}} \\ \\ v^2 = 3.873047 \\ \\ v = 1.968 \ m/s[/tex]
Thus, the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s
the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s
SpeedWhat all information we have?
Mass attached to the free end of the string, m = 423 g = 0.423 kg
Moment of inertia of pulley, I = 0.0352 kg m²
Radius of the pulley, r = 12.5 cm = 0.125 meters
Depth of fallen mass, h = 1.25 m
Acceleration due to gravity, g = 9.8 m/s²
Change in potential energyChange in potential energy = mgh
Change in potential energy = 0.423 × 9.8 × 1.25
Change in potential energy =5.18175 J
As such:
5.18175 J= mv²+1w²
where;
w is the angular velocity of the pulley
v is the velocity of the mass after falling 1.25 m
v=rw
w=v/r
Replacing into above equation;
5.18175 J=mv² + 1/2 (w/r²) ²
5.18175 = (m + /* (4) v²
v² = 3.873047
v² =1.968 m/s
Thus, the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s.
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The resistance of physiological tissues is quite variable. The resistance of the internal tissues of humans, primarily composed of salty solutions, is quite low. Here the resistance between two internal points in the body is on the order of 100 ohms. Dry skin, however, can have a very high resistance, with values ranging from thousands to hundreds of thousands of ohms. However, if skin is wet, it is far more conductive, and so even contact with small voltages can create large, dangerous currents though a human body. (For example, although there is no specific minimum current that is lethal, currents generally exceeding a couple tenths of Amps may be deadly.)
Assuming that electrocution can be prevented if currents are kept below 0.1 A, and assuming the resistance of dry skin is 100,000 ohms, what is the maximum voltage with which a person could come into contact while avoiding electrocution? (Of course, all bets are off and things become far more dangerous if this person's skin is wet, which can reduce the resistance by more than a factor of 100.)
Given Information:
Current = I = 0.1 A
Resistance = R = 100 kΩ
Required Information:
Voltage = V = ?
Answer:
Voltage = V = 1000 V
Step-by-step explanation:
We know that electrocution depends upon the amount of current flowing through the body and the voltage across the body.
V = IR
Where I is the current flowing through the body and R is the resistance of body.
If electrocution can be avoided when the current is below 0.1 A then
V = 0.1*10×10³
V = 1000 Volts
Therefore, 1000 V is the maximum voltage with which a person could come into contact while avoiding electrocution, any voltage more than 1000 V may result in fatal electrocution.
Also note that human body has very low resistance when the body is wet therefore, above calculated value would not be applicable in such case.
The switch in the circuit has been in the left position for a long time. At t=0 it moves to the right position and stays there.a. Write the expression for the capacitor voltage v(t), fort≥0. b. Write the expression for the current through the 2.4kΩ resistor, i(t), fort≥0+.
Answer:
Pls refer to attached file
Explanation:
Magnetic fields and electric fields are identical in that they both-
Answer:
Similarities between magnetic fields and electric fields: Electric fields are produced by two kinds of charges, positive and negative. Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges). Like poles repel; unlike poles attract.
they attract each other
Explanation:
8. In the procedure for measuring the frequency of oscillation, we are instructed to pull downward on the hanging mass about 10 cm. If we performed the experiment a second time and deliberately pulled down on the hanging mass by 15 cm, would our period change? Justify your answer. (5 points)
Answer:
No
Explanation:
The frequency of oscillation of spring mass system is independent of its amplitude.
The frequency of oscillation will not change because it is not a function of amplitude.
What is the spring-mass system?In a real spring-mass system, the spring has a non-negligible mass m. Since not all of the spring's length moves at the same velocity v as the suspended mass M,
To determine the frequency of oscillation, the effective mass of the spring is defined as the mass that needs to be added to M to correctly predict the behaviour of the system.
For vertical springs, however, we need to remember that gravity stretches or compresses the spring beyond its natural length to the equilibrium position. After we find the displaced position
Thus the frequency of oscillation will not change because it is not a function of amplitude.
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Pease circle the statements incompatible with the Kelvin-Planck Statement. (A) No heat engine can have a thermal efficiency of 100%. (B) It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work without rejecting waste heat to a cool reservoir. (C) The impossibility of having a 100% efficient heat engine is always due to friction or other dissipative effects such as the system is perfectly designed or the material needed for the system design is not available. (D) Any device that violates the Kelvin-Planck statement also violates the Clausius statement, and vice versa.
Answer:
(C) The impossibility of having a 100% efficient heat engine is always due to friction or other dissipative effects such as the system is perfectly designed or the material needed for the system design is not available.
Explanation:
The above option was never stated in the law
Point charges 1 mC and −2 mC are located at (3, 2, −1) and (−1, −1, 4), respectively. Calculate the electric force on a 10 nC charge located at (0, 3, 1) and the electric field intensity at that point.
Answer:
See attached handwritten document for answer
Explanation:
Answer:
Explanation:
a) You can compute the force by using the expression:
[tex]F=k\frac{q_1q_2}{[(x-x_o)^2+(y-y_o)^2+(z-z_o)^2]^{\frac{1}{2}}}[/tex]
where k=8.98*10^9Nm^2/C^2 and q1, q2 are the charges. By replacing for the forces you obtain:
[tex]F_T=k[\frac{(1*10^{-3}C)(10*10^{-9}C)}{[(3-0)^2+(2-3)^2+(-1-1)^2]}}][(3-0)\hat{i}+(2-3)\hat{j}+(-1-1)\hat{k}]\\\\ \ \ \ \ +k[\frac{(-2*10^{-3}C)(10*10^{-9}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}}][(-1-0)\hat{i}+(-1-3)\hat{j}+(4-1)\hat{k}]\\\\F_T=6.41*10^{-3}N[3i-j-2k]-6.9*10^{-3}N[-1i-4j+3k]\\\\=0.026N\hat{i}-0.034N\hat{j}-0.033\hat{k}[/tex]
b)
[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}]+k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}]\\\\E=641428.5N/C+690769.23N/C=1332197.73N/C[/tex]
[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}][3i-2j-2k]+\\\\k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}][-1i-4j+3k]\\\\E=641428.5N/C[3i-2j-2k]-690769.23N/C[-1i-4j+3k]=2615054.7i+1480219.92j-4973626.23k\\\\|E|=\sqrt{(E_x)^2+(E_y)^2+(E_z)^2}=5810896.56N/C[/tex]
where we you have used that E=kq/r^2
The power liens that run through your neighborhood carry alternating currents that reverse direction 120 times per second. As current changes so does magnetic field. If you put a loop of wire up near the power line to extract power by tapping the magnetic field, sketch how you would orient the coil of wire next to a power line to develop the max emf in the coilo If magnetic flux through a loop changes, induced emf is produced
o If the area of the loop is parallel to the field, the flux through the loop is minimum
• So no emf is produced
o To get the most of the flux through the loop you place it closer to the power lines and in orientations so the plane of the loop also contains power lines
• Flux would be max and result in greater induced emf as field goes from max to zero to max and then in other direction
Answer:
the normal to the area of the loop parallel to the wire to induce the maximum electromotive force
Explanation:
Faraday's law is
ε = - dΦ / dt
where Ф magnetic flow
the flow is
Ф = B. dA = B dA cos θ
therefore, to obtain the maximum energy, the cosine function must be maximum, for this the direction of the fluctuating magnetic field and the normal direction to the area must be parallel.
The magnetic field in a cable through which current flows is circular, so the loop must be perpendicular to the wire, this is the normal to the area of the loop parallel to the wire to induce the maximum electromotive force
an outline is shown in the attachment
the correct answer is b
In 1958, Meselson and Stahl conducted an experiment to determine which of the three proposed methods of DNA replication was correct. Identify the three proposed models for DNA replication. Conservative Semiconservative Dispersive Answer Bank The Meselson and Stahl experiment starts with E. coli containing 15 N / 15 N labeled DNA grown in 14 N media. Which result did Meselson and Stahl observe by sedimentation equilibrium centrifugation to provide strong evidence for the semiconservative model of DNA replication? Both the first and second generation have both 15 N / 15 N DNA and 14 N / 14 N DNA. No hybrid 15 N / 14 N DNA was observed. The first generation has hybrid 15 N / 14 N DNA and the second generation has both hybrid 15 N / 14 N DNA and 14 N / 14 N DNA. No 15 N / 15 N DNA was observed. The first generation has hybrid 15 N / 14 N DNA and the second generation has hybrid 15 N / 14 N DNA. No 15 N / 15 N DNA nor 14 N / 14 N DNA was observed.
Answer:
Explanation:
The original has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed. In this experiment:
Nitrogen is a significant component of DNA. 14N is the most bounteous isotope of nitrogen, however, DNA with the heavier yet non-radioactive and 15N isotope is likewise practical.
E. coli was developed for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates at which its density equals to the salt arrangement. The DNA of the cells developed in 15N medium had a higher density than cells developed in typical 14N medium. After that, E. coli cells with just 15N in their DNA were transferred to a 14N medium.
DNA was removed and compared to pure 14N DNA and 15N DNA. Immediately after only one replication, the DNA was found to have an intermediate density. Since conservative replication would result in equal measures of DNA of the higher and lower densities yet no DNA of an intermediate density, conservative replication was eliminated. Moreso, this result was consistent with both semi-conservative and dispersive replication. Semi conservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with the two strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.
The DNA from cells after two replications had been completed and found to comprise of equal measures of DNA with two different densities, one corresponding to the intermediate density of DNA of cells developed for just a single division in 14N medium, the other corresponding to DNA from cells developed completely in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, yet at the same time higher than cells become distinctly in 14N DNA medium, as the first 15N DNA would have been part evenly among all DNA strands. The result was steady with the semi-conservative replication hypothesis. The semi conservative hypothesis calculates that each molecule after replication will contain one old and one new strand. The dispersive model suggests that each strand of each new molecule will possess a mixture of old and new DNA.
In an alcohol-in-glass thermometer, the alcohol column has length 12.66 cm at 0.0 ∘C and length 22.49 cm at 100.0 ∘C. Part A What is the temperature if the column has length 18.77 cm ? Express your answer using four significant figures. T = nothing ∘C Request Answer Part B What is the temperature if the column has length 14.23 cm ? Express your answer using four significant figures.
Answer:
[tex]62.1566632757\ ^{\circ}C[/tex]
[tex]15.9715157681\ ^{\circ}C[/tex]
Explanation:
[tex]\Delta T[/tex] = Change in termperature
[tex]\Delta L[/tex] = Change in length
We have the relation
[tex]\dfrac{\Delta L}{\Delta T}=\dfrac{22.49-12.66}{100-0}=\dfrac{18.77-12.66}{t-0}\\\Rightarrow t=\dfrac{18.77-12.66}{0.0983}\\\Rightarrow t=62.1566632757\ ^{\circ}C[/tex]
The temperature is [tex]62.1566632757\ ^{\circ}C[/tex]
[tex]\dfrac{\Delta L}{\Delta T}=\dfrac{22.49-12.66}{100-0}=\dfrac{14.23-12.66}{t-0}\\\Rightarrow t=\dfrac{14.23-12.66}{0.0983}\\\Rightarrow t=15.9715157681\ ^{\circ}C[/tex]
The temperature is [tex]15.9715157681\ ^{\circ}C[/tex]
(A)
According to the question,
The change in length will be:
= [tex]l_2-l_1[/tex]
= [tex]22.49-12.66[/tex]
= [tex]9.83 \ cm[/tex]
The change per degree will be:
= [tex]\frac{Change \ in \ length}{Temperature}[/tex]
= [tex]\frac{9.83}{100}[/tex]
= [tex]0.0983 \ cm/deg[/tex]
Now,
The change in length,
= [tex]18.77-12.66[/tex]
= [tex]6.11 \ cm[/tex]
hence,
The temperature,
= [tex]\frac{6.11}{0.0983}[/tex]
= [tex]62.2^{\circ} C[/tex]
(B)
The change in length,
= [tex]14.23-12.66[/tex]
= [tex]1.57 \ cm[/tex]
hence,
The temperature will be:
= [tex]\frac{1.57}{0.0983}[/tex]
= [tex]15.98^{\circ} C[/tex]
Thus the above answers are correct.
Learn more about temperature here:
https://brainly.com/question/847609
Consider the steel spring in the illustration.
(a) Find the pitch, solid height, and number of active turns.
(b) Find the spring rate. Assume the material is A227 HD steel.
(c) Find the force Fs required to close the spring solid.
(d) Find the shear stress in the spring due to the force Fs.
Answer:
Explanation:
find the answer below
German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity ( Δ v ) Δ x ≥ h 4 π m Δ v where h is Planck's constant and m is the mass of the object. The mass of an electron is 9.11 × 10 − 31 kg. What is the uncertainty in the position of an electron moving at 2.00 × 10 6 m/s with an uncertainty of Δ v = 0.01 × 10 6 m/s ?
According to the information given, the Heisenberg uncertainty principle would be given by the relationship
[tex]\Delta x \Delta v \geq \frac{h}{4\pi m}[/tex]
Here,
h = Planck's constant
[tex]\Delta v[/tex] = Uncertainty in velocity of object
[tex]\Delta x[/tex] = Uncertainty in position of object
m = Mass of object
Rearranging to find the position
[tex]\Delta x \geq \frac{h}{4\pi m\Delta v}[/tex]
Replacing with our values we have,
[tex]\Delta x \geq \frac{6.625*10^{-34}m^2\cdot kg/s}{4\pi (9.1*10^{-31}kg)(0.01*10^6m/s)}[/tex]
[tex]\Delta x \geq 5.79*10^{-9}m[/tex]
Therefore the uncertainty in position of electron is [tex]5.79*10^{-9}m[/tex]
A 1.0-kg standard cart A collides with a 0.10-kg cart B. The x component of the velocity of cart A is +0.70 m/s before the collision and +0.50 m/s after the collision. Cart B is initially traveling toward cart A at 0.40 m/s , and after the collision the x component of its velocity is +1.6 m/s .
Part A: What is the x component of the change in the momentum of A?
Part B: What is the x component of the change in the momentum of B?
Part C: What is the sum of these two x components of the changes in momentum?
The x component of the change in momentum of cart A is -0.20 kg·m/s, while for cart B, it is +0.20 kg·m/s. The sum of these changes in momentum is 0 kg·m/s, showing conservation of momentum in the system.
Explanation:The student has provided information about a collision between two carts, which involves concepts of momentum and conservation of momentum from mechanics in physics.
Part A: Change in Momentum of Cart A
The x component of the change in momentum of cart A can be calculated using the formula Δp = m(v_final - v_initial), where m is mass and v is velocity. Here, Δp = 1.0 kg (0.50 m/s - 0.70 m/s) = -0.20 kg·m/s. The negative sign indicates that the momentum decreased.
Part B: Change in Momentum of Cart B
Similarly, for cart B, Δp = 0.10 kg (1.6 m/s - (-0.40 m/s)) = 0.20 kg·m/s. The positive sign signifies an increase in momentum.
Part C: Sum of Changes in Momentum
The total change in momentum for both carts is the sum of their individual momentum changes, which is -0.20 kg·m/s + 0.20 kg·m/s = 0 kg·m/s. This means that the total momentum of the system is conserved.
During a track and field event, a metal javelin (not infinitesimally thin) is thrown due east and parallel to the ground. At the field where the event took place, the magnetic field is running straight up and perpendicular to the ground. In which direction will the current flow in the metal object?
Answer:
There is no induced current
Explanation:
[Find the attachment]
A standard door into a house rotates about a vertical axis through one side, as defined by the door's hinges. A uniform magnetic field is parallel to the ground and perpendicular to the axis. Through what angle must the door rotate so that the magnetic flux that passes through it decreases from its maximum value to 1/4 of its maximum value?
Answer:
75.5degrees
Explanation:
Magnitude of magnetic flux= BA
If rotated through an angle= BAcos theta
= (0.25)BA=BAcos theta
= costheta= 0.25
= theta= cos^-1 0.25
=75.5degrees
Double the resistance of the resistor by changing it from 10 Ω to 20 Ω. What happens to the current flowing through the circuit?
Answer:
2 amps
Explanation:
you just divide
A soft-drink manufacturer purchases aluminum cans from an outside vendor. A random sample of 70 cans is selected from a large shipment, and each is tested for strength by applying an increasing load to the side of the can until it punctures. Of the 70 cans, 53 meet the specification for puncture resistance.
Answer:
lol..WHAT IS THE QUESTION?!
Explanation:
Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 260000 kg and a velocity of 0.32 m/s in the horizontal direction, and the second having a mass of 52500 kg and a velocity of -0.14 m/s in the horizontal direction. What is their final velocity?
Answer:
0.243 m/s
Explanation:
From law of conservation of motion,
mu+m'u' = V(m+m')................. Equation 1
Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.
make V the subject of the equation
V = (mu+m'u')/(m+m')................. Equation 2
Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s
Substitute into equation 2
V = (260000×0.32+52500×(-0.14))/(260000+52500)
V = (83200-7350)/312500
V = 75850/312500
V = 0.243 m/s
An electron follows a helical path in a uniform magnetic field given by:B =(20i^−50j^−30k^)mTAt time t = 0, the electron's velocity is given by:⃗v=(40i^−30j^+50k^)m/sa. What is the angle ϕ between v and B. The electron's velocity changes with time. Do b. its speed c. the angled. What is the radius of the helical path?
Answer:
a) 1.38°
b) 7.53*10^11 m/s/s
c) 6.52*10^-9m
Explanation:
a) to find the angle you can use the dot product between two vectors:
[tex]\vec{v}\cdot\vec{B}=vBcos\theta\\\\\theta=cos^{1}(\frac{\vec{v}\cdot\vec{B}}{vB})[/tex]
v: velocity of the electron
B: magnetic field
By calculating the norm of the vectors and the dot product and by replacing you obtain:
[tex]B=\sqrt{(20)^2+(50)^2+(30)^2}=61.64mT\\\\v=\sqrt{(40)^2+(30)^2+(50)^2}=70.71m/s\\\\\vec{v}\cdot\vec{B}=[(20)(40)+(50)(30)-(30)(50)]mTm/s=800mTm/s\\\\\theta=cos^{-1}(\frac{800*10^{-3}Tm/s}{(70.71m/s)(61.64*10^{-3}T)})=cos^{-1}(0.183)=1.38\°[/tex]
the angle between v and B vectors is 1.38°
b) the change in the speed of the electron can be calculated by the change in the momentum in the following way:
[tex]\frac{dp}{dt}=F_e=qvBsin\theta\\\\\frac{dp}{dt}=(1.6*10^{-19}C)(70.71m/s)(61.64*10^{-3}T)(sin(1.38\°))=6.85*10^{-19}N[/tex]
due to the mass of the electron is a constant you have:
[tex]\frac{dp}{dt}=\frac{mdv}{dt}=6.85*10^{-19}N\\\\\frac{dv}{dt}=\frac{6.85*10^{-19}N}{9.1*10^{-31}kg}=7.53*10^{11}(m/s)/s[/tex]
the change in the speed is 7.53*10^{11}m/s/s
c) the radius of the helical path is given by:
[tex]r=\frac{m_ev}{qB}=\frac{(9.1*10^{-31}kg)(70.71m/s)}{(1.6*10^{-19}C)(61.64*10^{-3}T)}=6.52*10^{-9}m[/tex]
the radius is 6.52*10^{-9}m
Rotational dynamics about a fixed axis: A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center. Starting with an angular velocity of 2.40 rev/s, it stops after turning through 18.2 rev with uniform acceleration. The net torque acting on this sphere as it is slowing down is closest to:A) 0.149 N m. B) 0.0620N m. C) 0.00593 N m. D) 0.0372 N m. E) 0.0466 N·m
Answer:
D) 0.0372 N m
Explanation:
r = 45/2 cm = 22.5 cm = 0.225 m
As 1 revolution = 2π rad we can convert to radian unit
2.4 rev/s = 2.4 * 2π = 15.1 rad/s
18.2 rev = 18.2 * 2π = 114.35 rad
We can calculate the angular (de)acceleration using the following equation of motion
[tex]-\omega^2 = 2\alpha \theta [/tex]
[tex]- 15.1^2 = 2*\alpha * 114.35[/tex]
[tex]\alpha = \frac{-15.1^2}{2*114.35} = -0.994 rad/s^2[/tex]
The moment of inertia of the solid uniform sphere is
[tex]2mr^2/5 = 2*1.85*0.225^2/5 = 0.0375 kgm^2[/tex]
The net torque acting on this according to Newton's 2nd law is
[tex]T = I\alpha = 0.0375 * 0.994 = 0.0372 Nm[/tex]
Answer:
(D) The net torque acting on this sphere as it is slowing down is closest to 0.0372 N.m
Explanation:
Given;
mass of the solid sphere, m = 1.85 kg
radius of the sphere, r = ¹/₂ of diameter = 22.5 cm
initial angular velocity, ω = 2.40 rev/s = 15.08 rad/s
angular revolution, θ = 18.2 rev = 114.37 rad
Torque on the sphere, τ = Iα
Where;
I is moment of inertia
α is angular acceleration
Angular acceleration is calculated as;
[tex]\omega_f^2 = \omega_i^2 +2 \alpha \theta\\\\0 = 15.08^2 + (2*114.37)\alpha\\\\\alpha = \frac{-15.08^2}{(2*114.37)} = -0.994 \ rad/s^2\\\\\alpha = 0.994 \ rad/s^2 \ (in \ opposite \ direction)[/tex]
moment of inertia of solid sphere, I = ²/₅mr²
= ²/₅(1.85)(0.225)²
= 0.03746 kg.m²
Finally, the net torque on the sphere is calculated as;
τ = Iα
τ = 0.03746 x 0.994
τ = 0.0372 N.m
Therefore, the net torque acting on this sphere as it is slowing down is closest to 0.0372 N.m