Final answer:
To find the transformed vector representations in a rotated coordinate system, the angle of vector A is adjusted by the rotation angle, and the components are calculated using trigonometric functions. Vector B's components in the rotated system are found using a rotation matrix.
Explanation:
The provided question pertains to transforming the representation of vectors in a rotated coordinate system in the subject of physics. The coordinate system is rotated counterclockwise, and the goal is to find the new representations of vectors A and B in unit-vector notation on the x'y' system. Given the initial magnitude and direction angle of vector A and the Cartesian components of vector B on the xy coordinate system, we can calculate their components on the rotated x'y' coordinate system.
The original vector A has a magnitude of 14.4 m and an angle of 51.6° from the positive x-axis. After rotation by 20°, the new angle becomes 51.6° - 20.0° = 31.6° from the new x'-axis. Using the formulas Ax' = A cos θ' and Ay' = A sin θ', where θ' is the new angle, we can find the rotated components of A.
The vector B is already given in Cartesian coordinates as ( 14.3 m )i + (8.52 m )j. To find the components of B in the rotated system, we use a rotation matrix, giving us new components Bx' and By'.
In conclusion, to find the transformed vectors in the rotated system, we apply the rotation to both the magnitude and angle of A, and use a rotation matrix for the components of B.
A perfectly spherical iron ball bearing weighs 21.91 grams. Derive the diameter of the ball bearing assuming an iron atom has an effective radius of 0.124nm and iron is BCC at room temperature. The answer should be in cm with 2 decimals of accuracy.
Final answer:
The diameter of the iron ball bearing, which weighs 21.91 grams and is composed of iron atoms organized in a BCC structure, is roughly 1.62 cm.
Explanation:
To derive the diameter of a spherical iron ball bearing weighing 21.91 grams, given that iron atoms have an effective radius of 0.124 nm and are arranged in a Body-Centered Cubic (BCC) structure at room temperature, we need to calculate the volume of the iron ball and then find the diameter using the volume of a sphere formula. First, we will use the density of iron (7.9 g/cm³) to find the volume of the ball bearing:
V = mass / density = 21.91 g / 7.9 g/cm³ = 2.77342 cm³
Next, we use the volume of a sphere formula V = (4/3)πr³, where V is the volume and r is the radius, to find the diameter (d = 2r):
r³ = V / ((4/3)π) = 2.77342 cm³ / ((4/3)π) ≈ 0.52733 cm³
r ≈ 0.8092 cm
d = 2 * r ≈ 2 * 0.8092 cm ≈ 1.6184 cm
Therefore, the estimated diameter of the iron ball bearing is approximately 1.62 cm.
A river flows due east with a speed of 3.00 m/s relative to earth. The river is 80.0 m wide. A woman starts at the southern bank and steers a motorboat across the river; her velocity relative to the water is 5.00 m/s due north. How far east of her starting point will she reach the opposite bank?
Answer:
48 m
Explanation:
As she travels at the rate of 5m/s due north, the amount of time it would take for her to cross the 80m wide river would be
t = 80 / 5 = 16 seconds
This is also the time it takes for the river to push her to the east side at the rate of 3m/s. So after 16 seconds, she would reach the opposite point at a horizontal distance from her starting of
s = 16*3 = 48 m
The electron-volt is a unit of A. charge. B. electric potential. C. electric field. D. electric force. E. energy.
Answer:
E. Energy
Explanation:
Electron volt is a unit of energy commonly used in various branches of physics.
It is defined as the energy gained by an electron when the electrical potential of the electron increases by one volt.
The electron volt = 1.602 × 10^−12 erg, or 1.602 × 10^−19 joule
Answer:
The electron-volt is a unit of A. charge. B. electric potential. C. electric field. D. electric force. E. energy.
The answer is option E (energy)
Explanation:
There are many forms of energy which are divided into Potential Energy, Kinetic energy and the major energy sources are nonrenewable and renewable sources. Energy makes change; it does things for us. Since energy is a fundamental physical quantity, it is a property of matter that can be converted into work, heat or radiation. Energy can be converted from one form to another, but it cannot be created, nor can it be destroyed.
Energy conversion is essential for energy utilization. Energy can be measured in many different units which includes joules, calories, electron-volts, kilowatt-hours, and so many more.
The electron-volt is a unit used to measure the energy of subatomic particles. The electron-volt, symbol eV, can be defined as the amount of energy gained by the charge of a single electron (a charged particle carrying unit electronic charge) moved across an electric potential difference of one volt. One electron-volt, eV is equal to 1.602176634×10−19 J. Where J is in joules.
The storage coefficient of a confined aquifer is 6.8x10-4 determined by a pumping test. The thickness of the aquifer is 50 m and the porosity is 25%. Determine the fractions of the storage attributable to the expansibility of water and compressibility of the aquifer skeleton in terms of percentages of the storage coefficient of the aquifer.
Answer
given,
storage coefficient, S = 6.8 x 10⁻⁴
thickness of aquifer, t = 50 m
porosity of the aquifer, n = 25 % = 0.25
Density of the water, γ = 9810 N/m³
Compressibilty of water,β = 4.673 x 10⁻¹⁰ m²/N
We know,
S = γ t(nβ + α)
where, α is the compressibility of the aquifer
6.8 x 10⁻⁴ =9810 x 50 x (0.25 x 4.673 x 10⁻¹⁰+ α)
α = 1.269 x 10⁻⁹ m²/N
Expansability of water
= n t β γ
= 0.25 x 50 x 4.673 x 10⁻¹⁰ x 9810
= 5.73 x 10⁻⁵
During an auto accident, the vehicle’s air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, air bags produce a maximum acceleration of 60 g that lasts for only 36 ms (or less). How far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60 g?
Answer:
d = 0.38 m
Explanation:
As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:
vf = v₀ -a*t
If vf = 0, we can solve for v₀:
v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s
With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.
Just for simplicity, we can use the following equation:
[tex]vf^{2} -vo^{2} = 2*a*d[/tex]
where vf=0, v₀ =21.2 m/s and a= -588 m/s².
Solving for d:
[tex]d = \frac{-vo^{2}}{2*a} = \frac{(21.2m/s)^{2} }{2*588 m/s2} =0.38 m[/tex]
⇒ d = 0.38 m
Answer:
A person travels 39 cm in coming to a complete stop in 36 ms at a constant acceleration of 60 g.
Explanation:
Hi there!
The equation of position of an object moving in a straight line at constant acceleration is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position at time t.
x0 = initial position.
v0 = initial velocity.
a = acceleration.
t = time.
So, let's see how much distance the person moves inside the car. Let's imagine that the person is initially at rest and suddenly is accelerated at 60 g (60 · 10 m/s² = 600 m/s²). In this case, x0 and v0 = 0 and the traveled distance will be:
x = 1/2 · 600 m/s² · (0.036)²
x = 0.39 m or 39 cm
Here, we have calculated the distance traveled by a person accelerated at 60 g from rest in 36 ms. Notice that the distance is the same if we calculate the traveled distance of a person that is brought to rest in 36 ms with an acceleration of 60 g.
A person travels 39 cm in coming to a complete stop in 36 ms at a constant acceleration of 60 g.
Suppose the rocket is coming in for a vertical landing at the surface of the earth. The captain adjusts the engine thrust so that rocket slows down at the rate of 2.05 m/s2 . A 6.50-kg instrument is hanging by a vertical wire inside a space ship.Find the force that the wire exerts on the instrument.
Answer:
R= 78.32 N
Explanation:
Given that
Acceleration ,a= 2.05 m/s²
Mass , m = 6.5 kg
The force due to acceleration
F= mass x Acceleration
F= ma
F= 6.5 x 2.05 N
F= 13.32 N
The force due to weight
F' = m g
F' = 6.5 x 10 N ( take g= 10 m/s²)
F'= 65 N
Therefore the net total force will be summation of force due to weight and force due to acceleration
R= F + F'
R= 65 + 13.32 N
R= 78.32 N
The force that the wire exerts on the instrument is R= 78.32 N
Calculation of force:
Given that
Acceleration ,a= 2.05 m/s²Mass , m = 6.5 kgThe force due to accelerationSince The force due to weight
[tex]F= ma\\\\F= 6.5 \times 2.05 N\\\\F= 13.32 N[/tex]
Now
[tex]F' = m g\\\\F' = 6.5 \times 10 N ( take\ g= 10 m/s^2)[/tex]
F'= 65 N
Now
R= F + F'
R= 65 + 13.32 N
R= 78.32 N
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magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet, she throws a rock straight upward with an initial velocity of 20 m/s. If the astronaut were instead on Earth, and threw a ball in the same way while standing on a 50.0 m high cliff, what would be the time difference (in s) for the rock to hit the ground below the cliff in each case
Using kinematics equations, we can calculate the time for a rock to fall from the cliff on Earth. Without the value of acceleration due to gravity on the extrasolar planet, we can't quantify the time difference. A difference would exist if the gravitational accelerations of the two planets differ.
Explanation:This question pertains to kinematics and physical laws regarding free fall. On earth, when the astronaut throws a rock straight up with a certain velocity, initially, gravity will slow down the rock and finally make it fall back toward the ground. This action is governed by the equation of motion: h = vit + 1/2gt² where h is the height, vi the initial velocity, t the time and g the acceleration due to gravity on earth (-9.81 m/s²).
In the extrasolar planet scenario, considering it is not mentioned, we could assume that the acceleration due to gravity might be similar to earth's. However, without specific details about the gravity, we can still calculate the time it takes for the rock to land from the 50m high cliff using the motion equation for a free-falling body. We can conclude that if the gravity in the two settings is different, then there would be a time difference. We can’t quantify the time difference without knowing the value of the gravity on the extrasolar planet.
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After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity, using a linear fit to the appropriate data; (d) the % elongation; (e) the % reduction in area;
Answer:
Part a: The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.
Part b: The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa
Part c: The value of Young's modulus at given point is 172 GPa.
Part d: The percentage elongation is 18.55%.
Part e: The percentage reduction in area is 15.81%
Explanation:
From the complete question the data is provided for various Loads in ductile testing machine for a sample of d0=20 mm and l0=40mm. The plot is drawn between stress and strain whose values are calculated using following formulae. The corresponding values are attached with the solution.
The engineering-stress is given as
[tex]\sigma=\frac{F}{A}\\\sigma=\frac{F}{\pi \frac{d_0^2}{4}}\\\sigma=\frac{F}{\pi \frac{(20 \times 10^-3)^2}{4}}\\\sigma=\frac{F}{3.14 \times 10^{-4}}[/tex]
Here F are different values of the load
Now Strain is given as
[tex]\epsilon=\frac{l-l_0}{l_0}\\\epsilon=\frac{\Delta l}{40}\\[/tex]
So the curve is plotted and is attached.
Part aThe value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.
Part bThe value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa
Part cYoung's Modulus is given as
[tex]E=\frac{\sigma}{\epsilon}\\E=\frac{238 /times 10^6}{0.00138}\\E=172,000 MPa\\E=172 GPa[/tex]
The value of Young's modulus at given point is 172 GPa.
Part dThe percentage elongation is given as
[tex]Elongation=\frac{l_f-l_0}{l_0} \times 100\\Elongation=\frac{47.42-40}{40}\times 100\\Elongation=18.55 \%\\[/tex]
So the percentage elongation is 18.55%
Part eThe reduction in area is given as
[tex]Reduction=\frac{A_0-A_n}{A_0} \times 100\\Reduction=\frac{\pi \frac{d_0^2}{4}-\pi \frac{d_n^2}{4}}{\pi \frac{d_0^2}{4}}\times 100\\Reduction=\frac{{d_0^2}-{d_n^2}}{{d_0^2}} \times 100\\Reduction=\frac{{20^2}-{18.35^2}}{{20^2}} \times 100\\Reduction=15.81\%[/tex]
So the reduction in area is 15.81%
To determine the engineering properties, knowledge of additional values is needed. These properties include 0.2% offset yield strength, tensile strength, modulus of elasticity, % elongation, and % reduction in area, derived through respective formulas. The engineering stress-strain curve can be plotted with these details.
Explanation:To calculate the engineering properties asked in your question some additional values such as the original length and diameter, the load at yield point, the maximum load sustained, and the length and diameter after fracture are required. However, the basic formulas for the calculations are as follows
0.2% offset yield strength = (Load at yield point/Area) * 0.002 Tensile strength = Maximum load sustained / Original cross-sectional area Modulus of elasticity = Stress/Strain = (Load/Area)/(deformation/Original Length) % Elongation = ((final length - original length)/original length) * 100 % Reduction in area = ((original area - final area)/original area) * 100
Please note that the engineering stress-strain curve should be plotted after obtaining these values with stress on the Y-axis and strain on the X-axis. The curve typically starts from the origin, goes linearly upwards till the yield point (Proportional limit), followed by a non-linear portion (elastic limit), and reaches maximum at tensile strength, after which it falls down to the fracture point.
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Astronomers analyze starlight to determine a star’s (a) temperature; (b) composition; (c) motion; (d) all of the above.
One of the characteristics of the luminous gas clouds is that they do not have direct affectation by some type of external electric or magnetic fields.
In addition, we must bear in mind that color is a variable that is depending on the gas in the mixture. Therefore its relationship with spectroscopy allows us to deduce that scientists take advantage of the wavelength spectrum to know the type of composition of one of the clouds. The speed of a cloud is measured by determining the Doppler shift of its spectral lines. From wine's law, wavelength of light emitting from the object depends on temperature of object
Therefore the correct option is D
Astronomers analyze starlight to obtain various pieces of information about stars, including their temperature, composition, and motion. The correct answer is (d) all of the above.
(a) Temperature: By examining the spectrum of starlight, astronomers can analyze the distribution of wavelengths or colors present in the light. The temperature of a star affects the intensity and distribution of light at different wavelengths.
(b) Composition: The spectrum of starlight also provides information about the chemical composition of stars. Different elements and molecules in a star's atmosphere absorb or emit light at specific wavelengths, creating characteristic absorption or emission lines in the spectrum.
(c) Motion: Through the analysis of starlight, astronomers can also determine the motion of stars. By studying the Doppler effect on spectral lines, which causes a shift in wavelength due to the motion of a star toward or away from Earth, astronomers can measure a star's radial velocity.
Therefore, by analyzing starlight, astronomers can gather information about a star's temperature, composition, and motion, making option (d) all of the above the correct choice
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A 55 kgkg meteorite buries itself 5.5 mm into soft mud. The force between the meteorite and the mud is given by F(x)F(x) = (630 N/m3N/m3 )x3x3, where xx is the depth in the mud. Find the work done on the meteorite by the mud.
Answer:
W = 1.44 10⁻⁷ J
Explanation:
The expression for the job is
W = ∫ F. dx
Where the point is the scalar product in this case the direction of the meteor and the depth is parallel, whereby the scalar product is reduced to the ordinary product
W = 630 ∫ x³ dx
W = 630 x⁴ / 4
Let's evaluate between the lower limit x = 0, w = 0 to the upper limite the point at x = 5.5 10⁻³ m
W = 157.5 ((5.5 10⁻³)⁴ -0)
W = 1.44 10⁻⁷ J
sinusoidal wave is described by the wave function y 5 0.25 sin (0.30x 2 40t) where x and y are in meters and t is in seconds. Determine for this wave (a) the amplitude, (b) the angular frequency, (c) the angular wave number, (d) the wavelength
For the sinusoidal wave described by y = 0.25 sin (0.30x - 40t), the amplitude is 0.25 meters, angular frequency is 40 rad/s, the angular wave number is 0.30 rad/m and the wavelength is approximately 20.94 meters.
Explanation:The given function for the sinusoidal wave is y = 0.25 sin (0.30x - 40t). We can extract the details of this wave from this function:
Amplitude (A): This is the maximum height of the wave, represented by the coefficient before the sin function. Here, A = 0.25 meters. Angular frequency (w): This value is associated with the 't' term in the function and represents how the wave frequency changes with time. So, w = 40 rad/s. Angular wave number (K): This is the coefficient of the 'x' term, providing a measure of the wave's spatial frequency. In this case, k = 0.30 rad/m.Wavelength (λ): It is connected to the angular wave number by the relation λ = 2π/k. Substituting the value of k, we get λ = 2π/0.30 = approximately 20.94 meters. Learn more about Sinusoidal Wave here:https://brainly.com/question/33443431
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If a spectral line from a distant star is measured to have a wavelength of 497.15 nm, but is normally at 497.22 nm how fast (speed, not velocity) with respect to the Earth is the star moving in m/s
Answer:
v = -4.22 x 10⁻⁴ m/s
Explanation:
given,
measured wavelength = 497.15 nm
Normally wavelength = 497.22 nm
Change in wavelength
Δ λ = 497.15 - 497.22
Δ λ = -0.07 nm
using Doppler's equation
[tex]\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}[/tex]
v is the speed of the star
c is the speed of light
[tex]\dfrac{-0.07\ nm}{497.22\ nm}=\dfrac{v}{3\times 10^8}[/tex]
v = -4.22 x 10⁻⁴ m/s
Speed of the star moving is equal to v = -4.22 x 10⁻⁴ m/s
The speed of the star with respect to the Earth is -4.22× 10⁻⁴ m/s m/s. The negative sign indicates the star is moving away.
Given:
Observed wavelength (λ) = 497.15 nm
Rest wavelength (λ₀) = 497.22 nm
To calculate the speed of a star with respect to Earth. Doppler effect technique can be used. The Doppler effect gives the relation between wavelength, speed, and speed of light.
The formula for the Doppler shift is given as:
Δλ / λ₀ = v / c
Δλ = λ - λ₀
Δλ = (497.15 x 10⁻⁹ m) - (497.22 x 10⁻⁹ m)
Δλ = -0.07 x 10⁻⁹ m
The speed of the star is evaluated as:
v = (Δλ / λ₀) x c
v = (-0.07 x 10⁻⁹ m / 497.22 x 10⁻⁹ m) x 299,792,458 m/s
v = -4.22× 10⁻⁴ m/s
Hence, the speed of the star with respect to the Earth is -4.22× 10⁻⁴ m/s m/s. The negative sign indicates the star is moving away.
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If we were to construct an accurate scale model of the solar system on a football field with the Sun at one end and Neptune at the other, the planet closest to the center of the field would be (a) Earth; (b) Jupiter; (c) Saturn; (d) Uranus.
Answer:
a) Earth
Explanation:
When constructing an accurate scale model of the solar system on a football field with the Sun at one end and Neptune at the other, the planet closest to the center of the field would be Earth.
From the given options the earth is the closest planet to the center of the field. The center of the field can be considered as the sun. In reality we have mercury as the closest planet in our solar system.
Position update: Initially the bottom of the block is at y = 0.12 m. Approximating the average velocity in the first time interval by the final velocity, what will be the new position of the bottom of the block at time t = 0.07 seconds? y = 1. m
Answer:
The new position is 0.1865 m
Explanation:
As the context of the data is not available, thus following data is utilized from the question as attached above
x_relax=0.32 m
x_stiff=0.13 m
spring stiffness k=9 N/m
mass of block =0.073 kg
t=0.07 s
Velocity of the block is to be estimated thus
Force due to compression in spring is given as
F_s=k Δx
F_s=9(0.32-0.13)
F_s=1.71 N
Force on the block is given as
F_m=mg
F_m=0.073 x 9.8
F_m=0.71 N
Net Force
F=F_s-F_m
F=1.71-0.71 N
F=1 N
As Ft=Δp
So
Δp=1x0.07=0.07 kgm/s
Δp=p_final-p_initial
0.07=p_final-0
p_final=0.07 kgm/s
p_final=m*v_f
v_f=(p_final)/(m)
v_f=0.07/0.073
v_f=0.95 m/s
So now the velocity of the block is 0.95 m/s
time is 0.07 s
y_new=y_initial+y_travel
y_new=0.12+(0.95 x 0.07)
y_new=0.12+0.065
y_new=0.1865 m
So the new position is 0.1865 m
Final answer:
The new position of the bottom of the block at time t = 0.07 seconds is 1 m.
Explanation:
To find the new position of the bottom of the block at time t = 0.07 seconds, we can use the concept of average velocity. The average velocity is given by the change in position divided by the change in time. In this case, if we approximate the average velocity in the first time interval by the final velocity, we can say that the change in position is equal to the average velocity multiplied by the change in time. The new position can then be calculated by adding this change in position to the initial position.
Given that the initial position of the bottom of the block is at y = 0.12 m and the final velocity is approximated to be y = 1 m, we can calculate the change in position as:
Change in position = (Final velocity - Average velocity) * Change in time = (1 m - 0.12 m) * (0.07 s - 0 s) = 0.88 m
Therefore, the new position of the bottom of the block at time t = 0.07 seconds is y = 0.12 m + 0.88 m = 1 m.
A frictionless piston-cylinder device contains air at 300 K and 1 bar and is heated until its volume doubles and the temperature reaches 600 K. Answer the following: A. You are interested in studying the air in the piston-cylinder device as a closed system. Draw a schematic of your device and the boundary that defines your system. Assume the cylinder is in horizontal position. B. Determine the final pressure of the air at the end of the process, in bar. Hint: use the ideal gas law equation. If you need the value for the universal gas constant ???????? ????in your textbook or in a chemistry book (or on-line). Just make sure your units are dimensionally correct. C. On a different occasion (different temperature and pressure), you find the piston-cylinder device contains 0.5 kmol of H2O occupying a volume of 0.009 m3. Determine the weight of the H2O in N. Hint: Start with the relationship between number of moles, molecular mass and mass. D. Determine the specific volume of the H2O (from Part C) in m3/kg.
Answer:
Part a: The schematic diagram is attached.
Part b: The pressure at the end is 1 bar.
Part c: Weight of 0.5kmol of water is 88.2 N.
Part d: The specific volume is 0.001 m^3/kg
Explanation:
Part aThe schematic is given in the diagram attached.
Part bPressure is given using the ideal gas equation as
Here
P_1=1 barP_2=? to be calculatedV_2=2V_1T_1=300KT_2=600K[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\frac{1\times V_1}{300}=\frac{P_2\times 2V_1}{600}\\P_2=\frac{600}{600}\\P_2=1 bar[/tex]
So the pressure at the end is 1 bar.
Part cMass of 0.5kmol is given as follows
[tex]Mass=n_{moles} \times Molar \, Mass\\Mass=0.5 \times 10^3 \times 18 \times 10^{-3}\\Mass=9.0 kg[/tex]
Weight is given as
[tex]W=mxg\\W=9 \times 9.8\\W=88.2 \, N[/tex]
So weight of 0.5kmol of water is 88.2 N.
Part dSpecific volume is given as
[tex]v=\frac{Volume}{Mass}\\v=\frac{0.009}{9}\\v=0.001 m^3/kg[/tex]
So the specific volume is 0.001 m^3/kg
A. Draw a schematic of the system with a boundary around the piston-cylinder device. B. The final pressure can be determined using the ideal gas law equation. C. The weight of H2O can be calculated using the relationship between moles, molecular mass, and mass. D. The specific volume of H2O can be determined by dividing the volume by the mass.
Explanation:A. To study the air in the piston-cylinder device as a closed system, we consider the device itself as the system and draw a boundary around it, including the air inside and excluding the surroundings. The schematic would show a cylindrical container with a piston separating the initial and final air volumes.
B. To determine the final pressure of the air, we can use the ideal gas law. The equation is PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature. Since the volume doubles and the temperature increases to 600 K, we can set up the equation (1 bar)(2V) = n(R)(600 K), and solve for the final pressure.
C. To determine the weight of H2O in the piston-cylinder device, we use the relationship between number of moles, molecular mass, and mass. The weight of H2O in N can be calculated as (0.5 kmol)(molecular mass of H2O)(Acceleration due to gravity).
D. The specific volume of H2O can be determined by dividing the volume (0.009 m3) by the mass of H2O (which we can calculate from the number of moles and molecular mass).
A step-up transformer has 22 turns on the primary coil and 800 turns on the secondary coil. If this transformer is to produce an output of 5300 V with a 16- mA current, what input current and voltage are needed?
Final answer:
For a step-up transformer with 22 primary turns and 800 secondary turns, to produce an output of 5300 V at 16 mA, the required input voltage is 145.75 volts and the input current needed is 0.5818 amperes.
Explanation:
Calculating Input Current and Voltage for a Step-Up Transformer
The student's question involves a step-up transformer with a known number of turns in the primary and secondary coils, a given secondary voltage, and a secondary current. To find the required input current and voltage, we can use the transformer equations that relate the primary and secondary sides of the transformer:
Primary voltage (VP) / Secondary voltage (VS) = Number of turns in the primary coil (NP) / Number of turns in the secondary coil (NS)
Primary current (IP) * Number of turns in the primary coil (NP) = Secondary current (IS) * Number of turns in the secondary coil (NS)
We're given:
NP = 22 turns
NS = 800 turns
IS = 16 mA = 0.016 A
VS = 5300 V
To find the input voltage VP:
VP = (NP / NS) * VS = (22 / 800) * 5300 V = 145.75 V
To find the input current IP:
IP = (NS / NP) * IS = (800 / 22) * 0.016 A = 0.5818 A
Therefore, the required input voltage is 145.75 volts, and the required input current is 0.5818 amperes.
A steam catapult launches a jet aircraft from the aircraft carrier john C. Stennis, giving it a peed of 175 mi/h in 2.50 . (a) Find the average acceleration of the plane. (b) Assuming the acceleralion is conslant, find lhe dislance the plane moves.
Answer
given,
Speed of the Aircraft,v = 175 mi/h
1 mi/h = 0.44704 m/s
175 mi/h = 78.232 m/s
time, t = 2.5 s
a) average acceleration = ?
[tex]a= \dfrac{v - u}{t}[/tex]
[tex]a= \dfrac{78.232 - 0}{2.5}[/tex]
a = 31.29 m/s²
b) Distance traveled by the Pane
using equation of motion
v² = u² + 2 a s
78.232² = 0² + 2 x 31.29 x s
s = 97.79 m
Distance moved by the plane is equal to 97.79 m
Consider two concentric conducting spheres. The outer sphere is hollow and initially has a charge Q1 = -10Q deposited on it. The inner sphere is solid and has a charge Q 2 = +1Q on it. 1)How much charge is on the outer surface
Answer:
Q_out,shell = 9Q
Explanation:
Given:
- Q_in,shell = -10 Q
- Q_sphere = +1Q
Find:
How much charge is on the outer surface?
Solution:
The electric field in the material of both the sphere and shell must be zero. The only way for this to occur is if the charge inside the
inner surface of the shell is such that its charge plus the solid's charge is zero. The rest of the excess charge from the shell moves to the outside of the shell.
Hence,
Q_out,shell + Q_in,shell + Q_sphere = 0
Q_out,shell -10 Q + 1 Q = 0
Q_out,shell = 9Q
Final answer:
The charge on the outer surface of the hollow conducting sphere, which initially had a charge of -10Q and contains an inner sphere with a charge of +1Q, would be -9Q, as determined by Gauss' Law and the conservation of charge.
Explanation:
The student is asking about the charge distribution on concentric conducting spheres when one sphere is placed inside another and they each have different charges. According to Gauss' Law, when a charge is placed inside a conducting shell, it induces an equal and opposite charge on the inner surface of the shell to maintain an electric field of zero inside the material of the conductor. In the given scenario, the inner solid sphere has a charge of +1Q and the outer hollow sphere has a charge of -10Q.
By Gauss' Law, since the electric field inside a conductor must be zero, we know that the inner surface of the hollow outer sphere must have a charge of -1Q to cancel out the electric field from the +1Q charge of the inner solid sphere.
Considering charge conservation, if the outer sphere initially had a total charge of -10Q and now there is -1Q on the inner surface, the outer surface of the hollow sphere must have the remainder, which is -10Q + 1Q = -9Q. Therefore, the charge on the outer surface of the outer hollow sphere is -9Q.
Why is the heat of vaporization of water greater at room temperature than it is at its boiling point?
Explanation:
The temperature of a fluid is proportional to the average kinetic energy of its molecules, since the room temperature is lower than the temperature in the boiling point, the energy that the water must overcome to become steam is greater. Therefore, the heat of vaporization will be greater.
Two Resistances R1 = 3 Ω and R2 = 6 Ω are connected in series with an ideal battery supplying a voltage of ∆ = 9 Volts. Sketch this circuit diagram. Now, replace the two resistors with an equivalent resistance R connected to the same battery. Sketch this circuit. (a) What is current I in R? (b) What is the potential difference V across R? Using this information, answer the following questions about the original, two-resistor circuit. (c) What is the current I1 in R1? (d) What is the current I2 in R2? (e) What is the potential difference V1 across R1? (f) What is in the potential difference V2 across R2? (g) How are V1 and V2 related to the battery voltage? Comparing the two circuits: (h) How are I1 and I2 related to I? (i) How are V1 and V2 related to ∆?
Two Resistances R1 = 3 Ω and R2 = 6 Ω are connected in parallel with an ideal battery supplying a voltage of ∆ =
9 Volts. Now, replace the two resistors with an equivalent resistance R connected to the same battery. Sketch this circuit. (a) What is current I in R? (b) What is the potential difference V across R? Using this information, answer the following questions about the original, two-resistor circuit. (c) What is the current I1 in R1? (d) What is the current I2 in R2? (e) What is the potential difference V1 across R1? (f) What is in the potential difference V2 across R2? (g) How are V1 and V2 related to the battery voltage? Comparing the two circuits: (h) How are I1 and I2 related to I? (i) How are V1 and V2 related to ∆?
Answer:
Explanation:
Check attachment for solution
Suppose that the resistance between the walls of a biological cell is 3.9 × 109 Ω. (a) What is the current when the potential difference between the walls is 84 mV? (b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.73 s?
Answer:
(a) 2.154×10⁻¹¹ A.
(b) 98300000.
Explanation:
(a)
Using Ohm's law,
V = IR ........................ Equation 1
Where V = Potential difference between the walls, I = current, R = Resistance between the walls.
Make I the subject of the equation
I = V/R...................... Equation 2
Given: V = 84 mV, = 0.084 V, R = 3.9×10⁹ Ω.
Substitute into equation 2
I = 0.084/(3.9×10⁹)
I = 2.154×10⁻¹¹ A.
(b)
I = q/t
q = It .................... Equation 1
Where q = quantity of electric charge, t = time.
Given: I = 2.154×10⁻¹¹ A, t = 0.73 s.
q = 2.154×10⁻¹¹×0.73
q = 1.572×10⁻¹¹ C.
The charge on an electron e = 1.6×10⁻¹⁹ C
n = q/e
where n = number of ions.
n = 1.572×10⁻¹¹/1.6×10⁻¹⁹
n = 9.83×10⁷
n = 98300000.
Now the same particle is removed from the thread and placed over the center of a charged plate. Are there any conditions under which it is possible for the particle to be suspended in the air above the plate? Show any relevant calculations and explain your reasoning.
Answer:
changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.
Explanation:
When the particle is removed from the wire, friction can be electrically charged, either with negative charges (extra electrons) or with positive charge by electron removal, in this case when the particle is between the condenser plates it experiences a force due to the electric field given by
ΔV = E d
Where ΔV is the potential difference, d the distance between the plates and E the electric field.
In these cases we can use Newton's second law, where the acceleration is zero
[tex]F_{e}[/tex] –W + B = 0
F_{e} = W –B
q E = mg - ρ_air g V
dodne B is the hydrostatic thrust
if we know the density of the particular
ρ_particle = m / V
m = ρ_particle V
We replace
q E = g v (ρ_particle - ρ_air)
Therefore, by changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.
You are 2m from one audio speaker and 2.1m from another audio speaker. Both generate the identical sine wave with a frequency of 680 Hz. At your location, what is the phase difference between the waves? Give the answer in radians, using 340m/s as the velocity of sound.
Answer:
the phase difference is 1.26 radian
Solution:
As per the question:
Distance, d = 2 m
Distance from the other speaker, d' = 2.1 m
Frequency, f = 680 Hz
Speed of sound, v = 340 m/s
Now,
To calculate the phase difference, [tex]\Delta \phi[/tex]:
Path difference, [tex]\Delta d = d' - d = 2.1 - 2 = 0.1\ m[/tex]
For the wavelength:
[tex]f\lambda = v[/tex]
where
c = speed of light in vacuum
[tex]\lambda [/tex] = wavelength
Now,
[tex]680\times \lambda = 340[/tex]
[tex]\lambda = 0.5\ m[/tex]
Now,
Phase difference, [tex]\Delta phi = 2\pi \frac{\Delta d}{\lambda}[/tex]
[tex]\Delta phi = 2\pi \frac{0.1}{0.5} = 1.26\ rad[/tex]
What angular speed (in revolutions per minute) is needed for a centrifuge to produce an acceleration of 759 times the gravitational acceleration 9.8 m/s 2 at a radius of 4.83 cm ?
Answer:
Angular velocity, [tex]\omega=3747.33\ rev/min[/tex]
Explanation:
In this case, we need to find the angular speed needed for a centrifuge to produce an acceleration of 759 times the gravitational acceleration.
Radius of the circular path, r = 4.83 cm
The acceleration acting on the particle in circular path is given by :
[tex]a=r\omega^2[/tex]
[tex]\omega[/tex] is the angular speed in rad/s
[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]
[tex]\omega=\sqrt{\dfrac{759\times 9.8}{4.83\times 10^{-2}}}[/tex]
[tex]\omega=392.42\ rad/s[/tex]
or
[tex]\omega=3747.33\ rev/min[/tex]
So, there are 3747.33 revolutions per minute that is needed. Hence, this is the required solution.
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation 3.0m above the ground. The ball lands 30m away.
What is his pitching speed? Vox=38 m/s
Answer:
His pitching speed is 38 m/s.
Explanation:
Hi there!
Please see the attached figure for a better understanding of the problem.
The position of the ball at any time t is given by the following vector:
r = (x0 + v0 · t, y0 + 1/2 · g · t²)
Where:
r = position vector of the ball at time t.
x0 = initial horizontal position.
v0 = initial horizontal velocity.
t = time.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Let's place the origin of the frame of reference at the throwing point so that x0 and y0 = 0.
When the ball reaches the ground, its position vector will be r1 (see figure). Using the equation of the vertical component of the position vector, we can find the time at which the ball reaches the ground. At that time, the horizontal component of the position is 30 m and the vertical component is -3.0 m (see figure):
y = y0 + 1/2 · g · t² (y0 = 0)
y = 1/2 · g · t²
-3.0 m = 1/2 · (-9.8 m/s²) · t²
-3.0 m / -4.9 m/s² = t²
t = 0.78 s
Now, knowing that at this time x = 30 m, we can find v0:
x = x0 + v0 · t (x0 = 0)
x = v0 · t
30 m = v0 · 0.78 s
v0 = 30 m / 0.78 s
v0 = 38 m/s
His pitching speed is 38 m/s.
The pitching speed can be calculated using the horizontal distance and the height of the ledge. By considering the horizontal motion and the effects of gravity, the time taken can be determined. Substituting the given values into the appropriate equations, the pitching speed can be calculated as 38 m/s.
Explanation:The pitching speed can be calculated using the horizontal distance and the height of the ledge. To find the initial velocity, we can use the equation:
Vox = d / t
where Vox is the horizontal component of velocity, d is the horizontal distance, and t is the time taken. Since the ball is thrown horizontally, the vertical component of velocity is zero, and the only force acting on the ball is gravity in the downward direction. Therefore, we can use the equation:
[tex]h = 0.5 * g * t^2[/tex]
where h is the height of the ledge, g is the acceleration due to gravity, and t is the time taken. By rearranging the equation, we can find the time taken:
t = sqrt(2 * h / g)
Substituting the values given in the question, we can calculate the pitching speed:
Vox = d / t = d * sqrt(g / (2 * h))
Using the values d = 30m, h = 3.0m, and g = 9.8m/s^2, we can find the pitching speed:
Vox = 30m * [tex]sqrt(9.8m/s^2[/tex]) = 38 m/s
What is the change in temperature of a 2.50 L system when its volume is reduced to 1.00 L if the initial temperature was 298 K?
Answer:
-178.8 K
Explanation:
From Charles law,
V₁/T₁ = V₂/T₂.................... Equation 1
Where V₁ = Initial volume, T₁ = Initial Temperature, V₂ = Final volume, T₂ = Final Temperature.
Making T₂ the subject of the equation
T₂ = V₂T₁/V₁............... Equation 2
Given: V₂ = 1.00 L, V₁ = 2.5 L, T₁ = 298 K.
Substitute into equation 2
T₂ = 1.00(298)/2.5
T₂ = 119.2 K.
But,
Change in temperature = T₂ - T₁ = 119.2-298
Change in temperature = -178.8 K.
Hence the change in temperature = -178.8 K
The change in temperature of a system when its volume is reduced is calculated by utilizing Charles's law which states the volume of a gas is directly proportional to its absolute temperature under constant pressure.
Explanation:
To answer your query about the change in temperature when a system's volume is reduced, we would use Charles's law. Initially, the volume (V₁) is 2.50 L and the temperature (T₁) is 298 K. When the volume is reduced to 1.00 L (V₂), we're required to find the new temperature (T₂). According to Charles's law, V₁/T₁=V₂/T₂.
By manipulating this formula, we calculate the final temperature (T₂) as T₂=(V₂ * T₁)/V₁. Substituting the given values into this equation, we find T₂=(1.00 L * 298 K)/2.50 L. After performing the calculation, you will have the final temperature in Kelvin (K) when the volume is reduced to 1.00 L.
Learn more about Charles's law here:https://brainly.com/question/16927784
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A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt 2+qt, with p = 0.36 m/s2and q = -1.10 m/s . Determine the mouse's average speed between t = 1.0 s and t = 4.0 s. I have tried everything and the answer is not 0.40 m/s
Answer: the average speed of the rat from the information given above is 0.7m/s
Explanation:
position is given as
x(t) = pt² + qt
finding the diffencial of x(t) with respect to t, we have
d(x(t))/dt = 2pt + q
we substitute the p = 0.36m/s² and q= -1.10 m/s
d(x(t))/dt = 2(0.36)t + (-1.10)
so, at t= 1s
d(x(t))/dt = 2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s
at t= 4s
d(x(t))/dt = 2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s
To find the average speed,
average speed = (V1 + V2)/ 2
average speed = (1.78 + (-0.38))/2 = 0.7m/s
The speed is defined as the distance per unit of time. The unit of speed is m/s. The speed is a scalar quantity which means it only depends on the magnitude.
According to the question, The average speed of the mouse is 0.7m/s
The solution of the question is as follows:-
The required equation is:-
[tex]x(t) = pt^2 + qt[/tex]
The Finding the differential of x(t) with respect to t, we have
[tex]\frac{dxt}{dt} = 2pt + q[/tex]
Put the value p = 0.36m/s² and q= -1.10 m/s
[tex]\frac{d(x(t)}{dt} = 2(0.36)t + (-1.10)[/tex]
so, at t= 1s
After solving it [tex]2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s[/tex]
so,at t= 4s
After solving it =[tex]2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s[/tex]
The formula of average speed = [tex]\frac{(V1 + V2)}{2}[/tex]
[tex]= \frac{(1.78 + (-0.38))}{2} = 0.7m/s[/tex]
Hence, the average speed is 0.7m/s
For more information, refer to the link:-
https://brainly.com/question/22610586
A star (not Barnard's star) at a distance of 10 pc is observed to have a proper motion of 0.5 arcsec / year. What is its transverse speed in AU / year?
Answer:
The star will have a transverse speed of 315950.9 AU/year
Explanation:
d = 1/p
d = distance to star, measured in parsecs
p = parallax, measured in arcseconds = 0.5 arcsec/year
So, d = 1/0.5 = 2 parce
1 parsec = 3.26 light years
2 parce = 6.52 light years
⇒Transverse speed in AU / year = Distance/parallax
distance = 10pc = 2060000 AU
Transverse speed in AU / year = 2060000 Au/6.52 light years
Transverse speed = 315950.9 AU/year
Therefore, A star (not Barnard's star) at a distance of 10 pc observed to have a proper motion of 0.5 arcsec / year. Will have a transverse speed of 315950.9 AU/year
In the Bohr model the hydrogen atom consists of an electron in a circular orbit of radius a 0 = 5.29 × 10 − 11 m around the nucleus. Using this model, and ignoring relativistic effects, what is the speed of the electron?
To solve this problem we will apply the concept of balance of Forces in the body. For such an effect the centripetal force must be equivalent to the electrostatic force of the body, therefore
[tex]F_c = F_e[/tex]
[tex]\frac{mv^2}{r} = k \frac{q_pq_e}{r^2}[/tex]
Here
m = Mass of electron
r = Distance between them
k = Coulomb's constant
[tex]q_p[/tex] = Charge of proton
[tex]q_e[/tex] = Charge of electron
v = Velocity
Rearranging to find the velocity we have that,
[tex]v^2 = \frac{kq_pq_e}{mr}[/tex]
[tex]v = \sqrt{\frac{kq_pq_e}{mr}}[/tex]
Replacing,
[tex]v = \sqrt{\frac{(9*10^9)(1.6*10^{-19})(1.6*10^{-19})}{(9.1*10^{-31})(5.29*10^{-11})}}[/tex]
[tex]v = 2.19*10^6m/s[/tex]
Therefore the speed of the electron is [tex]2.19*10^6m/s[/tex]
A charge Q is spread uniformly along the circumference of acircle of radius R. A point
particlewith charge q is placed at the center of this circle.The total force exerted on the
particle q can be calculated by Coulomb's law:
A) just use R for the distance D) result of the calculation iszero
B) just use 2R for the distance E) none of the above
C) just use 2πR for the distance
Answer:
D) result of the calculation is zero
Explanation:
Coulomb's Law is valid for only point-like particles. Since the ring is not a point-like, then we have to choose an infinitesimal portion (ds) of the ring, apply the Coulomb's Law to this portion and then integrate over the ring to find the total force.
The small portion (dq) will have the same charge density as the ring itself. Furthermore, the length of the infinitesimal portion is equal to the radius times the corresponding angle, dθ.
[tex]\lambda = \frac{Q}{2\pi R} = \frac{dq}{Rd\theta}\\dq = \frac{Qd\theta}{2\pi}[/tex]
Therefore, the force between the charge at the center and the small portion is
[tex]dF = \frac{1}{4\pi\epsilon_0}\frac{qdq}{R^2} = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}[/tex]
Since force is a vector, we have separate its x- and y-components,
[tex]dF_x = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}\cos(\theta)\\dF_y = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}\sin(\theta)[/tex]
Now, we can integrate both of them over the ring.
[tex]F_x = \int\limits^{2\pi}_0 dF_x = \frac{1}{4\pi\epsilon_0}\frac{qQ}{2\pi R^2}\int\limits^{2\pi}_0\cos(\theta)d\theta = 0\\F_y = \int\limits^{2\pi}_0 dF_y = \frac{1}{4\pi\epsilon_0}\frac{qQ}{2\pi R^2}\int\limits^{2\pi}_0\sin(\theta)d\theta = 0[/tex]
Since the integration from 0 to 2π for sine and cosine functions results as zero.
Therefore, the force on the charge at the center of a uniformly distributed ring is equal to zero.
The total force exerted on a charge q placed at the center of a circle with a uniformly distributed charge Q along the circumference is zero due to the symmetry of the charge distribution.
Explanation:When a charge Q is spread uniformly along the circumference of a circle with radius R, and a point particle with charge q is placed at the center of this circle, we must apply Coulomb's law to calculate the force exerted on the charge q. Thanks to the symmetry of the charge distribution, the forces exerted by individual segments of the charged circumference on the central charge will cancel each other out in every direction. Hence, while the distance from the charge q to any point on the circle is R, the resulting total force on charge q will be zero due to symmetry.
It's important not to confuse the circumference with other distances, such as the diameter (2R) or the circumferential length (2πR), as these are not relevant for calculating the force on the central charge in this symmetric setup. Therefore, the correct answer is that the result of the calculation is zero (Option D), because the uniform distribution of charge Q around the circle results in an equilibrium of forces.