Answer:
2.42631E-13m
Explanation:
First we find the mass defect
Which is m= 0-2(9.10939E-33kg)
= - 1.82188E-30kg
Now find the energy
S
E= mc²=( -18.82188E-30)(2.999792E8)²
= 1.63742E-13J
Thus energy per photon will be
1.63742E-13J/2= 8.18710E-14J
So wavelength is given as
Lambda= hc/E
= (6.62608E-34)(2.997E8)/8.18710J
= 2.42631E-13m
The wavelength of radiation used to annihilate a positron and an electron is required.
The wavelength of the electromagnetic radiation used is 2.42 pm.
The mass of positron and electron are equal
m = Mass = [tex]\dfrac{0.000549}{6.022\times 10^{26}}=9.11\times 10^{-31}\ \text{kg}[/tex]
c = Speed of light = [tex]3\times 10^{8}\ \text{m/s}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}\ \text{Js}[/tex]
Total energy in the collision is
[tex]E_T=mc^2+mc_2\\\Rightarrow E_T=2mc^2\\\Rightarrow E_T=2\times 9.11\times 10^{-31}\times (3\times 10^{8})^2\ \text{J}[/tex]
Energy released per photon is
[tex]E=\dfrac{E_T}{2}\\\Rightarrow E=\dfrac{2\times 9.11\times 10^{-31}\times (3\times 10^{8})^2}{2}\\\Rightarrow E=9.11\times 10^{-31}\times (3\times 10^{8})^2\ \text{J}[/tex]
Energy is given by
[tex]E=\dfrac{hc}{\lambda}\\\Rightarrow \lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{9.11\times 10^{-31}\times (3\times 10^{8})^2}\\\Rightarrow \lambda=2.42\times 10^{-12}\ \text{m}=2.42\ \text{pm}[/tex]
The wavelength of the electromagnetic radiation used is 2.42 pm.
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A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m. If the spring is stretched to a length of 0.5m and released with initial velocity 0, find the position of the mass at any time t. Here damping constant is zero.
Answer:
Explanation:
force constant of spring k = force / extension
= 35.6 / 0.5
k = 71.2 N / m
angular frequency ω of oscillation by spring mass system
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
where m is mass of the body attached with spring
Putting the values
[tex]\omega = \sqrt{\frac{71.2}{5} }[/tex]
ω = 3.77 radian / s
The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2
so the equation for displacement from equilibrium position that is middle point can be given as follows
x = .5 sin ( ω t + π / 2 )
= 0.5 cos ω t
= 0.5 cos 3.77 t .
x = 0.5 cos 3.77 t .
Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on each? A. 4.0x10-6 C 4.0x10-6 C B. 7.4x10-6 C 0.6x10-6 C C. 6.6x10-6 C 1.4x10-6 C D. 5.0x10-6 C 3.0x10-6 C
Answer:
B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C
Explanation:
From Coulomb's Law the electrostatic repulsive force is given by the following formula:
F = kq₁q₂/r²
where,
F = Repulsive Force = 0.15 N
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = Magnitude of 1st Charge = ?
q₂ = Magnitude of 2nd Charge = ?
r = Distance between Charges = 0.5 m
Therefore,
0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²
q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)
q₁q₂ = 4.17 x 10⁻¹²
q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)
The sum of charges is given as:
q₁ + q₂ = 8 μC
q₁ + q₂ = 8 x 10⁻⁶
using equation (1):
(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶
(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂
q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0
Solving this quadratic equation:
q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C
q₂ = 7.4 μC (OR) q₂ = 0.6 μC
Therefore,
q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)
q₁ = 0.6μC
Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.
Therefore, the correct option will be:
B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C
How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 36.0°C greater than when they were laid? Their original length is 26.0 m.
Answer:
11 mm
Explanation:
Original length L' = 26 m
Increase in temperature dT = 36.0°C
The gap l =?
The coefficient of linear expansion for steel & = 12 × 10^−6 per °C
The gap will be gotten from the equation of linear expansion.
l = L' x & x dT
substituting, we have
L = 26 x 12 × 10^−6 x 36
L = 0.011 m = 11 mm
A small spherical body is tied to a string of length 1 m and revolved in a vertical circle such that the tension in the string is zero at the highest point . Find the linear speed of the body in the 1) lowest position & 2) highest position
Explanation:
At the highest point, the tension force is 0, so the only force acting on the sphere is gravity. Sum of forces on the sphere in the centripetal direction:
∑F = ma
mg = mv²/r
v = √(gr)
v = √(9.8 m/s² × 1 m)
v = 3.13 m/s
If the speed is constant, then the linear speed at the lowest point is also 3.13 m/s. Otherwise, we would need to know the tension in the string at that point.
The rotating loop in an AC generator is a square 12.0 cm on a side. It is rotated at 70.0 Hz in a uniform field of 0.800 T. Calculate the following quantities as functions of time t, where t is in seconds. I need help with d and e.
(a) the flux through the loop
11.52cos(140πt) mT·m2
(b) the emf induced in the loop
5.067sin(140πt) V
(c) the current induced in the loop for a loop resistance of 2.50 Ω
2.03sin(140πt) A
(d) the power delivered to the loop
4.12sin2(140πt) W (wrong)
(e) the torque that must be exerted to rotate the loop
mN·m
Answer:
Explanation:
d )
power in an electrical loop = volt x current
= 5.067sin(140πt) x 2.03sin(140πt)
= 10.286 sin²(140πt) .
10.286 ( 1 - cos 280πt ) / 2
= 5.143 ( 1 - cos 280πt )
e )
Torque on a current carrying loop in a magnetic field
= M B sinθ
M is magnetic moment of coil , B is magnetic field , θ is angle between area vector of coil and direction of B .
Here magnetic moment of coil
= A i where A is area of coil and i is current .
= .12² x 2.03sin(140πt)
= .029 sin(140πt)
B = .8 T
MB = .8 x .029 sin(140πt)
= .0232 sin(140πt) x sin(140πt)
=.0232 sin²(140πt).
You are looking down on a single coil in a constant magnetic field B = 1.2 T which points directly into of the screen. The dimensions of the coil go from a = 8 cm and b = 17 cm, to a* = 16 cm and b* = 22 cm in t = 0.04 seconds. If the coil has resistance that remains constant at 1.2 ohms. What would be the magnitude of the induced current in amperes?
Answer:
The current is [tex]I = 0.5425 \ A[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B = 1.2 \ T[/tex]
The first length is [tex]a = 8 \ cm = 0.08 \ m[/tex]
The second length is [tex]a^* = 16 \ cm = 0.16 \ m[/tex]
The first width is [tex]b = 17 \ cm = 0.17 \ m[/tex]
The second width is [tex]b^* = 22 \ cm = 0.22 \ m[/tex]
The time interval is [tex]dt = 0.04 \ s[/tex]
The resistance is [tex]R = 1.2 \ \Omega[/tex]
Generally the first area is
[tex]A = a * b[/tex]
=> [tex]A = 0.08 * 0.17[/tex]
=> [tex]A = 0.0136 \ m^2[/tex]
The second area is
[tex]A^* = a^* * b^*[/tex]
=> [tex]A^* = 0.16 * 0.22[/tex]
=> [tex]A^* = 0.0352 \ m^2[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - \frac{ B * [A^* - A]}{dt}[/tex]
This negative show that it is moving in the opposite direction of the motion producing it
=> [tex]|\epsilon | = \frac{ 1.2 * [ 0.0352-0.0135]}{0.04}[/tex]
=> [tex]|\epsilon | = 0.651 \ V[/tex]
The induced current is
[tex]I = \frac{|\epsilon|}{R}[/tex]
=> [tex]I = \frac{ 0.651}{1.2}[/tex]
=> [tex]I = 0.5425 \ A[/tex]
Yang can focus on objects 150 cm away with a relaxed eye. With full accommodation, she can focus on objects 20 cm away. After her eyesight is corrected for distance vision, what will her near point be while wearing her glasses?
Answer:
Explanation:
Without wearing glasses , her near point is 20 cm .
for correction of eye
u = infinity ,
v = - 150 cm
f = ?
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f }[/tex]
[tex]\frac{1}{-150} -0 = \frac{1}{f }\\[/tex]
f = - 150 cm
He must be wearing glass of focal length of 150 cm .
If near point be x after wearing glass ,
u = x
v = - 20 cm
f = - 150 cm
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f }[/tex]
[tex]\frac{1}{-20} -\frac{1}{x} = \frac{1}{-150 }[/tex]
[tex]\frac{1}{-20} + \frac{1}{150 }= \frac{1}{x}[/tex]
x = 23 cm .
While wearing the glasses, Yang's near point will be 23.08 cm.
Given information:
Yang can focus on objects 150 cm away with a relaxed eye.
With full accommodation, she can focus on objects 20 cm away.
For correction, we have to use a concave lens such that it can make the image of a distant object at 150 cm.
So, the object distance will be infinity, and the image distance will be [tex]v=-150[/tex] cm.
So, the focal length of the lens can be calculated by lens formula as,
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-150}-\dfrac{1}{\infty}=\dfrac{1}{f}\\f=-150\rm\;cm[/tex]
Now, after using the lens, the image distance will be [tex]v=-20[/tex] cm. Let u be the near point.
The near point, after correction, can be calculated as,
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-20}-\dfrac{1}{u}=\dfrac{1}{-150}\\\dfrac{1}{u}=\dfrac{1}{150}-\dfrac{1}{20}\\u=23.08\rm\; cm[/tex]
Therefore, while wearing the glasses, Yang's near point will be 23.08 cm.
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The spaceship Intergalactic landed on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.44 m. She sets the pendulum swinging, and her collaborators carefully count 1.10×102 complete cycles of oscillation during 2.00×102 s. What is the result
Answer:
18808.7 m/s^2
Explanation:
Given
Length of the pendulum L = 1.44 m
Number of complete cycles of oscillation n = 1.10 x 10^2
total time of oscillation t = 2.00 x 10^2 s
The period of the T = n/t
T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s
The period of a pendulum is gotten as
T = [tex]2\pi \sqrt{\frac{L}{g} }[/tex]
where g is the acceleration due to gravity
substituting values, we have
0.55 = [tex]2\pi \sqrt{\frac{1.44}{g} }[/tex]
0.0875 = [tex]\sqrt{\frac{1.44}{g} }[/tex]
squaring both sides of the equation, we have
7.656 x 10^-3 = 144/g
g = 144/(7.656 x 10^-3) = 18808.7 m/s^2
Suppose a particle of mass m is confined to a one-dimensional box of length L. We can model this as an infinite square well in which the particle's potential energy inside the box is zero and the potential energy outside is infinite. For a particle in its first excited state, what is the probability Prob(center20%) of finding the particle within the center 20% of the box
Answer: P = 4.86 × 10⁻²
Therefore, the particle's quantum number is 4.86 × 10⁻²
Explanation:
The expression of wave function for a particle in one dimensional box is given as;
φ(x) = ( √2/L ) sin ( nπx/L )
now we input our given figures, the limit of the particle to find it within the center of the box is
xₓ = L/2 + 20% of L/2
xₓ = L/2 + (0.2)L/2
xₓ = 3L/5
And the lower limit is,
x₁ = L/2 - 20% of L/2
x₁ = L/2 - (0.2) L/2
x₁ = 2L / 5
The expression for the probability of finding the particle within the center of the box is
P = ∫ˣˣₓ₁ ║φ(x)║² dx
P = ∫ ³L/⁵ ₂L/₅║(√2/L) sin ( nπx/L)║²dx
= 2/L ( ∫ ³L/⁵ ₂L/₅║sin ( nπx/L)║²dx
= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2πnx/L)/2) dx)
The particle is in its first excited state, then
n =2
Then calculate the particle's quantum number as follows;
= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2π(2)x/L)/(2)) dx)
= 1/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 4πx/L)/2) dx)
= 1/L ( x - (L/4π)sin (4πx/L)) ³L/⁵ ₂L/₅
= 1/L ((3L/5) - (L/4π) sin (( 4π(3L/5)/L)) - (( 2L/5) - (L/4π)sin ( 4π(2L/5)/L)))
= 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))
Use the trigonometric formula to solve the above equation
sinA - sinB = 2sin ( A-B/2) cos (A+B/2)
Calculate the particle's quantum number as follows
P = 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))
= 1/5 + 1/4π ( 2sin( (8π/5 -12π/5 ) / 2 ) cos ( (8π/5 + 12π/5) / 2 ))
= 1/5 + 1/2π ( -sin(2π/5) cos2π
= 1/5 - 1/2π ( sin (2π/5)(1))
= 0.0486 (10⁻²)(10²)
= 4.86 × 10⁻²
Therefore, the particle's quantum number is 4.86 × 10⁻²
What is the absolute pressure at a depth of 9.91 m below the surface of a deep lake? Assume atmospheric pressure is 1.01×105 Pa .
Answer:
P = 198.118 kPa
Explanation:
Given:
Atmospheric pressure = P[tex]_{atm\\}[/tex] = 1.01×10⁵
depth = h = 9.91 m
To find:
Absolute pressure P[tex]_{abs}[/tex]
Solution:
Density of water = ρ = 1.000x10 ³kg/m ³
acceleration due to gravity = ρ = 9.8 m/s²
P[tex]_{abs}[/tex] = P[tex]_{atm\\}[/tex] + ρgh
= 1.01×10⁵ + 1.000x10 ³x 9.8 x 9.91
= 101000 + 1000(9.8)(9.91)
= 101000 + 97118
= 198118 Pa
= 198.118 kPa
P[tex]_{abs}[/tex] = 198.118 kPa
The absolute pressure at a depth below the surface of this deep lake is 198.118 kPa.
Given the following data:
Atmospheric pressure = [tex]1.01 \times 10^5 \;Pa[/tex]Height (depth) = 9.91 meters.Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]Density of water = 1000 [tex]kg/m^3[/tex]To calculate the absolute pressure at a depth below the surface of a deep lake:
Mathematically, absolute pressure is given by this formula:
[tex]P_{abs} = P + \rho gh[/tex]
Substituting the given parameters into the formula, we have;
[tex]P_{abs} = 1.01 \times 10^5 + 1000 \times 9.8 \times 9.91 \\\\ P_{abs} = 101000+ 97118 \\\\[/tex]
Absolute pressure = 198118 Pa
Note: 1 kPa = 1000 Pa
Absolute pressure = 198.118 kPa
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A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA. What amount of vars must be added to bring the pf to 0.85
Answer:
[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]
Explanation:
Given that:
A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA
Here:
the initial power factor i.e cos θ₁ = 0.7 lag
θ₁ = cos⁻¹ (0.7)
θ₁ = 45.573°
Active power P = 1500 watts
Apparent power S = 2100 VA
What amount of vars must be added to bring the pf to 0.85
i.e the required power factor here is cos θ₂ = 0.85 lag
θ₂ = cos⁻¹ (0.85)
θ₂ = 31.788°
However; the initial reactive power [tex]Q_1[/tex] = P×tanθ₁
the initial reactive power [tex]Q_1[/tex] = 1500 × tan(45.573)
the initial reactive power [tex]Q_1[/tex] = 1500 × 1.0202
the initial reactive power [tex]Q_1[/tex] = 1530.3 vars
The amount of vars that must therefore be added to bring the pf to 0.85
can be calculated as:
[tex]Q_{sh} = P( tan \theta_1 - tan \theta_2)[/tex]
[tex]Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)[/tex]
[tex]Q_{sh} = 1500( 1.0202 - 0.6197)[/tex]
[tex]Q_{sh} = 1500( 0.4005)[/tex]
[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]
With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 41 m
Answer:
The speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.
Explanation:
Given;
maximum vertical height of the throw, H = 41 m
Apply the following kinematic equation;
V² = U² + 2gH
where;
V is the final speed with which the ball will rise to a maximum height
U is the initial speed of the ball = 0
g is acceleration due to gravity = 0
V² = U² + 2gH
V² = 0² + 2gH
V² = 2gH
V = √2gH
V = √(2 x 9.8 x 41)
V = 28.35 m/s
Therefore, the speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.
. The Moon has an average distance from the Earth of 384,403 km and takes 27.32166 days to orbit the Earth. What is the velocity of the Moon in kilometers per hour
Answer:
Velocity of moon = 586.23 km/h
Explanation:
We are given;
Distance of moon from the Earth = 384403 km
Time taken to orbit earth;t = 27.32166 days
24 hours make 1 day, thus 27.32166 days = 27.32166 × 24 = 655.72 hours
Formula for velocity is distance/time
Thus,
Velocity of moon = distance from moon to earth/time taken to orbit the earth
Velocity of moon = 384403/655.72 = 586.23 km/h
If your front lawn is 21.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1350 new snowflakes every minute, how much snow, in kilograms, accumulates on your lawn per hour
Answer:
Mass of flake = 67.716 kg
Explanation:
Given:
Length of lawn = 21 ft
Width of lawn = 20 ft
Note:
Each snow flake mass = 1.90 mg
Find:
Weight of total flake in a minute
Computation:
Area of lawn = 21 × 22
Area of lawn = 440 ft²
Amount of flake per minute = 440 × 1350
Amount of flake per minute = 594000 flake/ minutes
Mass of flake = 594000 × 1.90 mg × 60 minutes
Mass of flake = 67,716,000 mg
Mass of flake = 67.716 kg
The temperature of a plastic cube is monitored while the cube is pushed 8.6 m across a floor at constant speed by a horizontal force of 19 N. The monitoring reveals that the thermal energy of the cube increases by 120 J. What is the increase in the thermal energy of the floor along which the cube slides
Answer:
Answer:
43.4J
Explanation:
We know that
Work done = total heat energy
But work done is force x distance
=> F = 19 x8.6 = 163.4 J
So the total heat. Will be Heat of cube + heat of floor = 163.4J
So that heat of floor will now be
floor = 163.4 J - 120 J = 43.4 Joules
Explanation:
How far can a person run in 0.25hr if he or she runs at a average speed of 16km/he?
Answer:
4km
Explanation:
d = st
d=0.25*16
d=4km
In the anatomical position, the gluteal and lumbar are on the ___
Answer:
Posterior of the body
Explanation:
Gluteal region is located at the proximal end of the femur and posterior to the pelvic girdle. The gluteal muscles help to move the lower limb at the hip joint. The gluteal region is divided into two groups: Deep lateral rotators and superficial abductors and extenders.
The lumbar is the lower region of the spine commonly known as lower back, it has five vertebrates. The lumbar contain tissue and nerves that control communication between legs and brain. In anatomical terms they are located inferior to the rib cage, at the bottom section of the vertebral column and superior to sacrum and pelvis.
Consider a vacuum-filled parallel plate capacitor (no dielectric material between the plates). What is the ratio of conduction current Jc to displacement current Jd at 10 MHz
Answer: 1798
Explanation:
Given that there is no dielectric material between the plates,
the permittivity of free space = 9 × 10^-12 f/m
The frequency F = 10 MHz
The ratio of conduction current JC to the displacement current Jd is also known as loss tangent.
Please find the attached file for the solution
A cheetah can accelerate from rest to 25.0 m/s in 6.22 s. Assuming constant acceleration, how far has the cheetah run in this time
Answer:
The distance covered by the cheetah during the motion is 77.75 m
Explanation:
Given;
initial velocity of cheetah, u = 0
final velocity of cheetah, v = 25 m/s
time of acceleration, t = 6.22 s
Apply kinematic equation;
[tex]s = (\frac{v+u}{2} )t[/tex]
where;
s is the distance covered by the cheetah during the motion
[tex]s = (\frac{v+u}{2} )t\\\\s = (\frac{25+0}{2} )6.22\\\\s = 77.75 \ m[/tex]
Therefore, the distance covered by the cheetah during the motion is 77.75 m
How many grams of water (H2O) have the same number of oxygen atoms as 7.0 mol of oxygen gas
Answer:
126 g of water, H2O.
Explanation:
First, we'll begin by calculating the number of atoms in 7 mole oxygen gas.
This is illustrated below:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This means that 1 mole of O2 also contains 6.02×10²³ atoms.
Now, if 1 mole of O2 contains 6.02×10²³ atoms, then 7 moles of O2 will contain = 7 × 6.02×10²³ = 4.214×10²⁴ atoms.
Finally, we shall determine the mass of H2O that will contain 4.214×10²⁴ atoms.
This is illustrated below:
6.02×10²³ atoms is present in 1 mole of any substance contains
1 mole of H2O = (2x1) + 16 = 18 g
6.02×10²³ atoms is present in 18 g of H2O.
Therefore, 4.214×10²⁴ atoms will be present in = (4.214×10²⁴ × 18)/6.02×10²³
= 126 g of water, H2O.
Therefore, 126 g of water, H2O will contain the same number of oxygen atoms as 7.0 mol of oxygen gas.
7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.
We want to know the mass of water (H₂O) that has the same number of oxygen atoms as 7.0 mol of oxygen gas (O₂).
First, we will calculate the number of oxygen atoms in 7.0 mol of O₂ considering the following relations.
1 mol of O₂ contains 6.02 × 10²³ molecules of O₂ (Avogadro's number).1 molecule of O₂ contains 2 atoms of O.[tex]7.0 mol O_2 \times \frac{6.02 \times 10^{23}molecule O_2 }{1molO_2} \times \frac{2atomO}{1molecule O_2} = 8.4 \times 10^{24} atomO[/tex]
Now, we want to calculate the mass of H₂O that contains 8.4 × 10²⁴ atoms of O. We will consider the following relations.
1 molecule of H₂O contains 1 atom of O.1 mol of H₂O contains 6.02 × 10²³ molecules of H₂O (Avogadro's number).The molar mass of H₂O is 18.02 g/mol.[tex]8.4 \times 10^{24} atomO \times \frac{1moleculeH_2O}{1atomO} \times \frac{1molH_2O}{6.02 \times 10^{23}moleculeH_2O } \times \frac{18.02gH_2O}{1molH_2O} = 250 gH_2O[/tex]
7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.
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Capacitor C1 is in series with capacitors C2 and C3 in parallel. Then three capacitor system is connected to battery with V0. Determine the charge stored by C1 when C1 = 20 μF, C2 = 10 μF, C3 = 30 μF, and V0 = 18 V.g
Answer:
Q₁ = 2.4 10⁻⁴ C
Explanation:
We have a circuit with several capacitors, let's find the equivalent capacitor of the parallel
[tex]C_{eq1}[/tex] = C₂ + C₃
C_{eq1} = (10 +30) 10⁻⁶
C_{eq1} = 40 10⁻⁶ F
There remains a series system between C₁ and C_{eq1}, let's find the equivalent capacitor
1/C_{eq2} = 1 / C₁ + 1 / C_{eq1}
1 /C_{eq2} = 1 / 20 10⁻⁶ + 1/40 10⁻⁶
1 / C_{eq2} = 0.075 10⁶
C_{eq2} = 13.33 10⁻⁶ F
let's use the relationship
V = Q / C_{eq2}
Q = V C_{eq2}
Q = 18 13.33 10⁻⁶
Q = 2.4 10⁻⁴ C
In a combination of capacitors in series the charge is constant, so the charge on C₁ is the same
Q₁ = 2.4 10⁻⁴ C
Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of the force on a 3.0-m length of one of the wires is equal to 8.0 μN, what is the greater of the two currents?
Answer:
The greater of the two currents is 0.692 A
Explanation:
Given;
distance between the two parallel wires; r = 6 mm = 6 x 10⁻³ m
let the current in the first wire = I₁
then, the current in the second wire = 2I₁
length of the wires, L = 3.0 m
magnitude of force on the wires, F = 8 μN = 8 x 10⁻⁶ N
The magnitude of force on the two parallel wires is given by;
[tex]F = \frac{\mu_o I_1(2I_1)}{2\pi r}\\\\F = \frac{\mu_o 2I_1^2}{2\pi r}\\\\I_1^2 = \frac{F*2\pi r}{2\mu_o} \\\\I_1^2 = \frac{8*10^{-6}*2\pi (6*10^{-3})}{2(4\pi*10^{-7})}\\\\I_1^2 = 0.12\\\\I_1 = \sqrt{0.12}\\\\ I_1 =0.346 \ A[/tex]
the current in the second wire = 2I₁ = 2 x 0.346 A = 0.692 A
Therefore, the greater of the two currents is 0.692 A
If the momentum of a system is to be conserved, which must be true of the net external force acting on the system?
A. nonzero but constant.
B. increasing
C. decreasing
D. zero
Answer:
D. zero
Explanation:
For momentum of an isolated or closed system to be conserved (initial momentum must equal final momentum), the net external force acting on the system must be zero.
There is always external forces acting on a system, for this system’s momentum to remain constant, all the external forces acting on the system must cancel out, so that the net external force on the system is zero.
[tex]F_{ext} = 0[/tex]
Therefore, the correct option is "D"
D. zero
Tech A says voltage drops can be measured as long as current is flowing. Tech B says voltage drops can be measured across components, connectors, or cables. Who is correct?
A. Tech A
B. Tech B
C. Both Techs A and B
D. Neither Tech A nor B
Answer:
C. Both Techs A and B
Explanation:
For voltage drop to be measured in the circuit, then there must be a voltage in the circuit. Once there is a voltage across the circuit, there will be current flowing through the the circuit, hence technician A is correct. Voltage drop is usually measured across components in the circuit. Components in a circuit are consumptive in the circuit, hence their is usually a voltage drop when current flows through them in a circuit. Technician B is correct.
Answer:
C
Explanation:
The length and width of a rectangle are 1.82 cm and 1.5 cm respectively. Calculate area of the rectangle and write in correct significant number.
Answer:
Hey mate ,
Area of rectangle = l×b
1.82×1.5
2.73cm2
When your scalpel gets dull or is broken, it should be disposed of by
a. Placing in the broken glass/sharps container
b. Thrown in the trash
c. Kept in your dissection kit
A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to accelerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him?
Answer:
70 N
Explanation:
Draw a free body diagram of the boy. There are four forces:
Weight force mg pulling down,
A 300 N normal force pushing up,
A 75 N applied force pulling right,
and a 5 N friction force pushing left.
The boy's acceleration in the y direction is 0, so the net force in the y direction is 0.
The net force in the x direction is 75 N − 5 N = 70 N.
You have been asked to design a can with a volume of 672cm3 that is shaped like a right circular cylinder. The can will have a closed top. What radius r and height h, in centimeters, would minimize the amount of material needed to construct this can
Answer:
r = 4.747 cm and h = 9.4925 cm
Explanation:
We know that volume of a cylinder is given as:
V = πr²h
Also, surface area is given as;
S = 2πr² + 2πrh
Where r is radius and h is height
Now, we are told that the volume is 672 cm³
Thus, πr²h = 672
Making h the subject gives;
h = 672/πr²
Putting 672/πr² for h in the surface area equation gives;
S = 2πr² + 2πr(672/πr²)
Factorizing gives;
S = 2π[r² + 672/πr]
Differentiating to get first derivative gives;
S' = 2π[2r - (672/πr²)]
Equating to zero gives;
2π[2r - (672/πr²)] = 0
4πr - 1344/r² = 0
4πr = 1344/r²
r³ = 1344/4π
r³ = 106.95212175775
r = ∛106.95212175775
r = 4.747 cm
So, since h = 672/πr²
Then, h = 672/π(4.747)²
h = 9.4925 cm
The velocity of any point on a rigid body is _________ to the relative position vector extending from the IC to the point. Group of answer choices
Answer:
The velocity of any point on a rigid body is ___Always perpendicular______ to the relative position vector extending from the IC to the point
Explanation:
This is because The instantaneous center (IC) of zero velocity for the rigid body is at the point in contact with ground. The velocity direction at any point on the body is always perpendicular to the line connecting the point to the IC.
The velocity of any point on a rigid body is : Perpendicular to the relative position vector extending from the IC to the point.
The IC ( instantaneous center ) is the point of the rigid body in contact with the ground when the rigid body is at zero velocity.
The velocity direction from any point of the rigid body will always be perpendicular to the ( IC ) since the IC is the reference point of the rigid body when the body is at rest ( zero velocity ).
Hence we can conclude that the velocity of any point on a rigid body will be perpendicular.
Learn more : https://brainly.com/question/13108309
Although the options related to your question is missing an accurate answer is provided within the scope of your question
A plane moves 100 meters to the right in two seconds. What is its velocity?
0 m/s
O 30 m/s
50 m/s
O 100 m/s
200 m/s
Answer:
50 m/s
Explanation:
[tex]Distance = 100m\\Time = 2 secs\\\\Velocity = \frac{Distance}{Time} \\\\V = \frac{100m}{2s} \\\\= 50m/s[/tex]