Answer:
y and length is directly relation
Explanation:
Given data
A single-slit diffraction pattern is formed on a distant scree
angles involved = small
to find out
what factor will the width of the central bright spot on the screen change
solution
we know that for single slit screen formula is
mass ƛ /area = sin θ and y/L = sinθ
so we can say mass ƛ /area = y/L
and y = mass length ƛ / area .................1
in equation 1 here we can see y and length is directly relation so we can say from equation 1 that the width of the central bright spot on the screen change if the distance from the slit to the screen is doubled
A +3.00); C charge with a mass of 4.00 x 10-3 kg hovers above a horizontal flat insulating surface, neither descending nor ascending, and thus in mechanical equilibrium. Taking g = 9.8 m/s2, what is the magnitude of the electric force on the charge? Hint: keep in mind that the weight force is given by F mg ? A. 0.0281 N O B. 0.0392 N C. 0.0473 N 0 D. 0.0671 N
Answer:
option (B)
Explanation:
q = 3 c, m = 4 x 10^-3 kg, g 9.8 m/s^2,
In the equilibrium condition, the weight of the charge particle is balanced by the electrostatic force.
q E = mg
Electrostatic force = m g = 4 x 10^-3 x 9.8 = 0.0392 N
Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags mutually exert a gravitational attraction F1 on each other. You now take two bricks from one bag and add them to the other bag, causing the bags to attract each other with a force F2. What is the closest expression for F2 in terms of F1?
The closest expression for [tex]\( F_2 \)[/tex] in terms of [tex]\( F_1 \)[/tex] is.
[tex]\[ \\F_2 = \frac{3}{4} F_1} \][/tex]
This means that after transferring 2 bricks from one bag to the other, the gravitational attraction [tex]\( F_2 \)[/tex] between the bags is [tex]\( \frac{3}{4} \)[/tex] of the initial attraction [tex]\( F_1 \).[/tex]
When initially each bag contains 4 bricks of mass [tex]\( M \)[/tex] each, the total mass in each bag is [tex]\( 4M \)[/tex]. The bags exert a gravitational attraction [tex]\( F_1 \)[/tex] on each other.
Let's denote:
-[tex]\( F_1 \):[/tex] Initial gravitational attraction between the bags when each bag has 4 bricks.
- [tex]\( F_2 \):[/tex] Gravitational attraction after transferring 2 bricks from one bag to the other.
Initial Situation Before Transfer
Each bag has 4 bricks so the mass of each bag is [tex]\( 4M \).[/tex]
Gravitational Attraction [tex]\( F_1 \)[/tex]
[tex]\[ F_1 = G \frac{(4M)(4M)}{d^2} \][/tex]
where [tex]\( G \)[/tex] is the gravitational constant and [tex]\( d \)[/tex] is the distance between the centers of the bags.
After Transferring 2 Bricks:
Now, one bag has 6 brick mass 6m and the other bag has 2 bricks mass2m
Gravitational Attraction[tex]\( F_2 \)[/tex]
[tex]\[ F_2 = G \frac{(6M)(2M)}{d^2} \]s[/tex]
Simplifying [tex]\( F_2 \)[/tex]
[tex]\[ F_2 = G \frac{12M^2}{d^2} \][/tex]
Relation between [tex]\( F_2 \)[/tex] and [tex]\( F_1 \)[/tex]
To find the relation between [tex]\( F_2 \)[/tex] and [tex]\( F_1 \)[/tex] we compare them
[tex]\[ \frac{F_2}{F_1} = \frac{G \frac{12M^2}{d^2}}{G \frac{16M^2}{d^2}} \][/tex]
[tex]\[ \frac{F_2}{F_1} = \frac{12M^2}{16M^2} \][/tex]
[tex]\[ \frac{F_2}{F_1} = \frac{3}{4} \][/tex]
An ion source is producing 6Li ions, which have charge +e and mass 9.99 × 10-27 kg. The ions are accelerated by a potential difference of 13 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude B = 1.0 T. Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the 6Li ions to pass through undeflected.
Answer:
6.45 x 10^5 N/C
Explanation:
q = + e = 1.6 x 10^-19 C
m = 9.99 x 10^-27 kg
V = 13 kV = 13000 V
B = 1 T
Let v be the speed and E be the strength of electric field.
1/2 mv^2 = eV
v^2 = 2 e v / m
v^2 = (2 x 1.6 x 10^-19 x 13000) / (9.99 x 10^-27)
v = 6.45 x 10^5 m/s
As the charge particle is undeflected, the force due to magnetic field is counter balanced by the force due to electric field.
q E = q v B
E = v B = 6.45 x 10^5 x 1 = 6.45 x 10^5 N/C
The result will be the magnitude of the smallest electric field (E) required for the 6Li ions to pass through the magnetic field undeflected.
Here's how to calculate the strength of the electric field required for the 6Li ions to pass through the magnetic field undeflected:
Force Balance:
For the ions to move undeflected, the magnetic force acting on them needs to be balanced by the electric force acting in the opposite direction.
Magnetic Force:
The magnetic force (F_magnetic) on a charged particle moving through a magnetic field is given by: F_magnetic = q * v * B
Where:
q is the charge of the particle (q = +e for 6Li ion)
v is the velocity of the particle
B is the magnetic field strength
Electric Force:
The electric force (F_electric) on the charged particle in an electric field (E) is: F_electric = q * E
Balancing Forces:
For undeflected motion: F_magnetic = F_electric
Substitute the expressions from steps 2 and 3: q * v * B = q * E
Solving for Electric Field (E):
Since the charge (q) of the ion appears on both sides, we can cancel it.
E = v * B
Finding Ion Velocity (v):
The ions are accelerated by a potential difference (V) of 13 kV (13000 V).
The potential difference is related to the ion's kinetic energy (KE) by: KE = q * V
Assuming the ion starts from rest, all the potential energy is converted to kinetic energy.
KE = 1/2 * m * v^2 (where m is the mass of the ion)
Solve for v:
Combine equations from steps 6: 1/2 * m * v^2 = q * V
Solve for v: v = sqrt( 2 * q * V / m )
Substitute v in the equation for E (from step 5):
E = sqrt( 2 * q * V / m ) * B
Plug in the values:
q = +e (elementary charge = 1.602 x 10^-19 C)
V = 13000 V
m = 9.99 x 10^-27 kg
B = 1.0 T
Calculate E using the above values and the constant for elementary charge.
A golf club strikes a 0.031-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball’s motion, has a magnitude of 6240 N, and is in contact with the ball for a distance of 0.011 m. With what speed does the ball leave the club
Answer:
66.5 m/s
Explanation:
m = mass of the golf ball = 0.031 kg
F = magnitude of force applied to the ball = 6240 N
Acceleration experienced by the ball is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{6240}{0.031}[/tex]
a = 201290.32 m/s²
d = distance for which the ball is in contact with the golf club = 0.011 m
v₀ = initial speed of the ball = 0 m/s
v = final speed of the ball = 0 m/s
Using the kinematics equation
v² = v₀² + 2 a d
v² = 0² + 2 (201290.32) (0.011)
v = 66.5 m/s
An inductor of 299 mH with a resistance of 51 Ω is connected to a power supply with a maximum voltage of 227 V and a frequency of 72 Hz. Find the current in the circuit. Answer in units of A.
Answer:
The answer is 1.1A
Explanation:
See the attached file
60. The label on a box of cereal gives the mass of cereal in two units: 978 grams and 34.5 oz. Use this information to find a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?
Answer : The mass of cereal in two units are, 34.815 oz and 978.075 grams respectively.
Explanation :
As there are two systems for measuring the units which are English and Metric system.
The Metric System measured the things in meters, grams, liters, etc and adds prefixes like kilo, milli and centi to the count orders of the magnitude.
The English system measured the things in feet, inches, pounds, mile, etc.
As we are given the mass of cereal in 978 grams that means 987 grams is in metric unit. Now we have to convert metric unit into English unit.
Conversion factor used : (1 oz = 28.350 g)
As, 28.350 grams = 1 oz
So, 987 grams = [tex]\frac{987\text{ grams}}{28.350\text{ grams}}\times 1oz[/tex]
= 34.815 oz
As we are given the mass of cereal in 34.5 oz that means 34.5 oz is in metric unit. Now we have to convert metric unit into English unit.
Conversion factor used : (1 oz = 28.350 g)
As, 1 oz = 28.350 grams
So, 34.5 oz = [tex]\frac{34.5\text{ oz}}{1\text{ oz}}\times 28.350\text{ grams}[/tex]
= 978.075 grams
Therefore, the mass of cereal in two units are, 34.815 oz and 978.075 grams respectively.
The conversion factor from ounces to grams, divide 978 grams by 34.5 ounces, resulting in approximately 28.3 g/oz when rounded to three significant figures.
To find a conversion factor between English and metric units using the information that a box of cereal has a mass of 978 grams and 34.5 ounces, you would divide the number of grams by the number of ounces.
Conversion factor = 978 g / 34.5 oz ≈ 28.34783 g/oz
However, we need to consider significant figures in our answer. The number 34.5 has three significant figures, and the number 978 has three significant figures as well. Therefore, we can justify three significant figures in our conversion factor, giving us 28.3 g/oz as the conversion with significant figures properly accounted for.
The International Space Station operates at an altitude of 350 km. When final construction is completed, it will have a weight (measured at the Earth’s surface) of 4.22 x 106 N. What is its weight when in orbit?
The International Space Station's operational weight in orbit is effectively zero due to its state of continuous free-fall around Earth, even though the gravitational force at its altitude is not significantly less than on Earth's surface.
Explanation:The question relates to the weight of the International Space Station (ISS) when in orbit. Weight in physics is defined as the force exerted on an object due to gravity, calculated as the product of mass and gravitational acceleration (g). On Earth's surface, g is approximately 9.81 m/s2, but this value decreases with altitude due to the equation g = GME / r2, where G is the gravitational constant, ME is Earth's mass, and r is the distance from Earth's center. At the ISS's altitude (> 350 km), g is about 8.75 m/s2. However, it's crucial to understand that while the ISS has a significant mass, leading to a large weight calculation on Earth, its apparent weight in orbit is effectively zero due to it being in a continuous free-fall state around Earth, experiencing microgravity. This explains why astronauts appear weightless, even though the actual gravitational force at that altitude is not much less than on Earth's surface. Therefore, while the ISS has a calculable weight based on its mass and Earth's gravitational pull at its altitude, its operational weight in orbit, in terms of the experience within it, is zero.
A container is filled with water to a depth of 26.2 cm. On top of the water floats a 16.1 cm thick layer of oil with a density of 0.85 × 103 kg/m3. The acceleration of gravity is 9.81 m/s2 . a) What is the pressure at the surface of the water? Answer in units of Pa.
Answer:
1.34 x 10^3 Pa
Explanation:
density of oil = 0.85 x 10^3 kg/m^3
g = 9.81 m/s^2
height of oil column = 16.1 cm = 0.161 m
Pressure on the surface of water = height of oil column x density of oil x g
= 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa
Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.
An amplifier has a power output of 100 mW when the input power is 0.1 mW. The amplifier gain is_________ dB.
a. 10
b. 20
c. 30
d. 40
Answer:
The amplifier gain is 30 dB.
(c) is correct option.
Explanation:
Given that,
Output power = 100 mW
Input power = 0.1 mW
We need to calculate the amplifier gain in dB
Using formula of power gain
[tex]a_{p}=10\ log_{10}(A_{p})[/tex]....(I)
We calculate the [tex]A_{p}[/tex]
[tex]A_{p}=\dfrac{P_{o}}{P_{i}}[/tex]
[tex]A_{p}=\dfrac{100}{0.1}[/tex]
[tex]A_{p}=1000[/tex]
Now, put the value of [tex]A_{p}[/tex] in equation (I)
[tex]a_{p}=10\ log_{10}(1000)[/tex]
[tex]a_{p}=10\times log_{10}10^{3}[/tex]
[tex]a_{p}=10\times 3log_{10}10[/tex]
[tex]a_{p}=30\ dB[/tex]
Hence, The amplifier gain is 30 dB.
The gain of an amplifier, given a power output of 100 mW and an input power of 0.1 mW, can be calculated using the gain formula in decibels, which results in a gain of 30 dB.
Explanation:In this context, the gain of the amplifier can be calculated using the formula for Gain in decibels (dB), which is 10 times the log base 10 of the output power divided by the input power. Therefore, we first divide 100 mW by 0.1 mW, which gives us 1000. Taking the log base 10 of 1000 returns 3, and multiplying 3 by 10 gives us a gain of 30 dB.
So the correct answer to your question: 'An amplifier has a power output of 100 mW when the input power is 0.1 mW, what is the amplifier gain?' is option c which states that the gain is 30 dB.
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An artificial satellite is in a circular orbit around a planet of radius r = 5.25 Ã 10^3 km at a distance d = 450.0 km from the planet's surface. The period of revolution of the satellite around the planet is T = 2.15 hours. What is the average density of the planet?
Answer:
3020.68 kg/m^3
Explanation:
r = 5.25 x 10^3 km = 5.25 x 10^6 m, d = 450 km = 450 x 10^3 m
T = 2.15 hours = 2.15 x 3600 second = 7740 second
Let the density of the planet is ρ and M be the mass of planet.
The formula for the orbital velocity is
[tex]v = \sqrt{\frac{GM}{r+d}}[/tex]
Time period is given by
[tex]T = {\frac{2\pi (r+d)}{v}}[/tex]
[tex]T = \frac{2\pi (r +d)^{1.5}}{\sqrt{GM}}[/tex]
[tex]7740= \frac{2\pi (5700\times 1000)^{1.5}}{\sqrt{6.67\times 10^{-11}M}}[/tex]
M = 1.83 x 10^24 kg
Density = mass / Volume
ρ = 1.83 x 10^24 / (4/3 x 3.14 x (5.25 x 10^6)^3)
ρ = 3020.68 kg/m^3
An electric field of 8.20 ✕ 105 V/m is desired between two parallel plates, each of area 25.0 cm2 and separated by 2.45 mm. There's no dielectric. What charge must be on each plate?
Answer:
q = 1.815 \times 10^{-8} C
Charge on one plate is positive in nature and on the other plate it is negative in nature.
Explanation:
E = 8.20 x 10^5 V/m, A = 25 cm^2, d = 22.45 mm
According to the Gauss's theorem in electrostatics
The electric field between the two plates
[tex]E = \frac{\sigma }{\varepsilon _{0}}[/tex]
[tex]{\sigma }= E \times {\varepsilon _{0}}[/tex]
[tex]{\sigma }= 8.20 \times 10^{5} \times {8.854 \times 10^{-12}[/tex]
[tex]{\sigma }= 7.26 \times 10^{-6} C/m^{2}[/tex]
Charge, q = surface charge density x area
[tex]q = 7.26 \times 10^{-6} \times 25 \times 10^{-4}[/tex]
q = 1.815 \times 10^{-8} C
The average distance from Earth to the Moon is 384,000 km. In the late 1960s, astronauts reached the Moon in about 3 days. How fast (on average) must they have been traveling (in km/h) to cover this distance in this time? Compare this speed to the speed of a jet aircraft (800 km/h).
Answer:
They must have been traveling at 5333.33 km/h to cover that distance in 3 days.
That speed are 6,66 times higher than the speed of an aircraft jet.
Explanation:
d= 384000 km
t= 3 days = 3*24hr = 72hr
V= 384000km/72hr
V= 5333.33 km/h
comparison:
V1/V2= 5333.33/800
V1/V2= 6.66
An empty, free-moving box car with a mass of 22,509 kg is coasting along at 4.21 m/s, when it runs into a second, stationary loaded box car with a mass of 31,647 kg. What is the speed of the two cars after they collide and attach?
Answer:
Final velocity, v = 1.74 m/s
Explanation:
Given that,
Mass of car 1, m₁ = 22509 kg
Velocity of car 1, v₁ = 4.21 m/s
Mass of car 2, m₂ = 31647 kg
It is stationary, v₂ = 0
Let v be the velocity of the two cars after they collide and attach. It can be calculated using law of conservation of momentum as :
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]v=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}[/tex]
[tex]v=\dfrac{22509\ kg\times 4.21\ m/s+0}{22509\ kg+31647\ kg}[/tex]
v = 1.74 m/s
So, the velocity of two cars after the collision is 1.74 m/s. Hence, this is the required solution.
A grasshopper floating in water generates waves at a rate of three per second with a wavelength of two centimeters. (a) What is the period of these waves? (b) What is the wave velocity?
Answer:
(a) 0.33 second
(b) 6 cm/s
Explanation:
Frequency, f = 3 waves per second
wavelength, λ = 2 cm = 0.02 m
(a) The period of wave is defined as the time taken by the wave to complete one oscillation. It is the reciprocal of frequency.
T = 1 / f = 1 / 3 = 0.33 second
(b) the relation between wave velocity, frequency and wavelength is given by
v = f x λ
v = 3 x 0.02 = 0.06 m /s
v = 6 cm /s
Final answer:
The period of the waves generated by a grasshopper in water is 0.333 seconds, and the wave velocity is 0.06 m/s.
Explanation:
Calculating the Period and Wave Velocity
When dealing with waves generated by a grasshopper floating in water, two key properties to determine are the period of the waves and the wave velocity.
(a) The Period of the Waves
The period (T) of a wave is the amount of time it takes for one complete wave cycle to pass. It can be calculated as the inverse of the frequency (f), which is the rate at which waves are generated. The formula to find the period is:
T = 1/f
In this case, the grasshopper generates waves at a frequency of 3 waves per second (3 Hz). Therefore, the period is:
T = 1/3 Hz = 0.333 seconds
(b) The Wave Velocity
The velocity (v) of a wave is determined by multiplying the frequency (f) of the wave by its wavelength (λ). The formula for wave velocity is:
v = f × λ
Here, the wavelength given is 2 cm, which we need to convert to meters (since the SI unit for velocity is m/s). Thus:
λ = 2 cm = 0.02 m
The velocity of the waves generated by the grasshopper is then:
v = 3 Hz × 0.02 m = 0.06 m/s
A commuter train blows its 200-Hz horn as it approaches a crossing. The speed of sound is 335 m/s. An observer waiting at the crossing receives a frequency of 207 Hz. What is the speed of the train?
Answer:
346m/s
Explanation:
v1=335m/s
f1=200hz
f2=207hz
v2=?
v1/f1=v2/f2
335/200=v2/207
v2=335*207/200
v2=346m/s
A 5.0-kg crate is on an incline that makes an angle 30° with the horizontal. If the coefficient of static friction is 0.5, what is the maximum force that can be applied parallel to the plane without moving the crate?
Answer:
[tex]F_{applied} = 45.8 N[/tex]
Explanation:
When maximum force is applied on the crate along the plane so that it will not move then in that case friction force and component of the weight of the crate is along the plane opposite to the applied force
So here we will have
[tex]mgsin\theta + F_f = F_{applied}[/tex]
now we know that
[tex]F_f = \mu F_n[/tex]
also we know that
[tex]F_n = mg cos\theta[/tex]
so we will have
[tex]F_f = \mu ( mg cos\theta)[/tex]
now we will have
[tex]mg sin\theta + \mu (mg cos\theta) = F_{applied}[/tex]
[tex](5)(9.81)sin30 + (0.5)(5)(9.81)cos30 = F_{applied}[/tex]
so we will have
[tex]F_{applied} = 45.8 N[/tex]
The maximum force that can be applied parallel to the plane without moving the 5.0-kg crate on a 30° incline, considering a static friction coefficient of 0.5, is 45.7N.
Explanation:To determine the maximum force that can be applied without moving the crate, we need to consider the effect of static friction, gravity, and the angle of the incline. The weight of the crate (W) is its mass (m) times acceleration due to gravity (g), which equals 5kg*9.8m/s² = 49N. However, this is the weight vertically down, so the force from gravity parallel to the incline is less, and we should multiply W by sin30⁰, getting 24.5N. The normal force (N) on the incline is W*cos30⁰, equal to 42.4N. Therefore, the maximum static friction force (fs) is coef. of static friction (μs) times N, which equals 0.5*42.4N = 21.2N. The max force applied to keep the crate from moving is the sum of the force of gravity and the static friction forces on the incline, which equals 24.5N + 21.2N = 45.7N.
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The rms speed of an oxygen molecule (O2) in a container of oxygen gas is 650 m/s. What is the temperature of the gas?
Answer:
542.06 K
Explanation:
v = rms speed of oxygen molecule = 650 m/s
M = molecular mass of the oxygen molecule = 32 g = 0.032 kg
R = universal gas constant = 8.314 J/(mol K)
T = temperature of the gas
Rms speed of oxygen molecule is given as
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
[tex]650 = \sqrt{\frac{3(8.314)T}{0.032}}[/tex]
T = 542.06 K
Calculate the speed (in m/s) of an electron and a proton with a kinetic energy of 1.25 electron volt (eV). (The electron and proton masses are me = 9.11 ✕ 10−31 kg and mp = 1.67 ✕ 10−27 kg. Boltzmann's constant is kB = 1.38 ✕ 10−23 J/K.)
The speed of an electron with a kinetic energy of 1.25 eV is approximately 1.57 x 10⁶ m/s, and the speed of a proton with the same kinetic energy is approximately 5.29 x 10⁵ m/s.
To calculate the speed of an electron and a proton with a kinetic energy of 1.25 electron volts (eV), we can use the kinetic energy formula and relate it to the speed of the particles. The kinetic energy (KE) of a particle is given by:
KE = (1/2) * m * v²
Where:
KE = kinetic energy
m = mass of the particle
v = speed of the particle
We are given the kinetic energy in electron volts (eV), but we need to convert it to joules (J) since the mass is given in kilograms (kg). The conversion factor is 1 eV = 1.60219 x 10⁻¹⁹ J.
So, the kinetic energy in joules is:
KE = 1.25 eV * 1.60219 x 10⁻¹⁹ J/eV = 2.0027375 x 10⁻¹⁹ J
Now, we can rearrange the kinetic energy formula to solve for the speed (v):
v = √((2 * KE) / m)
For an electron:
Mass of electron (mₑ) = 9.11 x 10⁻³¹ kg
v(electron) = √((2 * 2.0027375 x 10⁻¹⁹ J) / (9.11 x 10⁻³¹ kg))
Calculating this gives us the speed of the electron.
For a proton:
Mass of proton (m_p) = 1.67 x 10⁻²⁷ kg
v(proton) = √((2 * 2.0027375 x 10⁻¹⁹ J) / (1.67 x 10⁻²⁷ kg))
Calculating this gives us the speed of the proton.
Now, let's calculate these speeds.
After performing the calculations, the speed of the electron is approximately 1.57 x 10⁶ m/s, and the speed of the proton is approximately 5.29 x 10⁵ m/s.
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The work function for metallic caesium is 2.24 eV. Calculate the kinetic energy and speed of electrons ejected if a light source of a) 250 nm and b) 600 nm is used.
Answer:
a) [tex]4.362580048\times 10^{-19}\ Joule[/tex]
b)[tex]0.27566898\times 10^{-19}\ Joule[/tex]
Explanation:
a) When, wavelength=λ=250 nm
[tex]\text{Work function of metallic caesium}=2.24\ eV\\=2.24\times 1.6021\times 10^{-19}\\=3.588704\times 10^{-19}\ Joule \ \text{(converting to SI units)}\\\lambda =\text {Wavelength of light}=250\ nm\\Energy\\E=\frac{hc}{\lambda}\\\text{Where h=Plancks constant}=6.62607004\times 10^{-34} m^2kg / s\\\text{c=speed of light}=3\times 10^8\ m/s\\\Rightarrow E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{250\times 10^{-9}}\\\Rightarrow E=0.07951284048\times 10^{-17} Joule\\[/tex]
[tex]\text{Kinetic energy}=\text{E - Work function}\\K.E.=(7.951284048\times 10^{-19})-(3.588704\times 10^{-19})\\K.E.=4.362580048\times 10^{-19}\ Joule\\[/tex]
b) When, λ=600 nm
[tex]E=\frac{hc}{\lambda}\\E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{600\times 10^{-9}}\\\Rightarrow E=3.313030502\times 10^{-19}\\\text{Kinetic energy}=\text{E - Work function}\\K.E.=(3.31303502\times 10^{-19})-(3.588704\times 10^{-19})\quad \text{Work function remains constant}\\K.E.=-0.27566898\times 10^{-19}\ Joule[/tex]
An automobile starter motor draws a current of 1.7 A from a 17.2 V battery when operating at normal speed. A broken pulley locks the armature in position, and the current increases to 12.3 A. What was the back emf of the motor when operating normally? Answer in units of V.
Answer:
14.82 Volts
Explanation:
[tex]V_{back}[/tex] = Back emf of the motor
[tex]V_{battery}[/tex] = battery voltage = 17.2 Volts
[tex]i_{locked}[/tex] = Current in locked condition = 12.3 A
R = resistance
In locked condition, using ohm's law
[tex]R = \frac{V_{battery}}{i_{locked}}[/tex]
[tex]R = \frac{17.2}{12.3}[/tex]
R = 1.4 Ω
[tex]i_{normal}[/tex] = Current in normal condition = 1.7 A
Back emf of the motor is given as
[tex]V_{back}[/tex] = [tex]V_{battery}[/tex] - [tex]i_{normal} R[/tex]
[tex]V_{back}[/tex] = 17.2 - (1.7 x 1.4)
[tex]V_{back}[/tex] = 14.82 Volts
A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle B moves straight away from A to make the distance between them 18.2 mm. What vector force does particle B then exert on A?
Answer:
[tex]F_2 = 1.10 \mu N[/tex]
Explanation:
As we know that the electrostatic force is a based upon inverse square law
so we have
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now since it depends inverse on the square of the distance so we can say
[tex]\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}[/tex]
now we know that
[tex]r_2 = 18.2 mm[/tex]
[tex]r_1 = 12.2 mm[/tex]
also we know that
[tex]F_1 = 2.45 \mu N[/tex]
now from above equation we have
[tex]F_2 = \frac{r_1^2}{r_2^2} F_1[/tex]
[tex]F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)[/tex]
[tex]F_2 = 1.10 \mu N[/tex]
The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Figure below.. Determine the gage pressure of air in the tank if hl -0.2 m, h2 = 0.3 m, and h3 = 0.4 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively.
Answer:
Spongebob: Bye Mr. Krabs! Bye Squidward! BYE SQUIDWARD!
Patrick: (clearly triggered) Why'd you say "bye squidward" twice?
Spongebob: I LiKe SqUiDwArD
Determine the total impedance of an LRC circuit connected to a 10.0- kHz, 725-V (rms) source if L = 36.00 mL, R = 10.00 kΩ, and C = 5.00 nF.
Answer:
10042.6 ohm
Explanation:
f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm
C = 5 nF = 5 x 10^-9 F
XL = 2 x π x f x L
XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm
Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)
Xc = 3184.7 ohm
Total impedance is Z.
Z^2 = R^2 + (XL - Xc)^2
Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2
Z = 10042.6 ohm
A large bottle contains 150 L of water, and is open to the atmosphere. If the bottle has a flat bottom with an area of 2 ft, calculate the absolute pressure at the bottom of the bottle. Give your answer in Pa. Assume the water has a density of 1000 kg/m, and assume the system is at sea level on Earth. Answer: over 100,000 Pa
Answer:
Total pressure exerted at bottom = 119785.71 N/m^2
Explanation:
given data:
volume of water in bottle = 150 L = 0.35 m^3
Area of bottle = 2 ft^2
density of water = 1000 kg/m
Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water
Pressure due to water P = F/A
F, force exerted by water = mg
m, mass of water = density * volume
= 1000*0.350 = 350 kg
F = 350*9.8 = 3430 N
A = 2 ft^2 = 0.1858 m^2
so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2
Atmospheric pressure
At sea level atmospheric pressure is 101325 Pa
Total pressure exerted at bottom = 18460.71 + 101325 = 119785.71 N/m^2
Total pressure exerted at bottom = 119785.71 N/m^2
True False Suppose I have a resistor of some resistance R. If I were to double the length and double the cross-sectional area of the resistor, what is the new resistance?
Explanation:
The resistance of a wire is given by :
[tex]R=\rho\dfrac{l}{A}[/tex]
Where
[tex]\rho[/tex] is the resistivity of the wire
l = initial length of the wire
A = initial area of cross section
If length and the area of cross section of the wire is doubled then new length is l' and A', l' = 2 l and A' = 2 A
So, new resistance of the wire is given by :
[tex]R'=\rho\dfrac{l'}{A'}[/tex]
[tex]R'=\rho\dfrac{l}{A}[/tex]
R' = R
So, the resistance of the wire remains the same on doubling the length and the area of wire.
A sewing machine needle moves in simple harmonic motion with a frequency of 2.5 Hz and an amplitude of 1.27 cm. (a) How long does it take the tip of the needle to move from the highest point to the lowest point in its travel? (b) How long does it take the needle tip to travel a total distance of 11.43 cm?
Answer:
The tip of the needle to move from the highest point to the lowest point in 0.4 sec and the needle tip to travel a total distance in 0.9 sec.
Explanation:
Given that,
Frequency = 2.5 Hz
Amplitude = 1.27 cm
(a). We need to calculate the time
The frequency is the reciprocal of the time.
[tex]f=\dfrac{1}{T}[/tex]
[tex]T=\dfrac{1}{f}[/tex]
[tex]T=\dfrac{1}{2.5}[/tex]
[tex]T=0.4\ sec[/tex]
The time taken from highest point to lowest point
[tex]T'=\dfrac{T}{2}[/tex]
[tex]T'=\dfrac{0.4}{2}[/tex]
[tex]T'=0.2\ sec[/tex]
(b). We need to calculate the time
The time taken in one cycle = 0.4 sec
The distance covered in one sec= 4 times x amplitude
[tex]d=4\times1.27[/tex]
[tex]d=5.08\ m[/tex]
We need to calculate the speed
Using formula of speed
[tex]v=\dfrac{5.08}{0.4}[/tex]
[tex]v=12.7[/tex]
We need to calculate the time
[tex]t=\dfrac{11.43}{12.7}[/tex]
[tex]t= 0.9 sec[/tex]
Hence, The tip of the needle to move from the highest point to the lowest point in 0.4 sec and the needle tip to travel a total distance in 0.9 sec.
The amount of time it took the tip of the needle to move from the highest point to the lowest point in its travel is equal to 0.2 seconds.
Given the following data:
Frequency, F = 2.5 HzAmplitude, A = 1.27 cmHow to calculate the time.Mathematically, the frequency of an object in simple harmonic motion is give by:
[tex]F=\frac{1}{T} \\\\T=\frac{1}{F}\\\\T=\frac{1}{2.5}[/tex]
T = 0.4 seconds.
Now, we can calculate the time it took the needle to move from the highest point to the lowest point:
[tex]t = \frac{T}{2} \\\\t = \frac{0.4}{2}[/tex]
Time, t = 0.2 seconds.
How to calculate the time w.r.t a distance of 11.43 cm.The distance traveled by the needle per seconds is given by:
[tex]d=4A\\\\d=4 \times 1.27\\\\[/tex]
Distance, d = 5.08 cm.
For the speed:
[tex]Speed = \frac{distance}{time} \\\\Speed =\frac{5.08}{0.4}[/tex]
Speed = 12.7 cm/s.
For the time:
[tex]Time = \frac{distance}{speed} \\\\Time =\frac{11.43}{12.7}[/tex]
Time = 0.9 seconds.
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All neutron stars are things that produce intense gravity. All neutron stars are extremely dense objects. Therefore, all extremely dense objects are things that produce intense gravity. Identify the major, minor, and middle terms.
Answer:
Major term is 'things that provide intense gravity'
Minor term is 'extremely dense objects'
Middle term is 'neutron stars'
Explanation:
Major term is given by the predicate part of the conclusionMinor term is given by the subject part of sentence in conclusionMiddle term is given by the subject part and not the conclusionYou have just landed on Planet X. You take out a 100-g ball, release it from rest from a height of 10.0 m, and measure that it takes 2.2 s to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the 100-g ball weigh on the surface of Planet X?
Answer:
Weight in planet X = 0.413 N
Explanation:
Weight = Mass x Acceleration due to gravity.
W = mg
Mass, m = 100 g = 0.1 kg
We have equation of motion s = ut + 0.5 at²
Displacement, s = 10 m
Initial velocity, u = 0 m/s
Time, t = 2.2 s
Substituting
s = ut + 0.5 at²
10 = 0 x 2.2 + 0.5 x a x 2.2²
a = 4.13 m/s²
Acceleration due to gravity, a = 4.13 m/s²
W = mg = 0.1 x 4.13 = 0.413 N
Weight in planet X = 0.413 N
In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.
What is the gravitational acceleration (g)?In physics, gravitational acceleration is the acceleration of an object in free fall within a vacuum (and thus without experiencing drag).
Step 1. Calculate the gravitational acceleration of Planet X.A 100-g (m) ball is released from rest from a height of 10.0 m (s) and it takes 2.2 s (t) for it to reach the ground. We can calculate the gravitational acceleration using the following kinematic equation.
s = 1/2 × g × t²
g = 2 s / t² = 2 (10.0 m) / (2.2 s)² = 4.1 m/s²
Step 2. Calculate the weight (w) of the ball on the surface of Planet X.We will use Newton´s second law of motion.
w = m × g = 0.100 kg × 4.1 m/s² = 0.41 N
In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.
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A capacitor is being charged from a battery and through a resistor of 10 kΩ. It is observed that the voltage on the capacitor rises to 80% of its maximal value in 4 seconds. Calculate the capacitor's capacitance.
Answer:
[tex]C = 2.48 \times 10^{-4} Farad[/tex]
Explanation:
As per the equation of voltage on capacitor we know that
[tex]V = V_{max}(1 - e^{-\frac{t}{\tau}})[/tex]
now we know that voltage reached to its 80% of maximum value in 4 second time
so we will have
[tex]0.80 V_{max} = V_{max}(1 - e^{-\frac{4}{\tau}})[/tex]
[tex]0.20 = e^{-\frac{4}{\tau}}[/tex]
[tex]-\frac{4}{\tau} = ln(0.20)[/tex]
[tex]-\frac{4}{\tau} = -1.61[/tex]
[tex]\tau = 2.48[/tex]
as we know that
[tex]\tau = RC[/tex]
[tex](10 k ohm)(C) = 2.48[/tex]
[tex]C = 2.48 \times 10^{-4} Farad[/tex]
Let's begin with the angular acceleration of a compact disk (CD). To play music the CD must rotate at high speed while a laser reads data encoded in a spiral pattern on the disk. The disc has radius R=6.0cm=0.060m; when data are being read, it spins at 7200 rev/min. What is the CD's angular velocity ω in radians per second? How much time is required for it to rotate through 90∘? If it starts from rest and reaches full speed in 4.0 s, what is its average angular acceleration?
The angular velocity of the CD is 754.4 rad/s. The time it takes for the CD to rotate through 90 degrees is 0.0021 seconds. The average angular acceleration of the CD is 188.6 rad/s².
Explanation:To calculate the angular velocity of the CD, we can convert the given 7200 rev/min to radians per second. Since one revolution is equal to 2π radians, we can use the conversion factor to find the angular velocity. Thus, the angular velocity of the CD is 754.4 rad/s.
To calculate the time it takes for the CD to rotate through 90 degrees, we need to find the fraction of the total rotation that corresponds to 90 degrees. Since a full rotation is 360 degrees or 2π radians, 90 degrees is equal to π/2 radians. We can then use the formula Δθ = ωΔt to find the time it takes, where Δθ is the angle in radians, ω is the angular velocity, and Δt is the time. Rearranging the formula, we have Δt = Δθ/ω. Substituting the values, we get Δt = π/2 / 754.4 = 0.0021 seconds.
The average angular acceleration can be found using the formula α = (ωf - ωi) / Δt, where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and Δt is the time interval. The CD starts from rest and reaches full speed in 4 seconds, so the initial angular velocity is 0. Using the given full speed of 7200 rev/min, we can convert it to radians per second and use it as the final angular velocity. Thus, the average angular acceleration is α = (754.4 rad/s - 0 rad/s) / 4 s = 188.6 rad/s².
Learn more about Angular velocity here:https://brainly.com/question/29557272
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