Answer:
gasoline zone P_net = 6860 Pa
Water zone P_net = 26460 Pa
Force direction is out of tank
Explanation:
The pressure is defined
P = F / A
F = P A
let's write Newton's equation of force
F_net = F_int - F_ext
P_net A = (P_int - P_ext) A
The P_ext is the atmospheric pressure
P_ext = P₀
the pressure inside is
gasoline zone
P_int = P₀ + ρ' g h'
water zone
P_int = P₀ + ρ' g h' + ρ_water h_water
we substitute
Zone with gasoline
P_net = ρ' g h'
P_net = 700 9.8 1
P_net = 6860 Pa
Water zone
P_net = rho’ g h’ + rho_water g h_water
P_net = 6860 + 1000 9.8 2
P_net = 26460 Pa
To find the explicit value of the force, divide by a specific area.
Force direction is out of tank
2. As a pendulum swings, its energy is constantly converted between kinetic
energy and potential energy. At which point in its swing does a pendulum have
the greatest potential energy and the least kinetic energy? *
O
A. at point F
O
B. at point G
O
C. at point H
O
D. at point J
answer fast
Answer:
at point F
Explanation:
To know the point in which the pendulum has the greatest potential energy you can assume that the zero reference of the gravitational energy (it is mandatory to define it) is at the bottom of the pendulum.
Then, when the pendulum reaches it maximum height in its motion the gravitational potential energy is
U = mgh
m: mass of the pendulum
g: gravitational constant
The greatest value is obtained when the pendulum reaches y=h
Furthermore, at this point the pendulum stops to come back in ts motion and then the speed is zero, and so, the kinetic energy (K=1/mv^2=0).
A) answer, at point F
Answer:
at point F
Explanation:
If 1.0 joule of work is required to move a charge of 1.0 coulomb between two points is an electric field the potential difference between these two point is?
Answer:1V
Explanation: simply put it
V=W/q
=1/1
=1V
The potential difference between these two point is 1V which can be determined from the equation of potential difference that involves work done and charge.
What is Potential difference?Potential difference is the work done per unit charge. A potential difference of 1 V means that 1 joule of work is done per coulomb of charge. Potential difference in a circuit is measured using a voltmeter which is placed in parallel with the component of interest in the circuit.
Given:
Work = 1.0 Joule
Charge = 1.0 coulomb
V=W/q
V=1/1
V=1 V
Thus, the potential difference is one Volts.
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An infinitely long wire carrying a current of 4.5 A is bent at the origin and extends in both the y-direction and x-direction. (In other words, the wire makes a 90 degree angle with the x axis and y axis and has a bend at the origin) Find the magnetic field at the point x = 3 cm and y = 2 cm.
Answer:
6.444 × [tex]10^{-5}[/tex] T
Explanation:
Find the given attachments
The magnitude of the magnetic field at the point [tex](x = 3 \text{ cm}, y = 2 \text{ cm})[/tex] is approximately [tex]\( 1.08 \times 10^{-5} \text{ T} \)[/tex].
The magnitude of the magnetic field [tex]\( B \)[/tex] at a distance [tex]\( r \)[/tex] from a long straight wire carrying a current [tex]\( I \)[/tex] is given by:
[tex]\[ B = \frac{\mu_0 I}{2\pi r} \][/tex]
where [tex]\( \mu_0 \)[/tex] is the permeability of free space [tex]\( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \)[/tex].
For the horizontal wire, the distance from the wire to the point is [tex]\( y = 2 \text{ cm} = 0.02 \text{ m} \)[/tex]. Thus, the magnetic field due to the horizontal wire is:
[tex]\[ B_x = \frac{\mu_0 (4.5 \text{ A})}{2\pi (0.02 \text{ m})} \][/tex]
Next, consider the vertical section of the wire (along the y-axis). Similarly, the magnetic field due to this section at a perpendicular distance [tex]\( r \)[/tex] is given by the same formula. Here, the distance from the wire to the point is [tex]\( x = 3 \text{ cm} = 0.03 \text{ m} \)[/tex].
The magnetic field due to the vertical wire is:
[tex]\[ B_y = \frac{\mu_0 (4.5 \text{ A})}{2\pi (0.03 \text{ m})} \][/tex]
The total magnetic field [tex]\( B \)[/tex] at the point is the vector sum of [tex]\( B_x \)[/tex] and [tex]\( B_y \)[/tex]. We can use Pythagoras' theorem to find the magnitude of the resultant magnetic field:
[tex]\[ B = \sqrt{B_x^2 + B_y^2} \][/tex]
Substituting the expressions for [tex]\( B_x \)[/tex] and [tex]\( B_y \)[/tex] we get:
[tex]\[ B = \sqrt{\left(\frac{\mu_0 (4.5 \text{ A})}{2\pi (0.02 \text{ m})}\right)^2 + \left(\frac{\mu_0 (4.5 \text{ A})}{2\pi (0.03 \text{ m})}\right)^2} \][/tex]
Plugging in the value of [tex]\( \mu_0 \)[/tex] and simplifying, we find:
[tex]\[ B = \sqrt{\left(\frac{4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} (4.5 \text{ A})}{2\pi (0.02 \text{ m})}\right)^2 + \left(\frac{4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} (4.5 \text{ A})}{2\pi (0.03 \text{ m})}\right)^2} \][/tex]
[tex]\[ B = \sqrt{\left(\frac{18 \times 10^{-7} \text{ T}\cdot\text{m}}{0.02 \text{ m}}\right)^2 + \left(\frac{18 \times 10^{-7} \text{ T}\cdot\text{m}}{0.03 \text{ m}}\right)^2} \][/tex]
[tex]\[ B = \sqrt{\left(9 \times 10^{-6} \text{ T}\right)^2 + \left(6 \times 10^{-6} \text{ T}\right)^2} \][/tex]
[tex]\[ B = \sqrt{81 \times 10^{-12} \text{ T}^2 + 36 \times 10^{-12} \text{ T}^2} \][/tex]
[tex]\[ B = \sqrt{117 \times 10^{-12} \text{ T}^2} \][/tex]
[tex]\[ B \approx 1.08 \times 10^{-5} \text{ T} \][/tex]
A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity ratio of 0.5. The center-to-center distance is 9 ft. The angular speed of the small pulley is 1750 rev/min as it delivers 2 hp. The service is such that a service factor Ks of 1.25 is appropriate. (a) Find Fc, Fi, (F1)a, and F2, assuming operation at the maximum tension limit. (b) Find Ha, nfs, and belt length.
Answer:
a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,
b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm
Explanation:
Given that:
γ= 9.5 kN/m³ = 9500N/m3
b = 6 inches = 0.1524 m
t = 0.0013 mm
d = 2 inches = 0.0508 m
n = 1750 rpm
[tex]H_{nom}=2hp=1491.4W[/tex]
L = 9 ft = 2.7432 m
Ks = 1.25
g = 9.81 m/s²
a)
[tex]w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m[/tex]
[tex]V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s[/tex]
[tex]F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N[/tex]
[tex](F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N[/tex]
[tex]T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm[/tex]
[tex]F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N[/tex]
[tex]F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N[/tex]
b)
[tex]H_a=1491*1.25=1863.75W[/tex]
[tex]n_f_s=\frac{H_a}{H_{nom}K_S }=1[/tex]
dip = [tex]\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm[/tex]
A) The values of Fc, Fi, ( F1)a and F2
Fc = 4.15 N Fi = 435.65 N( F1 )a = 640 N F2 = 239.6 NB ) The values of Ha, nfs and belt length
Ha = 1863.75 Nnfs = 1belt length = 11.8 mmGiven data :
Angular velocity ratio of large pulley = 0.5
width of polyamide ( b ) = 6 inches = 0.1524 m
pulley width ( d ) = 2 inches = 0.0508 m
center to center distance ( L ) = 9 ft = 2.7432 m
angular speed of small pulley ( n ) = 1750 rev/min
power of small pulley ( Hnom ) = 2 hp = 1491.4 Watts
Ks = 1.25
γ = 9500 N/m³
g = 9.81 m/s²
t = 0.0013 mm
A) Determine the values of Fc, Fi, F1a and F2
w = γ * b * t
= 9500 * 0.1524 * 0.0013 = 1.88 N/m
V = [tex]\frac{\pi dn}{60}[/tex] = ( π * 0.0508 * 1750 ) / 60 = 4.65 m/s
T = 10.17 Nm ( calculated value )
i) Fc = [tex]\frac{wV^2}{g}[/tex] ---- ( 1 )
Insert values into equation ( 1 )
Fc = ( 1.88 * 4.65² ) / 9.81
= 4.15 N
ii) Fi = [tex]\frac{(F1)a+F2 }{2}[/tex] - Fc ----- ( 2 )
where ( F1 )a = 0.1524 * 6000 * 0.7 * 1 = 640 N
F2 = ( F1 )a - [tex]\frac{2T}{D}[/tex] = 640 - [tex]\frac{2 *10.17}{0.0508}[/tex] = 239.6 N
back to equation ( 2 )
Fi = [ ( 640 + 239.6 ) / 2 ] - Fc
= 435.65 N
iii) ( F1 )a = b * Fa * Cp * Cv
= 0.1524 * 6000 * 0.7 * 1 = 640 N
iv) F2 = ( F1 )a - [tex]\frac{2T}{D}[/tex]
= 640 - [tex]\frac{2 *10.17}{0.0508}[/tex] = 239.6 N
B) Find the values of Ha, nfs and belt length
i) Ha = 1491 * 1.25
= 1863.75 W
ii) nfs = [tex]\frac{Ha}{Hnom*Ks } = 1[/tex]
iii) belt length ( dip ) = [tex]\frac{L^2w}{8fi}[/tex]
= ( 2.7432 * 1.88 ) / 435.65 = 11.8 mm
Hence we can conclude that the values of the variables in the question is as listed above
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The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts of kinetic energy loss by the electrons. Suppose instead a similar experiment is done with a very cold gas of atomic hydrogen, so that all of the hydrogen atoms are initially in the ground state. If the kinetic energy of an electron is 11.1 eV just before it collides with a hydrogen atom, how much kinetic energy will the electron have just after it collides with and excites the hydrogen atom? Kfinal = eV
Answer:
the final kinetic energy is 0.9eV
Explanation:
To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:
[tex]E_n=\frac{-13.6eV}{n^2}[/tex]
you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is
[tex]E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV[/tex]
-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:
[tex]E_{k}=11.1eV-10.2eV=0.9eV[/tex]
The final kinetic energy the electron will have just after it collides with and excites the hydrogen atom is; E_k = 0.9 eV
The formula for the energy of the electron of hydrogen atom in its nth orbit is given by the formula;
E_n = -13.6eV/n²
Where n is the principal quantum number.
Thus;
For n = 1;
E_1 = -13.6/1²
E_1 = -13.6 eV
At n = 2;
E_2 = -13.6/2²
E_2 = -3.4 eV
Thus, the energy that the electron looses is;
E_2 - E_1 = -3.4 - (-13.6)
E_2 - E_1 = 10.2 eV
We are told that the kinetic energy of an electron is 11.1 eV just before it collides with a hydrogen atom.
Thus, the final kinetic energy the electron will have just after it collides with and excites the hydrogen atom is;
E_k = 11.1 - 10.2
E_k = 0.9 eV
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When a conducting plate swings through a magnetic field, which is true?
a. Energy is transferred from mechanical energy to thermal energy.
b. Energy is transferred from mechanical energy to the magnetic field.
c. Energy is transferred from the magnetic field to thermal energy.
d. Energy is transferred from mechanical energy to thermal energy and then back to mechanical energy.
e. Energy is transferred from the magnetic field to mechanical energy.
Answer:
Energy is transferred from mechanical energy to thermal energy and then back to mechanical energy ( D )
Explanation:
when conducting plate swings through a magnetic field, energy is transferred from mechanical energy to thermal energy and then back to mechanical energy and this is due to the induction of eddy currents during the process of that the conducting plate swings through the magnetic field.
An eddy current is induced in conductor when the magnetic field in which the conductor is located varies.
Answer:
Energy is transferred from mechanical energy to thermal energy. (A)
Explanation:
As the conducting plate is swinging, we know it posseses some form of mechanical energy (either K or U). Because the plate is conducting, charge is transfered, yielding energy transfer to thermal.
A 10.0-g bullet is fired into a stationary block of wood having mass 5.00 kg. The bullet embeds 10 pts into the block and the speed of the block-and-bullet after the collision is 0.600 m/s. Find a) the original speed of the bullet, b) the mechanical energies of the block-bullet system before and after the collision, c) the percentage of mechanical energy lost to heat.
Answer:
A. The initial velocity of the bullet is [tex]= 300.6m/s[/tex]
B. Mechanical energy of the system before and after collision: 451.80 J, 0.9018 J
C. Percentage of K.E lost to heat is = 99.8 %
Explanation:
From conservation of linear momentum,
[tex](m_{1}v_{1} +m_{2}v_{2})= (m_{1}+m_{2})v[/tex]
let the mass of the block be m1 and velocity = v1
let the mass of the bullet be m2 and velocity = v2
Let the final velocity of the system be v.
A. Plugging our parameters into the equation, we have:
[tex][(5 \times 0) +(0.01\times v_{2})]= 5.01 \times 0.6[/tex]
[tex]v_{2}=\frac{3.006}{0.01}= 300.6m/s[/tex]
Hence, the initial velocity of the bullet is [tex]= 300.6m/s[/tex]
B. The mechanical energies of the system exist in form of kinetic energy.
I. Kinetic energy of the system before collision:
[tex]0.5 \times 5\times 0^{2} + 0.5 \times 0.01 \times 300.6^{2}= 451.80 J[/tex]
II. Kinetic energy after collision:
[tex]0.5\times 5.01 \times 0.6^{2}= 0.9018 J[/tex]
C. Change in Mechanical Energy = [tex]451.8 - 0.9018 J= 450.9J[/tex]
[tex]\frac{450.9}{451.8} \times 100 =99.8%[/tex]
Percentage of K.E lost to heat is = 99.8 %
Suppose a large power plant generates electricity at 12.0 kV. Its old transformer once converted this voltage to 315 kV. The secondary coil of this transformer is being replaced so that its output can be 730 kV for more efficient cross-country transmission on upgraded transmission lines.
Randomized Variables
Vi = 315 kV
V2 = 730 kV
(a) What is the ratio of turns in the new secondary to the number of turns in the old secondary?
(b) What is the ratio of new current output to the old current output for the same power input to the transformer?
Answer:
2.320.43Explanation:
12.0 kv primary voltage
315 kv secondary voltage ( converted voltage ) V1 or Vo
v2 (Vn)= 730 kv new secondary voltage
a) Ratio of turns in 730 kv to turns in 315 kv
[tex]\frac{Vn}{Vo} = \frac{Nn}{No}[/tex] = [tex]\frac{730}{315}[/tex] therefore the ratio of turns = 2.317 ≈ 2.32
B) ratio of the new current output to the old current output for the same power input to the transformer
since the power input is the same
[tex]\frac{In}{Io} = \frac{\frac{Vp}{Vn} }{\frac{Vp}{Vo} }[/tex] equation 1
Vp = primary voltage, Vo = old secondary voltage, Vn = new secondary voltage, In = new secondary current, Io = old secondary current
therefore equation 1 becomes
[tex]\frac{In}{Io} = \frac{Vo}{Vn}[/tex] = 315 / 730 = 0.43
Final answer:
The ratio of turns in the new secondary compared to the old secondary is 2.32, and the ratio of the new current output to the old current output for the same power input to the transformer is approximately 0.432.
Explanation:
Transformer Coil Turn Ratio and Current Output Ratio
In a transformer, the ratio of the voltage across the secondary coil to the voltage across the primary coil is equal to the ratio of the number of turns in the secondary coil to the number of turns in the primary coil. For the same power input to a transformer, the current output will be inversely proportional to the voltage output. Thus, when the voltage output is increased, the current output decreases proportionally.
(a) Ratio of Turns in the New Secondary to the Old Secondary
To find the ratio of turns in the new secondary compared to the old secondary, we use the ratio of the new voltage to the old voltage:
New ratio of turns (N2) / Old ratio of turns (N1) = V2 / V1 = 730 kV / 315 kV = 2.32.
(b) Ratio of the New Current Output to the Old Current Output
For the same power input, P = VI, where P is the power, V is the voltage and I is the current. Keeping the power constant and increasing the voltage output results in a decrease in current. Therefore, the ratio of the new current to the old current is inversely proportional to the ratio of the new voltage to the old voltage:
New current (I2) / Old current (I1) = V1 / V2 = 315 kV / 730 kV = 0.432.
Betelgeuse is 100,000 times more luminous than our sun, which means that it releases an estimated 3.846 x 1031 W of luminous light. If an exoplanet existed with the same mass (5.972 x 10^24 kg) and twice the radius of earth (1.27 x 10^7 m) and half the distance (7.5 x 10^10 m) from Betelgeuse what would the intensity of the light at the surface of that "earth" look like? a) What is the intensity of Betelgeuse at the "earth’s" surface?
Answer:
5.4 × 10⁸ W/m²
Explanation:
Given that:
The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W
the mass of the exoplanet = 5.972 × 10²⁴ kg
radius of the earth = 1.27 × 10⁷ m
half the distance (i.e radius r ) = 7.5 × 10¹⁰ m
a) What is the intensity of Betelgeuse at the "earth’s" surface?
The Intensity of Betelgeuse can be determined by using the formula:
[tex]Intensity \ I = \frac{P}{4 \pi r^2}[/tex]
[tex]I = \frac{3.846*10^{31}}{4 \pi (7.5*10^{10})^2}[/tex]
I = 544097698.8 W/m²
I = 5.4 × 10⁸ W/m²
Answer:
1.97*10^14 W/m^2
Explanation:
To find the intensity of the light emitted by Betelgeuse you taken into account that the light from Betelgeuse expands spherically in space.
You use the following formula:
[tex]I=\frac{P}{A}=\frac{P}{4\pi r^2}[/tex] (1)
I: intensity of light
r: distance from Betelgeuse to the exoplanet = (1/2)*7.5*10^10m
P: power = 100,000*3.486*10^31 W
By replacing the values of the parameters in (1) you obtain:[tex]I=\frac{100000(3.486*10^{31}W}{4\pi (0.5*7.5*10^{10}m)^2}=1.97*10^{14}\frac{W}{m^2}[/tex]
6) A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5 × 10−3 s, radius 10.0 km, and mass 2.8 × 1030 kg. The pulsar’s rotational period will increase over time due to the release of electromagnetic radiation, which doesn’t change its radius but reduces its rotational energy. (a) What is the angular momentum of the pulsar? (b) Suppose the angular velocity decreases at a rate of 10−14 rad/s2 . What is the torque on the pulsar?
Answer:
a) L = 2.10x10⁴⁰ kg*m²/s
b) τ = 1.12x10²⁴ N.m
Explanation:
a) The angular momentum (L) of the pulsar can be calculated using the following equation:
[tex] L = I \omega [/tex]
Where:
I: inertia momentum
ω: angular velocity
First we need to calculate ω and I. The angular velocity can be calculated as follows:
[tex] \omega = \frac{2 \pi}{T} [/tex]
Where:
T: is the period = 33.5x10⁻³ s
[tex] \omega = \frac{2 \pi}{T} = \frac{2 \pi}{33.5 \cdot 10^{-3} s} = 187.56 rad/s [/tex]
The inertia moment of the pulsar can be calculated using the following relation:
[tex] I = \frac{2}{5}mr^{2} [/tex]
Where:
m: is the mass of the pulsar = 2.8x10³⁰ kg
r: is the radius = 10.0 km
[tex] I = \frac{2}{5}mr^{2} = \frac{2}{5}2.8\cdot 10^{30} kg*(10\cdot 10^{3} m)^{2} = 1.12 \cdot 10^{38} kg*m^{2} [/tex]
Now, the angular momentum of the pulsar is:
[tex] L = I \omega = 1.12 \cdot 10^{38} kg*m^{2}*187.56 rad/s = 2.10 \cdot 10^{40} kg*m^{2}*s^{-1} [/tex]
b) If the angular velocity decreases at a rate of 10⁻¹⁴ rad/s², the torque of the pulsar is:
[tex] \tau = I*\alpha [/tex]
Where:
α: is the angular acceleration = 10⁻¹⁴ rad/s²
[tex]\tau = I*\alpha = 1.12 \cdot 10^{38} kg*m^{2} * 10^{-14} rad*s^{-2} = 1.12 \cdot 10^{24} N.m[/tex]
I hope it helps you!
Final answer:
To find the angular momentum of the pulsar, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Explanation:
To find the angular momentum of the pulsar, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid sphere can be calculated using the formula I = (2/5)mr², where m is the mass and r is the radius.
Using the given radius and mass of the pulsar, we can calculate the moment of inertia.
Once we have the moment of inertia, we can use the given period of the pulsar to calculate the angular velocity.
To find the torque on the pulsar, we can use the formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
The angular acceleration can be calculated using the given rate at which the angular velocity is decreasing.
Using these formulas and the given values, we can find the angular momentum and torque of the pulsar.
The density of air under normal conditions is about 3 1.2 kg/ m . For a wind speed of 10 m/s, find (a) the actual power carried in the wind; (b) the maximum power (Betz limit) extractable by a wind turbine with a blade area of 10 m2; and (c) the actual power extracted by a wind turbine with a blade area of 10 m2 and a power coefficient of 0.46.
[100 POINTS] Which of the following do not make their own energy through nuclear fusion? Select all that apply.
A. giant star
B. protostar
C. dwarf star
D. black hole
E. neutron star
F. main sequence star
Answer:
The correct answer is :
A . Giant star
B . proto star
F .main sequence star
Explanation:
hope this helps
Which of the following statements is NOT true? When a light wave encounter a boundary where they must bounce off of a surface with a larger index of refraction, the phase of the wave flips by pi A monochromatic light source is one that consists of many colors The intensity of the center maximum is the largest in a single slit interference pattern The intensity of the center maximum is the largest in a double slit interference pattern
Answer: A monochromatic light source is one that consists of many colors
Explanation:
Monochromatic comes from "mono - one, chromatic - color", so a monochromatic wave has only one wavelength, this means that it has only one color (while it may have different shades or tones). So the correct option is the second one "A monochromatic light source is one that consists of many colors "
The other 3 statements are true.
The incorrect statement is that 'A monochromatic light source is one that consists of many colors', because a monochromatic light source indeed emits light of a single wavelength or color. The other claims regarding phase shift on reflection, and the location of highest intensity in single and double slit interference patterns are accurate.
Explanation:The statement that is NOT true among the given options is: 'A monochromatic light source is one that consists of many colors'. In fact, a monochromatic light source emits light of a single wavelength or color.
Further addressing the related concepts: When a light wave encounters a boundary where it must reflect off a surface with a larger index of refraction, the phase of the wave indeed flips by pi. This is an important concept in understanding wave reflection and refraction.
Regarding the intensity of the central maximum in both single and double slit interference patterns, it is accurate to say that the intensity of the center maximum is the most significant. This consequence is a direct outcome of the interference pattern produced within these experiments. In the central maximum, most wave paths align constructively, resulting in the highest intensity.
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TRY
An electric oven of resistance 2002
drawn a current of 10A. it works 3
hours a daily. Calculate the weekly
bill when energy is $0.48 per kwh
Hint: P=1?R
Answer:
$2018.016
Explanation:
Applying,
P = I²R.............. Equation 1
Where P = power, drawn by the oven, I = current drawn by the oven, R = resistance of the oven.
Given: I = 10 A, R = 2002 Ω
Substitute this values into equation 1
P = 10²(2002)
P = 100(2002)
P = 200200 W.
P = 200.2 kW
If the electric oven works 3 hours daily,
Then the total time it works in a week = 3×7 = 21 hours.
E(kWh) = P(kW)×t(h)
E = 200.2×21
E = 4204.2 kWh.
If 1 kWh of energy cost $0.48,
Then, 4204.2 kWh = $4204.2(0.48) = $2018.016
What needs to be emphasized in this step when reading?
"Turn the two-bottle assembly over, and place the assembly on a table with the filled bottle on top. Watch the water slowly drip down into the lower bottle.... Now, rapidly rotate the bottles in a horizontal circle a few times. Observe the formation of a funnel-shaped vortex..."
a.
rapidly rotate the bottles in a horizontal circle
c.
both A and B
b.
Watch the water
d.
neither A nor B
Answer:
I would say A since it's important part of the instructions.
Explanation:
Answer:
A
Explanation:
A distant object will form what type of image on the retina of an emmetropic eye?
Which of the following statements ARE correct?
Group of answer choices:
A) If net torque is zero, the object will be moving at constant velocity or remain static.
B) If the net torque is zero, the object will not rotate or rotate at constant angular velocity.
C) If net torque is zero, the angular momentum is conserved.
D) If net force is zero, the object will be rotate at constant angle velocity or not rotate at all.
Answer:
Option A, C and D are correct statements.
Explanation:
Torque is the twisting effect of a force which causes an object to acquire angular acceleration.
The direction of the torque depends on the direction of the force on the axis. The SI unit for torque is the Newton-meter.
The net torque is the sum of the individual torques.
If the net torque on a rotatable object is zero then:
1. It will be in rotational equilibrium and the angular velocity of the body remains constant.
2. It will be in rotational equilibrium and not able to acquire angular acceleration. Thus, it will not be able to acquire angular velocity.
3. Angular momentum will be conserved.
If the forces on an object are balanced, the net force is zero.
If there net force on an object is zero, its speed and direction of motion do not change, including if it is at rest, the object will not accelerate and the velocity will remain constant.
By considering the basic fundamentals of Torque, the statements (A), (C) and (D) are correct.
The given problem is based on the concept and fundamentals of Torque. Torque is the twisting effect of a force which causes an object to acquire angular acceleration.
The direction of the torque depends on the direction of the force on the axis. The SI unit for torque is the Newton-meter. The net torque is the sum of the individual torques.
If the net torque on a rotatable object is zero then:
It will be in rotational equilibrium and the angular velocity of the body remains constant. It will be in rotational equilibrium and not able to acquire angular acceleration. Thus, it will not be able to acquire angular velocity. Angular momentum will be conserved. If the forces on an object are balanced, the net force is zero. If there net force on an object is zero, its speed and direction of motion do not change, including if it is at rest, the object will not accelerate and the velocity will remain constant.Thus, we can conclude that by considering the basic fundamentals of Torque, the statements (A), (C) and (D) are correct.
Learn more about the fundamentals of Torque here:
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A what is a negatively charged Atoms
Answer:
Negatively charged or positively charged atom is generally termed as ANION/CATION. Short Explanation: If an atom loses electrons or gains protons, it will have a net positive charge and is called a Cation. If an atom gains electrons or loses protons, it will have a net negative charge and is called an Anion.
Explanation:
You throw a small toy ball that is covered with suction cups at the center of your glass patio door. When it hits it sticks to the glass, and because the door was not latched, it causes the door to swing open with an angular velocity of 0.19 rad/s. If the ball has a mass of 120 g and the patio door can be treated as a uniform box that is 2.3 m high, 1.0 m wide, and 0.030 m thick with a mass of 7.5 kg what speed did you throw the ball at
Answer:
Explanation:
moment of inertia of the door M = 1 /3 m ( l² + b² + d² )
= 1 / 3 x 7.5 x ( 2.3² + 1² + .03² )
= 1 / 3 x 7.5 x 6.2909
I = 15.7272.
moment of inertia of the door + small toy
15.7272 + .12 x .5²
= 15.7272 + .03
= 15.7572
Applying conservation of angular momentum law
mvr = M ω
m is mass of the toy thrown with velocity v at distance r from the axis , M is moment of inertia of door + toy and ω is angular velocity ith which door opens.
.12 x v x .5 = 15.7572 x .19
v = 49.89 m /s .
Problem 9) If you had light of wavelength 700nm when emitted, what would the wavelength be for an observer who is a red shift z = 0.05 from that source?
Answer:
The answer is 735 Nanometer
Explanation:
Solution to this problem
Given that:
Solution)
The Redshift is categorized by the difference related between wavelength(emitted) and wavelength (observer).
Now,
The wavelength can be computed as :
1+z=\lambda (observer)\div \lambda (emitted)
hence,
1 + 0.05 = λ(observer) / 700 nm
Gives
= 1.05*700 = 735 nm
Therefore, The wavelength from that source for an observer would be 735 nanometer
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.80m away from the slits.
Part A
Which laser has its first maximum closer to the central maximum?
Part B
What is the distance \Delta _y__max-max_ between the first maxima (on the same side of the central maximum) of the two patterns?
Express your answer in meters.
Part C
What is the distance \Delta _y__max-max_ between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
Express your answer in meters.
Answer:
A) first laser
B) 0.08m
C) 0.64m
Explanation:
To find the position of the maximum you use the following formula:
[tex]y=\frac{m\lambda D}{d}[/tex]
m: order of the maximum
λ: wavelength
D: distance to the screen = 4.80m
d: distance between slits
A) for the first laser you use:
[tex]y_1=\frac{(1)(d/20)(4.80m)}{d}=0.24m\\[/tex]
for the second laser:
[tex]y_2=\frac{(1)(d/15)(4.80m)}{d}=0.32m[/tex]
hence, the first maximum of the first laser is closer to the central maximum.
B) The difference between the first maximum:
[tex]\Delta y=y_2-y_1=0.32m-0.24m=0.08m=8cm[/tex]
hence, the distance between the first maximum is 0.08m
C) you calculate the second maximum of laser 1:
[tex]y_{m=2}=\frac{(2)(d/20)(4.80m)}{d}=0.48m[/tex]
and for the third minimum of laser 2:
[tex]y_{minimum}=\frac{(m+\frac{1}{2})(\lambda)(D)}{d}\\\\y_{m=3}=\frac{(3+\frac{1}{2})(d/15)(4.80m)}{d}=1.12m[/tex]
Finally, you take the difference:
[tex]1.12m-0.48m=0.64m[/tex]
hence, the distance is 0.64m
3. A hydraulic lift is used to jack a 950-kg car 45 cm off the floor. The diameter of the output piston is 23 cm, and the input force is 375 N. a. What is the area of the input piston? (10pts) b. What is the work done in lifting the car 45 cm? (10pts) c. If the input piston moves 15 cm in each stroke, how high does the car move up for each stroke? (10pts) d. How many strokes are required to jack the car up 45 cm? (10pts) e. Show that the energy is conserved. (10pts)
Answer:
a. A input = 0.001669 m²
b. W= 4.193775 KJ
c. h output = 0.60357 cm
d. n = 74.55638 strokes
e. 4.1937 N = 4.1937 N
Explanation:
Hydraulic jacks lift loads using the force created by the pressure in the cylinder chamber by applying small effort. It works on Pascal's principles which explains that the pressure at a certain level and through a mass of fluid at rest is the same in all the directions.
Parameters given:
Diameter of the output piston, d = 0.23 m
Mass of the car, m= 950 kg
Force applied at the input piston, f = 375 N
Height, h = 0.45 m
(a) Finding the area of the input piston:
First, we use Pascal's principle to find the area
(f ÷ A output) ÷ (f ÷ A input)
Where A= area
g = 9.81
A output = πd² ÷ 4
(f ÷ (πd² ÷ 4)) = (f ÷ A input)
[(950 x 9.81) ÷ ((3.14 x 0.23²) ÷ 4) ] = 375 ÷ A input
9319.5 ÷ 0.0415 = 375 ÷ A input
A input = 0.001669 m²
(b) Finding the work done in lifting the car 45 cm
Work done, W = force, f x distance ( which in this case is height, h)
= (950 x 9.81) x 0.45
W= 4.193775 KJ
(c) Finding how the car move up for each stroke if the input piston moves 15 cm in each stroke.
W output = W input
F x h output = F x h input
= (950 x 9.81) x h output = 375 x 0.15
h output = 0.0060357 m
h output = 0.60357 cm
(d) Finding the number of strokes that are required to jack the car up 45 cm
n = h ÷ h output
n = 45 ÷ 0.60357 cm
n = 74.55638 strokes
(e) How the energy is conserved
W output = W input
F x h output = F x h input x n
(950 x 9.81) x 0.45 = 375 x 0.15 x 74.55638
4.1937 N = 4.1937 N
The area of the input piston will be 0.00167 square meters, the work done in lifting the car will be 4193.78 J, the height moved by car in each stroke will be 0.6035 cm, the number of strokes required to lift the car by 45 cm will be 74.57, and the energy conservation is justified.
Given information:
Mass of the car is [tex]m=950[/tex] kg
The lift of the car is [tex]h=45[/tex] cm.
The diameter of the output piston D is 23 cm.
The input force f is 375 N.
Let the output force be F and the input piston diameter be d.
Hydraulic lift follows Pascal's principle.
(a)
So, the area of the input piston will be calculated as,
[tex]\dfrac{F}{\frac{\pi}{4}D^2}=\dfrac{f}{\frac{\pi}{4}d^2}\\\dfrac{\pi}{4}d^2=\dfrac{f\frac{\pi}{4}D^2}{mg}\\\dfrac{\pi}{4}d^2=A=0.00167\rm\;m^2[/tex]
(b)
The work done in lifting the car will be,
[tex]W=mgh\\W=950\times 9.81\times0.45\\W=4193.78\rm\;J[/tex]
(c)
The input piston moves 15 cm in each stroke.
The height h' moved by car in each stroke will be calculated as,
[tex]mgh'=f\times 0.15\\h'=0.006035\rm\;m=0.6035\rm\;cm[/tex]
(d)
The number of strokes required to lift the car by 45 cm will be,
[tex]n=\dfrac{h}{h'}\\n=\dfrac{45}{0.6035}\\n=74.57[/tex]
(e)
The energy is conserved when output work is equal to input work.
So, the energy conservation can be verified as,
[tex]W_i=W_o\\f\times0.15\times n=mgh\\375\times 0.15\times74.57=950\times 9.81\times0.45\\4194.28\approx4194.1[/tex]
The input and output works are approximately equal. The value isn't exactly equal because of the rounding-off.
So, energy conservation is justified.
Therefore, the area of the input piston will be 0.00167 square meters, the work done in lifting the car will be 4193.78 J, the height moved by car in each stroke will be 0.6035 cm, the number of strokes required to lift the car by 45 cm will be 74.57, and the energy conservation is justified.
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What phase difference between two otherwise identical traveling waves, moving in the same direction along a stretched string, will result in the combined wave having an amplitude 1.5 times that of the common amplitude of the two combining waves? Express your answer in (a) degrees, (b) radians, and (c) as a fraction of the wavelength.
Answer:
a) 82.8°
b)1.44 rad
c)0.23λ
Explanation:
Wave function form for each wave will be
-> wave 1:
[tex]y_{1}[/tex](x,t)=[tex]y_{m[/tex]sin( kx - ωt)
-> wave 2:
[tex]y_{2[/tex](x,t)=[tex]y_{m[/tex]sin( kx - ωt +φ)
The summation of above two functions is the resultant wave.
Y(x,t)=[tex]y_{m[/tex]sin( kx - ωt)+[tex]y_{m[/tex]sin( kx - ωt +φ)
Using the trigonometric addition formula for sin i.e
sin(A)+ sin(B) = 2 sin(A+B/2) cos(A-B/2)
Y(x,t)=2[tex]y_{m[/tex]cos(φ/2) sin(kx - ωt +φ)
When comparing to the general wave form, the amplitude is 2[tex]y_{m[/tex]cos(φ/2)m
Also, [tex]y_{m[/tex]cos(φ/2)= 1.5 [tex]y_{m[/tex]
φ/2=[tex]cos^{-1}[/tex](1.5/s)
(a) φ= 82.8°
(b)φ= 82.8π/180 =>1.44 rad
(c) one complete wave is 2π radians
Therefore, for two waves, φ/2π= 1.44/2π => 0.23 fraction of a complete wave from each other i.e they are separated by 0.23λ
The switch is moved to position A at t = 0. a) What is the time constant in position A b ) How long does it take the capacitor voltage to reach 9V? Assume the initial voltage of the capacitor is 0 V. c) What is the time constant when Switch is moved to position B? d) How long after moving the switch to position B does it take for the current to reach -6 ma (miliamp)?
Answer:
b
Explanation:
A ball of mass 1.00 kg is thrown at a door with velocity v1x = 12.0 m/s. The rectangular door has a mass of 30.0 kg and is 1.00 m wide. The ball strikes the door perpendicular to it, a horizontal distance 75.0 cm away from the hinges, and causes the door to swing open with an angular speed of 1.20 rad/s. What is the velocity v2x of the ball after striking the door? (The moment of inertia of a thin rectangular plate with mass M and dimensions a × b and the axis along the edge of length b is I = 1/3Ma2.)"
Answer:
The velocity of the ball after collision is [tex]v_{2x} = -4 m/s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m_b = 1.00\ kg[/tex]
The velocity of the ball is [tex]v_{1x}= 12.0 \ m/s[/tex]
The mass of the rectangular door is [tex]m_d = 30 \ kg[/tex]
The width of the door is [tex]a = 1.00 \ m[/tex]
The distance of impact from the hinge is [tex]L = 75 \ cm = \frac{75}{100} = 0.75 \ m[/tex]
The angular speed of the door is [tex]w = 1.20 \ rad/s[/tex]
So the moment of inertia of the door is given from the question as
[tex]I = \frac{1}{3} M a^2[/tex]
substituting values
[tex]I = \frac{1}{3} * 30 * (1)[/tex]
[tex]I = 10 \ kg \cdot m^2[/tex]
According to the law of angular momentum conservation
[tex]L_i = L_f[/tex]
Where [tex]L_i[/tex] is the initial angular momentum of the system(the door and the ball) which is mathematically represented as
[tex]L_i = m_b * v_{1x} + Iw_i[/tex]
so [tex]w_i[/tex] is the initial angular speed of the door which is zero
So
[tex]L_i = m_b * v_{1x}[/tex]
[tex]L_f[/tex] is the final angular momentum of the system(the door and the ball) which is mathematically represented as
[tex]L_f = I w + m_b v_{2x} * L[/tex]
So
[tex]m_b * v_{1x} = I w + m_b v_{2x} * L[/tex]
Substituting values
[tex]1 * 12 * 0.75 = 10* 1.2 * v_{2x} * 0.75[/tex]
[tex]v_{2x} = -4 m/s[/tex]
The negative sign show a reversal in the balls direction
A single slit of width 0.50 mm is illuminated with light of wavelength 500 nm, and a screen is placed 120 cm in front of the slit. Find the widths (lengths, not angles)of(a) the central maximum,(b) the first maximum beyond the central maximum, and(c) the second maximum beyond the central maximum.
Answer:
a) 2.4 mm
b) 1.2 mm
c) 1.2 mm
Explanation:
To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:
[tex]m\lambda=asin\theta[/tex]
a: width of the slit
λ: wavelength
m: order of the minimum
for little angles you have:
[tex]y=\frac{m\lambda D}{a}[/tex]
y: height of the mth minimum
a) the width of the central maximum is 2*y for m=1:
[tex]w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm[/tex]
b) the width of first maximum is y2-y1:
[tex]w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm[/tex]
c) and for the second maximum:
[tex]w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm[/tex]
A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?
Complete Question
A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 [tex]\Omega[/tex]. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300[tex]\Omega[/tex] . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?
Answer:
a
The additional resistance is [tex]R_z = 4.4 \Omega[/tex]
b
The rate at which internal energy increase at the supply is [tex]Z_1 = 32 W[/tex]
c
The rate at which internal energy increase in the battery is [tex]Z_1 = 32 W[/tex]
d
The rate at which internal energy increase in the added series resistance is [tex]Z_3 = 70.4 W[/tex]
e
the increase rate of the chemically energy in the battery is [tex]C = 48 W[/tex]
Explanation:
From the question we are told that
The open circuit voltage is [tex]V = 40.0V[/tex]
The internal resistance is [tex]R = 2 \Omega[/tex]
The emf of each battery is [tex]e = 6.00 V[/tex]
The internal resistance of the battery is [tex]r = 0.300V[/tex]
The charging current is [tex]I = 4.00 \ A[/tex]
Let assume the the additional resistance to to added to the circuit is [tex]R_z[/tex]
So this implies that
The total resistance in the circuit is
[tex]R_T = R + 2r +R_z[/tex]
Substituting values
[tex]R_T = 2.6 +R_z[/tex]
And the difference in potential in the circuit is
[tex]E = V -2e[/tex]
=> [tex]E = 40 - (2 * 6)[/tex]
[tex]E = 28 V[/tex]
Now according to ohm's law
[tex]I = \frac{E}{R_T}[/tex]
Substituting values
[tex]4 = \frac{28}{R_z + 2.6}[/tex]
Making [tex]R_z[/tex] the subject of the formula
So [tex]R_z = \frac{28 - 10.4}{4}[/tex]
[tex]R_z = 4.4 \Omega[/tex]
The increase rate of internal energy at the supply is mathematically represented as
[tex]Z_1 = I^2 R[/tex]
Substituting values
[tex]Z_1 = 4^2 * 2[/tex]
[tex]Z_1 = 32 W[/tex]
The increase rate of internal energy at the batteries is mathematically represented as
[tex]Z_2 = I^2 r[/tex]
Substituting values
[tex]Z_2 = 4^2 * 2 * 0.3[/tex]
[tex]Z_2 = 9.6 \ W[/tex]
The increase rate of internal energy at the added series resistance is mathematically represented as
[tex]Z_3 = I^2 R_z[/tex]
Substituting values
[tex]Z_3 = 4^2 * 4.4[/tex]
[tex]Z_3 = 70.4 W[/tex]
Generally the increase rate of the chemically energy in the battery is mathematically represented as
[tex]C = 2 * e * I[/tex]
Substituting values
[tex]C = 2 * 6 * 4[/tex]
[tex]C = 48 W[/tex]
A hollow metal ball hangs on a long massless string. The device is driven by a small side-to-side oscillator with a constant very small amplitude and an adjustable frequency. The driver begins to oscillate at a very low frequency. The driver's frequency is then made to increase slowly and steadily. The frequency of the driver continues to increase regardless of what happens to the ball, eventually reaching very high frequencies. Predict what will happen to the ball over time as the driver frequency slowly increases.
Answer:
If the frequency is further increased (in a gradual manner), the amplitude of the oscillation of the ball would decrease.
Explanation:
As the frequency of driven force increases, the amplitude of the oscillation of the ball increases. This type of oscillation is referred to as forced oscillation.
At a particular higher frequency, the amplitude of oscillation becomes maximum. This frequency is referred to as the resonant frequency.
If the frequency is further increased (in a gradual manner), the amplitude of the oscillation of the ball would decrease.
Choose the appropriate word choices below that correctly complete the blanks in the following three sentences: At the beginning, the object is at the origin and moving to the left. In the first half of its motion, the object moves to the left with ________ velocity. Then, the motion of the object changes. In the second half of its motion, the object________ due to a ________ acceleration that is ________ in magnitude.
Answer:
Constant
Accelerates
Constant
Positive
An engineer is designing a small toy car that a spring will launch from rest along a racetrack. She wants to maximize the kinetic energy of the toy car when it launches from the end of a compressed spring onto the track, but she can make only a slight adjustment to the initial conditions of the car. The speed of the car just as it moves away from the spring onto the track is called the launch speed. Which of the following modifications to the car design would have the greatest effect on increasing the kinetic energy of the car? Explain your reasoning.
Decrease the mass of the car slightly.
Increase the mass of the car slightly.
Decrease the launch speed of the car slightly.
Increase the launch speed of the car slightly.
Answer:
we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically
Explanation:
Kinetic energy is
K = ½ m v²
the speed of the expensive we can find it r
v² = v₀² + 2 a x
we can find acceleration with Newton's second law
F = m a
a = F / m
F= cte
substitute in the velocity equation
v² = v₀² + 2 F/m x
let's substitute in the kinetic energize equation
K = ½ m (v₀² + 2 F/m x)
K = ½ m v₀² + f x
we see that the kinetic energy depends on two tomines
in January in these systems the force for launching is constant, which is why decreasing the mass increases the speed of the vehicle and therefore increases the kinetic energy
As the launch speed increases the initial energy increases quadratically
we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically