Explanation:
Given that,
Focal length of a converging lens, f = +25 cm
Size of the object, h = 10 cm
Object distance, u = -30 cm
We need to find the image distance and the size of the image.
Using lens formula, [tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex], v is image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{25}+\dfrac{1}{(-30)}\\\\v=150\ cm[/tex]
Magnification,
[tex]m=\dfrac{v}{u}=\dfrac{h'}{h}[/tex]
h' is the size of image
[tex]h'=\dfrac{vh}{u}\\\\h'=\dfrac{150\times 10}{(-30)}\\\\h'=-50\ cm[/tex]
So, the image distance is 150 cm the size of image is 50 cm.
Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of LaTeX: 190~mm^2190 m m 2. How much charge must be transferred from one plate to the other if 1.1 nJ of energy are to be stored in the plates
Answer:
5.5x 10^-11 C
Explanation:
Pls see attached file
A model airplane has momentum given by p=[(-0.75kg.m/s3)t2 + (3.0kg.m/s)] i + (0.25kg.m/s2)t j. Find the components Fx, Fy, and Fz of the net force on the airplane.
Answer:
[tex]F_x[/tex] = -1.5t
[tex]F_y[/tex] = 0.25
[tex]F_{z}[/tex] = 0
Explanation:
Given equation;
p = [(-0.75 kgm/s³)t² + (3.0 kgm/s)] i + (0.25 kgm/s²)t j.
From Newton's law, the rate of change of momentum of a body is the net force acting on that body. i.e
∑F = [tex]\frac{dp}{dt}[/tex] -----------(i)
Substitute the equation of p into equation (i) and differentiate with respect to t as follows;
∑F = [tex]\frac{dp}{dt}[/tex] = [tex]\frac{d| [(-0.75)t^{2} + (3.0)] i + (0.25)t j|}{dt}[/tex]
∑F = [tex]\frac{dp}{dt}[/tex] = [tex][-1.5t + 0]i + 0.25j[/tex]
∑F = [tex][-1.5t + 0]i + 0.25j[/tex]
But
∑F = [tex]F_xi + F_yj + F_zk[/tex]
Where;
[tex]F_x, F_y, F_z[/tex] are the components of the net force in the x, y and z direction respectively.
=> [tex]F_xi + F_yj + F_zk[/tex] = [tex][-1.5t + 0]i + 0.25j[/tex] = [tex]-1.5ti + 0.25j[/tex]
=> [tex]F_x[/tex] = -1.5t
=> [tex]F_y[/tex] = 0.25
=> [tex]F_{z}[/tex] = 0
A cube has one corner at the origin and the opposite corner at the point (L,L,L)(L,L,L). The sides of the cube are parallel to the coordinate planes. The electric field in and around the cube is given by
Answer:
Net charge = E• b • L^3.
Explanation:
NB: here, the symbol representation of the flux is "p" = electric Field • Area(dot Product).
So, we will take a look at the flux through -x face, through x face and through -y face, through y face and through - z face and through z face.
(1). Starting from -z and z faces which are the back and front faces of the cube:
Thus, We have that the flux,p = 0 for -z and z.
(2). Recall that we are given that E = =(a+bx)i^+cj^.
Thus, p_-y = (a + bx)i + cj (-j) (L^2)
Where y = 0
p_-y = -cL^2.
Obviously for p_j, we will have cL^2 and y = L
(3). For p_-x = =(a + bx)i + cj (-i) (L^2).
p_-x = -aL^2
Where x = 0.
When x = L and p_x = (a + bL)L^2.
This, adding all together gives Net charge = E • b • L^3.
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current of 26.0 A in the +x direction. The second wire lies along the x-axis. The wires exert attractive forces on each other, and the force per unit length on each wire is 295 µN/m. What is the y-value (in m) of the line in the xy-plane where the total magnetic field is zero?
Answer:
The y-value is z = 0.759 m
Explanation:
From the question we are told that
The position of the first y-axis is [tex]y_1 = 0.300 \ m[/tex]
The current on the first wire is [tex]I_ 1 = 26.0 \ A[/tex]
The force per unit length on each wire is [tex]\frac{F}{l} = 295 \mu N/m = 295 * 10^{-6} \ N/m[/tex]
Generally the force per unit length on first wire is mathematically represented as
[tex]\frac{F}{l} = \frac{\mu_o * I_1 * I_2 }{2*\pi* y_1}[/tex]
Where [tex]\mu _o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values
[tex]295 *10^{-6} = \frac{ 4\pi * 10^{-7} * 26.0 * I_2 }{2 *3.142* 0.300}[/tex]
[tex]I_2 = \frac{295 *10^{-6 } * 0.300 * 2* 3.142 }{ 4\pi * 10^{-7} * 26 }[/tex]
[tex]I_2 = 17.0 \ A[/tex]
Now the at the point where the magnetic field is zero the magnetic field of each wire are equal , let that point by z meters from the second wire on the y-axis so
[tex]\frac{\mu_o I_2}{2 * \pi * y_1} = \frac{\mu_o I_1}{2 * \pi * (y_1-z)}[/tex]
[tex]I_2 (y_1 - z) = I_1 * y_1[/tex]
substituting values
[tex]17.0 ( 0.300 - z) = 26 * 0.300[/tex]
z = 0.759 m
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1950 K. (kb is Boltzmann's constant, 1.38x10-23 J/K).
Answer:
The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m
The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m
Explanation:
Thermal kinetic energy of electron or proton = KE
∴ KE = 3kbT/2
given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K
so we substitute
KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2
kE = 4.0365 × 10⁻²⁰ ( is the kinetic energy for both electron and proton at temperature T )
Now we know that
mass of electron M'e = 9.109 × 10⁻³¹
mass of proton M'p = 1.6726 × 10⁻²⁷
We also know that
KE = p₂ / 2m
from the equation, p = √ (2mKE)
{ p is momentum, m is mass }
de Broglie wavelength = β
so β = h / p = h / √ (2mKE)
h = Planck's constant = 6.626 × 10⁻³⁴
∴ βe = h / √ (2m'e × KE)
βe = 6.626 × 10⁻³⁴ / √ (2 × 9.109 × 10⁻³¹ × 4.0365 × 10⁻²⁰ )
βe = 6.626 × 10⁻³⁴ / √ 7.3536957 × 10⁻⁵⁰
βe = 6.626 × 10⁻³⁴ / 2.71176984642871 × 10⁻²⁵
βe = 2.443422 × 10⁻⁹ m
βp = h / √ (2m'p ×KE)
βp = 6.626 × 10⁻³⁴ / √ (2 × 1.6726 × 10⁻²⁷ × 4.0365 × 10⁻²⁰ )
βp = 6.626 × 10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶
βp = 6.626 × 10⁻³⁴ / 1.16201978468527 × 10⁻²³
βp = 5.702140 × 10⁻¹¹ m
Sergio has made the hypothesis that "the more time that passes, the farther away a person riding a bike will be." Do the data in the table below support his hypothesis? A. Yes, the data support the hypothesis. B. No, the data support the opposite of the hypothesis. C. The data show no relationship between the time passed and the distance.
Answer:
Option A
Explanation:
Given that
Distance = Speed / Time
So, they are in inverse relation.
Such that when the time passes, the distance from the reacing point will become less and vice versa.
So, Yes! The more time that passes, the farther away a person riding a bike will be.
Two gliders with different masses move toward each other on a frictionless air track. After they colllide, glider B has a final v of 2 m/s. What is the final velocity of glider A
Answer:
2m/s
Explanation:
According to conservation of momentums, it states that the sum of collision of bodies before collision is equal to the sum of their momentum after collision. Both objects will move with the same velocity after collision.
According to the question, we were told that after they collide, glider B has a final velocity of 2 m/s. Since both bodies (Glider A and B) will move with the same velocity after collision according to the conservation of momentum, this means glider A will also have a final velocity of 2m/s like. Glider B.
An RC circuit is connected across an ideal DC voltage source through an open switch. The switch is closed at time t = 0 s. Which of the following statements regarding the circuit are correct?
a) The capacitor charges to its maximum value in one time constant and the current is zero at that time.
b) The potential difference across the resistor and the potential difference across the capacitor are always equal.
c) The potential difference across the resistor is always greater than the potential difference across the capacitor.
d) The potential difference across the capacitor is always greater than the potential difference across the resistor
e) Once the capacitor is essentially fully charged, There is no appreciable current in the circuit.
Answer:
e)
Explanation:
In an RC series circuit, at any time, the sum of the voltages through the resistor and the capacitor must be constant and equal to the voltage of the DC voltage source, in order to be compliant with KVL.
At= 0, as the voltage through the capacitor can't change instantaneously, all the voltage appears through the resistor, which means that a current flows, that begins to charge the capacitor, up to a point that the voltage through the capacitor is exactly equal to the DC voltage, so no current flows in the circuit anymore, and the charge in the capacitor reaches to its maximum value.
An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is 11) held in place, how fast will the electron be moving when it is 3.00 cm from the proton?
Answer:
Vf = 1.43 m/s
Explanation:
From Coulomb's Law, the electrostatic force between electron and proton is given as:
F = kq₁q₂/r²
F = Electrostatic force = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
q₂ = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
r = distance between electron and proton = 9 cm = 0.09 m
Therefore,
F = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(0.09 m)²
F = 2.84 x 10⁻²⁶ N
but, from Newton's second law:
F = 2.84 x 10⁻²⁶ N = ma
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
a = acceleration of electron = ?
Therefore,
2.84 x 10⁻²⁶ N = (1.67 x 10⁻²⁷ kg)(a)
a = 2.84 x 10⁻²⁶ N/1.67 x 10⁻²⁷ kg
a = 17.03 m/s²
Now, we apply 3rd equation of motion to the motion of electron from a distance of 9 cm to 3 cm near to the proton:
2as = Vf² - Vi²
where,
s = distance traveled = 9 cm - 3 cm = 6 cm = 0.06 m
Vf = speed of electron when it is 3 cm from proton = ?
Vi = Initial speed of electron = 0 m/s
Therefore,
2(17.03 m/s²)(0.06 m) = Vf² - (0 m/s)²
Vf = √2.04 m²/s²
Vf = 1.43 m/s
Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface temperature 1000K and emissivity 0.6:
Answer:
6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex]
Explanation:
From Wien's displacement formula;
Q = e A[tex]T^{4}[/tex]
Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.
The emissive intensity = [tex]\frac{Q}{A}[/tex] = e[tex]T^{4}[/tex]
Given from the question that: e = 0.6 and T = 1000K, thus;
emissive intensity = 0.6 × [tex](1000)^{4}[/tex]
= 0.6 × 1.0 × [tex]10^{12}[/tex]
= 6.0 × [tex]10^{11}[/tex] [tex]\frac{W}{m^{2} }[/tex]
Therefore, the emissive intensity coming out of the surface is 6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex].
Three m^3 of air in a rigid, insulated container fitted with a paddle wheel is initially at 295 K, 200 kPa. The air receives 1546 kJ of work from the paddle wheel. Assuming the ideal gas model, determine for the air the mass, in kg, final temperature, in K, and the amount of entropy produced, in KJ/K
Answer:
1. 7.08Kg
2. 311K
3. 0.268KJ/K
Explanation:
See attached file
A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns of wire. When running, the solenoid produced a field of 1.4 TT in the center. Given this, how large a current does it carry?
Answer:
The current is [tex]I = 2042\ A[/tex]
Explanation:
From the question we are told that
The length of the solenoid is [tex]l = 2.2 \ m[/tex]
The radius is [tex]r_i = 30 \ cm = 0.30 \ m[/tex]
The number of turn is [tex]N = 1200 \ turns[/tex]
The magnetic field is [tex]B = 1.4 \ T[/tex]
The magnetic field produced is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{l }[/tex]
making [tex]I[/tex] the subject
[tex]I = \frac{B * l}{\mu_o * N }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with values [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]
substituting values
[tex]I = \frac{1.4 * 2.2 }{4\pi *10^{-7} * 1200 }[/tex]
[tex]I = 2042\ A[/tex]
The pressure and temperature at the beginning of compression of an air-standard Diesel cycle are 95 kPa and 300 K, respectively. At the end of the heat addition, the pressure is 7.2 MPa and the temperature is 2150 K. Determine a) the compression ratio. b) the cutoff ratio. c) the thermal efficiency of the cycle. d) the mean effective pressure, in kPa.
Answer:
A.33.01
B.2.081
C.66%
Explanation:
See attached file pls
The phenomenon of magnetism is best understood in terms ofA) the existence of magnetic poles.B)the magnetic fields associated with the movement of charged particles.C)gravitational forces between nuclei and orbital electrons.D) electrical fluid
Answer:
A) the existence of magnetic poles.Explanation:
Magnetism is defined as the ability of a magnet to attract magnetic substance to itself. Such magnet has the ability of being magnetized. A magnet is known to possess poles which are the north poles and south poles. The presence of this poles is what makes them possess the properties of a magnet. An ordinary steel bar doesn't have the properties of a magnet unless it is magnetized and when you are trying to magnetize a steel bar, you are invariably introducing the magnetic poles.
According to the law of magnetism, like poles repel but unlike poles attract. From the above explanation, it can be concluded that the phenomenon of magnetism is best understood interns of existence of magnetic poles. This poles are called the north and the south poles.
Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies. Express your answer in terms of Ki and Kf.
The net work done (W) on a particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies is equal to [tex]W = K_f - K_i[/tex]
The net work done (W) can be defined as the work done in moving an object by a net force, which is the vector sum of all the forces acting on the object.
According to Newton's Second Law of Motion, the net work done (W) on an object or physical body is equal to the change in the kinetic energy possessed by the object or physical body.
Mathematically, the net work done (W) on an object or physical body is given by the formula:
[tex]W =\Delta K_E\\\\W = K_f - K_i[/tex]
Where:
W is the net work done.[tex]K_f[/tex] is the initial kinetic energy.[tex]K_i[/tex] is the final kinetic energy.Read more: https://brainly.com/question/22599382
g When a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A. What is the resistance of the glowing thin-filament bulb
Answer:
R = 42.67 ohms
Explanation:
It is given that, a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A.
It means that when two batteries are connected in series, then the total voltage is 3.2 volts
Let R is the resistance of the glowing thin-filament bulb. So, using Ohm's law we get :
[tex]V=IR\\\\R=\dfrac{V}{I}[/tex]
So,
[tex]R=\dfrac{3.2}{0.075}\\\\R=42.67\ \Omega[/tex]
So, the resistance of the bulb is 42.67 ohms.
"A parcel moving in a horizontal direction with speed v0 = 13 m/s breaks into two fragments of weights 1.4 N and 1.9 N, respectively. The speed of the larger piece remains horizontal immediately after the separation and increases to v1.9 = 29 m/s. Find the necessary speed and direction of the smaller piece immediately after the separation. (Assume the initial direction of the parcel is positive. Indicate the direction with the sign of your answer.)"
Answer:
the smaller particle moves with speed of 8.706 m/s in the opposite direction to the bigger particle.
Explanation:
Speed of the original particle = 13 m/s
We designate particles as A and B
The final weights of the component particles are
Particle A = 1.4 N
particle B = 1.9 N
The speed of the larger piece (particle B) = 29 m/s
We know that weight is the product of a body's mass and acceleration due to gravity g which is equal to 9.81 m/s^2, therefore, masses of the particles are
particle A = 1.4/9.81 = 0.143 kg
Particle B = 1.9/9.81 = 0.194 kg
The momentum of a body is the product of its mass and its velocity i.e
P = mv
This means that the mass of the particle before splitting is
0.143 kg + 0.194 kg = 0.337 kg
Momentum of the initial whole particle = mv
==> 0.337 x 13 = 4.381 kg-m/s
The bigger particle B remains horizontal, and has a momentum of
mv = 0.194 x 29 = 5.626 kg-m/s
According to the conservation of momentum, the total initial momentum of a system must be equal tot the total final momentum of the system.
Initial total momentum of the system = 4.381 kg-m/s (momentum of original particle before splitting)
Final total momentum of the system = Total momentum of the particles after splitting = 5.626 kg-m/s + ( 0.143 kg x [tex]V_{B}[/tex])
where [tex]V_{B}[/tex] is the velocity of smaller particle A
final total momentum of the system = 5.626 + 0.143[tex]V_{B}[/tex]
Equating the two momenta of the system, we'll have
4.381 = 5.626 + 0.143[tex]V_{B}[/tex]
4.381 - 5.626 = 0.143[tex]V_{B}[/tex]
-1.245 = 0.143[tex]V_{B}[/tex]
[tex]V_{B}[/tex] = -1.245/0.143 = -8.706 m/s
The negative sign indicates that the smaller particle moves in the opposite direction to the bigger particle
A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter together have mass m = 159 kg.
(a) What speed in m/s does the scooter have if the driver feels weightlessness (i.e., has an apparent weight of zero) at the top of the hill?
(b) If the driver is traveling at the speed above and encounters a hill with a radius 2r,
Answer:
68.585m/sec , 779.1 N
Explanation:
To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.
From centripetal motion.
F =( mv^2)/2
But since we are dealing with weightlessness
r = 480m
g = 9.8m/s^2
M also cancels, so forget M.
V^2 = Fr
V = √ Fr
V =√ (9.8 x 480) = 4704
= 68.585m/sec.
b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960
= 4.9m/sec^2.
Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)
159kg × ( 9.8-4.9)
159kg × 4.9
= 779.1N
A) The speed of the scooter at which the driver will feel weightlessness is;
v = 68.586 m/s
B) The apparent weight of both the driver and the scooter at the top of the hill is;
F_net = 779.1 N
We are given;
Mass; m = 159 kg
Radius; r = 480 m
A) Since it's motion about a circular hill, it means we are dealing with centripetal force.
Formula for centripetal force is given as;
F = mv²/r
Now, we want to find the speed of the scooter if the driver feels weightlessness.
This means that the centripetal force would be equal to the gravitational force.
Thus;
mg = mv²/r
m will cancel out to give;
v²/r = g
v² = gr
v = √(gr)
v = √(9.8 × 480)
v = √4704
v = 68.586 m/s
B) Now, he is travelling with speed of;
v = 68.586 m/s
And the radius is 2r
Let's first find the centripetal acceleration from the formula; α = v²/r
Thus; α = 4704/(2 × 480)
α = 4.9 m/s²
Now, since he has encountered a hill with a radius of 2r up the slope, it means that the apparent weight will now be;
F_app = m(g - α)
F_net = 159(9.8 - 4.9)
F_net = 779.1 N
Read more at; https://brainly.com/question/9017432
The dimension of a room has 5.31m by 7.6m. Find the limits of accuracy for the area of the room
Explanation:
Se supone que si es 5.31 x 7.6 los límites son 38.98 ahora si fuera en suma mueves los puntos dos veces a la izquierda la sumatoria seria la siguiente .00531 + .0076 la respuesta seria
.00607
Receiver maxima problem. When the receiver moves through one cycle, how many maxima of the standing wave pattern does the receiver pass through
The number of maxima of the standing wave pattern is two.
Maxima problem:At the time when the receiver moves via one cycle so here two maximas should be considered. At the time when the two waves interfere by traveling in the opposite direction through the same medium so the standing wave pattern is formed.
learn more about the waves here: https://brainly.com/question/3004869?referrer=searchResults
A commercial aircraft is flying westbound east of the Sierra Nevada Mountains in California. The pilot observes billow clouds near the same altitude as the aircraft to the south, and immediately turns on the "fasten seat belt" sign. Explain why the aircraft experiences an abrupt loss of 500 meters of altitude a short time later.
Answer:
Billow clouds provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents.
Explanation:
Billow clouds are created in regions that are not stable in a meteorological sense. They are frequently present in places with air flows, and have marked vertical shear and weak thermal separation and inversion (colder air stays on top of warmer air). Billow clouds are formed when two air currents of varying speeds meet in the atmosphere. They create a stunning sight that looks like rolling ocean waves. Billow clouds have a very short life span of minutes but they provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents, which although may not affect us on the ground but is a concern to aircraft pilots. The turbulence due to the Billow wave is the only logical explanation for the loss of 500 m in altitude of the plane.
Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K.
Complete Question
Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).
Answer:
The intensity is [tex]I = 3.535 *10^{-6} \ W/m^2[/tex]
Explanation:
From the question we are told that
The temperature is [tex]T = 2.81 \ K[/tex]
Now According to Stefan's law
[tex]Power(P) = \sigma * A * T^4[/tex]
Where [tex]\sigma[/tex] is the Stefan Boltzmann constant with value [tex]\sigma = 5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}[/tex]
Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as
[tex]I = \frac{P}{A}[/tex]
=> [tex]I = \frac{\sigma * A * T^4}{A}[/tex]
=> [tex]I = \sigma * T^4[/tex]
substituting values
[tex]I = 5.67 *10^{-8} * (2.81)^4[/tex]
[tex]I = 3.535 *10^{-6} \ W/m^2[/tex]
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.25 m/s2. Determine the orbital period of the satellite.
Answer:
118 minutes( 2 hours approximately )
Explanation:
Here, we are interested in calculating the orbital period of the satellite
Please check attachment for complete solution
Answer:
T = 7101 s = 118.35 mins = 1.9725 hrs
Explanation:
To solve the question, we apply the formula for gravitational acceleration
a = GM/r², where
a = acceleration due to gravity
G = gravitational constant
M = mass of the earth
r = distance between the satellite and center of the earth
Now, if we make r, subject of formula, we have
r = √(GM/a)
Recall also, that
a = v²/r, making v subject of formula
v = √ar
If we substitute the equation of r into it, we have
v =√a * √r
v =√a * √[√(GM/a)]
v = (GM/a)^¼
Again, remember that period,
T = 2πr/v, we already have v and r, allow have to do is substitute them in
T = 2π * √(GM/a) * [1 / (GM/a)^¼]
T = 2π * (GM/a³)^¼
T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼
T = 6.284 * [(3.982*10^14) / 244.140]^¼
T = 6.284 * (1.63*10^12)^¼
T = 6.284 * 1130
T = 7101 s
T = 118.35 mins
T = 1.9725 hrs
11. A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length
Answer:
The frequency is [tex]f_n = 257.1 \ Hz[/tex]
Explanation:
From the question we are told that
The third harmonic frequency of the tight guitar string is [tex]f_3 = 540 \ Hz[/tex]
Let the original length be L
Then the length at which it is fingered is 0.7 L
Generally the fundamental is mathematically represented as
[tex]f = \frac{v_s}{ 2L}[/tex]
Now when it finger at 70% it original length is
[tex]f_n = \frac{v}{2 * (0.7 L)}[/tex]
[tex]f_n = \frac{v}{1.4 L}[/tex]
Here v the velocity of sound
So
[tex]\frac{f_n}{f} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]
Also the fundamental frequency for the original length can also be represented as
[tex]f = \frac{f_3}{3}[/tex]
substituting values
[tex]f = \frac{540}{3}[/tex]
[tex]f = 180 \ Hz[/tex]
So
[tex]\frac{f_n}{180} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]
=> [tex]f_n =\frac{180}{0.7}[/tex]
=> [tex]f_n = 257.1 \ Hz[/tex]
The fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1 Hz.
What is the frequency?Frequency is defined as the number of repetitions of a wave occurring waves in 1 second.
f is the frequency of tight guitar string = 540 Hz
Let's call the original length L.
The amount of time it is fingered is then 0.7 L.
In general, the fundamental frequency is expressed mathematically as;
[tex]\rm f = \frac{v_0}{2L} \\\\[/tex]
For the given conditions;
[tex]\rm f_n=\frac{v}{2 \times 0.7L} \\\\ \rm f_n=\frac{v}{1.4L}[/tex]
The ratio of the frequency is;
[tex]\rm \frac{f_n}{f} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }[/tex]
Also, the fundamental frequency for the original length can also be represented as;
[tex]\rm f= \frac{f'}{3} \\\\ f=\frac{540}{3} \\\\ \rm f=180\ Hz[/tex]
On putting the given data;
[tex]\rm \frac{f_n}{180} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }\\\\ \rm f_n=\frac{180}{0.7}\\\\\ \rm f_n=257.1\ Hz[/tex]
Hence the fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1 Hz.
To learn more about the frequency reference the link;
https://brainly.com/question/14926605
if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c
Answer:
please brainliest!!!
Explanation:
V1/√T1 =V2/√T2
V1 = 331m/s
T1 = 0°C = 273k
V2 = ?
T2 = 35°c = 308k
331/√273 = V2/√308331/16.5 = V2/17.520.06 = V2/17.5V2 = 20.06 x 17.5 V2 = 351.05m/sIn a shot-put competition, a shot moving at 15m/s has 450J of mechanical kinetic energy. What is the mass of the shot? Please help, and include the formula for the answer and a step by step explanation
Answer:
Mass of shot (m) = 4 kg
Explanation:
Given:
Velocity (v) = 15 m/s
Mechanical kinetic energy (K.E) = 450 J
Find:
Mass of shot (m) = ?
Computation:
Mechanical kinetic energy (K.E) = 1/2mv²
Mechanical kinetic energy (K.E) = [1/2](m)(15)²
450 = [1/2](m)(15)²
900 = 225 m
Mass of shot (m) = 4 kg
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)
Answer:
1/4F
Explanation:
We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.
So F α Qq
But if it is now half the initial charges, then
F α (1/2)Q *(1/2)q
F α (1/4)Qq
Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.
Thus the answer will be 1/4F
A sinusoidal electromagnetic wave emitted by a mobile phone has a wavelength of 34.8 cm and an electric-field amplitude of 5.70×10−2 V/m at a distance of 210 m from the phone.
Calculate
(a) the frequency of the wave;
(b) the magnetic-field amplitude;
(c) the intensity of the wave.
Answer:
a) [tex] f = 8.62 \cdot 10^{8} Hz [/tex]
b) [tex] B = 1.9 \cdot 10^{-10} T [/tex]
c) [tex] I = 4.30 \cdot 10^{-6} W/m^{2} [/tex]
Explanation:
a) The frequency (f) of the wave can be found as follows:
[tex] f = \frac{c}{\lambda} [/tex]
Where:
c: is the speed of light = 3x10⁸ m/s
λ: is the wavelength = 34.8 cm
[tex] f = \frac{3 \cdot 10^{8} m/s}{0.348 m} = 8.62 \cdot 10^{8} Hz [/tex]
b) The magnetic-flied amplitude (B) is:
[tex] B = \frac{E}{c} [/tex]
Where:
E: is the electric field amplitude = 5.70x10⁻² V/m
[tex] B = \frac{E}{c} = \frac{5.70 \cdot 10^{-2} V/m}{3 \cdot 10^{8} m/s} = 1.9 \cdot 10^{-10} T [/tex]
c) The intensity of the wave (I) is the following:
[tex] I = \frac{E*B}{2\mu_{0}} [/tex]
Where:
μ₀: is the permeability of free space = 1.26x10⁻⁶ m*kg/(s²A²)
[tex] I = \frac{E*B}{2\mu_{0}} = \frac{5.70 \cdot 10^{-2} V/m*1.9 \cdot 10^{-10} T}{2*1.26 \cdot 10^{-6} m*kg/((s^{2}A^{2})} = 4.30 \cdot 10^{-6} W/m^{2} [/tex]
I hope it helps you!
The frequency of the wave is [tex]8.62\times 10^8\rm\;Hz[/tex], the magnetic-field amplitude is [tex]1.9\times 10^{-10}\rm\;T[/tex], and the intensity of the wave is [tex]4.298\rm\;W/m^2[/tex].
Given information:
A mobile phone emits electromagnetic radiation.
The wavelength of the wave is [tex]\lambda=34.8[/tex] cm.
The electric-field amplitude is [tex]5.70\times10^{-2}[/tex] V/m.
Phone is at a distance of 210 m.
The speed of the electromagnetic wave is [tex]c=3\times 10^8[/tex] m/s.
(a)
Now, the frequency of the wave will be calculated as,
[tex]f=\dfrac{c}{\lambda}\\f=\dfrac{3\times 10^8}{0.348}\\f=8.62\times 10^8\rm\;Hz[/tex]
(b)
The magnetic-field amplitude can be calculated as,
[tex]B=\dfrac{E}{c}\\B=\dfrac{5.70\times10^{-2}}{3\times 10^8}\\B=1.9\times 10^{-10}\rm\;T[/tex]
(c)
[tex]\mu_0[/tex] is the permeability of the vacuum. [tex]\mu_0=1.26\times10^{-6} \rm\;\frac{kg-m}{(A^2s^2)}[/tex]
The intensity of the wave can be calculated as,
[tex]I=\dfrac{BE}{2\mu_0}\\I=\dfrac{1.9\times10^{-10 }\times5.7\times10^{-2}}{2\times1.26\times10^{-6}}\\I=4.298\rm\;W/m^2[/tex]
Therefore, the frequency of the wave is [tex]8.62\times 10^8\rm\;Hz[/tex], the magnetic-field amplitude is [tex]1.9\times 10^{-10}\rm\;T[/tex], and the intensity of the wave is [tex]4.298\rm\;W/m^2[/tex].
For more details, refer to the link:
https://brainly.com/question/1393179
A raspberry has a red color because it _____ red light. A. emits B. reflects C. absorbs D. transmits
Answer:
B. reflects
Explanation:
Red objects appear red because they reflect red light.
Answer:
B
Explanation:
just did the quiz
A plane is flying horizontally with a constant speed of 55 .0 m/s when it drops a
rescue capsule. The capsule lands on the ground 12.0 s later.
c) How would your answer to part b) iii change if the constant speed of the plane is
increased? Explain.
Answer:
therefore horizontal displacement changes increasing with linear velocity
Explanation:
Since the plane flies horizontally, the only speed that exists is
v₀ₓ = 55.0 m / s
the time is the time it takes to reach the floor, which we can find because the speed on the vertical axis is zero
y =y₀ + v₀ t - ½ g t2
0 = I₀ + 0 - ½ g t2
t = √ 2y₀o / g
time is that we use to calculate the x-axis displacement
The distance it travels to reach the floor is
x = v t
x = 55 12
x = 660 m
When the speed horizontally the time remains the same and 120
x ’= v’ 12
therefore horizontal displacement changes increasing with linear velocity