Complete Question
A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.
(a)
What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding
(b) What wavelength is detected by a person on the platform as the train approaches?
Answer:
a
[tex]\Delta f = 81.93 \ Hz[/tex]
b
[tex]\lambda_1 = 0.867 \ m[/tex]
Explanation:
From the question we are told that
The speed of the train is [tex]v_t = 39.6 m/s[/tex]
The frequency of the train horn is [tex]f_t = 350 \ Hz[/tex]
Generally the speed of sound has a constant values of [tex]v_s = 343 m/s[/tex]
Now according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as
[tex]f_1 = f * \frac{v_s}{v_s - v_t}[/tex]
substituting values
[tex]f_1 = 350 * \frac{343 }{343-39.6}[/tex]
[tex]f_1 = 395.7 \ Hz[/tex]
Now according to dopplers equation when the train(source) moves away from the person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as
[tex]f_2 = f * \frac{v_s}{v_s +v_t}[/tex]
substituting values
[tex]f_2 = 350 * \frac{343}{343 + 39.6}[/tex]
[tex]f_2 = 313.77 \ Hz[/tex]
The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as
[tex]\Delta f = f_1 - f_2[/tex]
[tex]\Delta f = 395.7 - 313.77[/tex]
[tex]\Delta f = 81.93 \ Hz[/tex]
Generally the wavelength detected by the person as the train approaches is mathematically represented as
[tex]\lambda_1 = \frac{v}{f_1 }[/tex]
[tex]\lambda_1 = \frac{343}{395.7 }[/tex]
[tex]\lambda_1 = 0.867 \ m[/tex]
A rigid tank A of volume 0.6 m3 contains 5 kg air at 320K and the rigid tank B is 0.4 m3 with air at 600 kPa, 360 K. They are connected to a piston cylinder initially empty with closed valves. The pressure in the cylinder should be 800 kPa to float the piston. Now the valves are slowly opened and the entire process is adiabatic. The internal energy of the mixture at final state is:_____.
a. 229 k/kg.
b. 238 kJ/kg
c. 257 kg
d. cannot be determined.
Answer:
the internal energy of the mixture at final state = 238kJ/kg
Explanation:
Given
V= 0.6m³
m=5kg
R=0.287kJ/kg.K
T=320 K
from ideal gas equation
PV = nRT
where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.
Recall, mole = mass/molar mass
attached is calculation of the question.
If a sample of 346 swimmers is taken from a population of 460 swimmers,
the population mean, w, is the mean of how many swimmers' times?
Answer:
It is the mean of 460 swimmers
Explanation:
In this question, we are concerned with knowing the mean of the population w
Now, according to the question at hand, we have a total population of 460 swimmers and we have taken out 346 swimmers for a study
The population mean in this case is simply the mean of the swimming times of the 460 swimmers
There is another related thing here called the sample mean. For the sample mean, we only make a reference to the mean of the 346 swimmers who were taken out from the population to conduct a separate study
So conclusively, the population mean w is simply the mean of the total 460 swimmers
Copper wire of diameter 0.289 cm is used to connect a set of appliances at 120 V, which draw 1850 W of power total. The resistivity of copper is 1.68×10−8Ω⋅m.
A. What power is wasted in 26.0 m of this wire?
B. What is your answer if wire of diameter 0.417 cm is used?
Answer:
(a) The power wasted for 0.289 cm wire diameter is 15.93 W
(b) The power wasted for 0.417 cm wire diameter is 7.61 W
Explanation:
Given;
diameter of the wire, d = 0.289 cm = 0.00289 m
voltage of the wire, V = 120 V
Power drawn, P = 1850 W
The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m
Area of the wire;
A = πd²/4
A = (π x 0.00289²) / 4
A = 6.561 x 10⁻⁶ m²
(a) At 26 m of this wire, the resistance of the is
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 6.561 x 10⁻⁶
R = 0.067 Ω
Current in the wire is calculated as;
P = IV
I = P / V
I = 1850 / 120
I = 15.417 A
Power wasted = I²R
Power wasted = (15.417²)(0.067)
Power wasted = 15.93 W
(b) when a diameter of 0.417 cm is used instead;
d = 0.417 cm = 0.00417 m
A = πd²/4
A = (π x 0.00417²) / 4
A = 1.366 x 10⁻⁵ m²
Resistance of the wire at 26 m length of wire and 1.366 x 10⁻⁵ m² area;
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 1.366 x 10⁻⁵
R = 0.032 Ω
Power wasted = I²R
Power wasted = (15.417²)(0.032)
Power wasted = 7.61 W
The potential difference between two parallel conducting plates in vacuum is 165 V. An alpha particle with mass of 6.50×10-27 kg and charge of 3.20×10-19 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 40.0 cm.
Answer:
kinetic energy (K.E) = 5.28 ×10⁻¹⁷
Explanation:
Given:
Mass of α particle (m) = 6.50 × 10⁻²⁷ kg
Charge of α particle (q) = 3.20 × 10⁻¹⁹ C
Potential difference ΔV = 165 V
Find:
kinetic energy (K.E)
Computation:
kinetic energy (K.E) = (ΔV)(q)
kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)
kinetic energy (K.E) = 528 (10⁻¹⁹)
kinetic energy (K.E) = 5.28 ×10⁻¹⁷
toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, taken as the x-y plane. The 4.00 kg puck has a velocity of 3.00 i m/s at one instant. Eight seconds later, its velocity is (8.00 i 10.00 j) m/s. Assuming the rocket engine exerts a constant force, find (a) the components of the force and (b) its magnitude.
Answer:
Fx = 2.5 N
Fy = 5 N
|F| = 5.59 N
Explanation:
Given:-
- The mass of puck, m = 4.0 kg
- The initial velocity of puck, u = 3.00 i m/s
- The final velocity of puck, v = ( 8.00 i + 10.00 j ) m/s
- The time interval for the duration of force, Δt = 8 seconds
Find:-
the components of the force and (b) its magnitude.
Solution:-
- We will set up a coordinate system ( x - y ) plane. With unit vectors i and j along x and y axes respectively.
- To model the situation we will seek help from Newton's second law of motion. Defined by the rate of change of linear momentum of the system.
[tex]F_net = \frac{m*( v - u ) }{dt}[/tex]
Where,
Fnet: The net force that acts on the puck-rocket system
- Here we will assume that the mass of rocket is negligible compared to the mass of the puck. The only force ( F ) acting on the puck is due to the thrust produced of the rocket. The dry and air frictions are both neglected for the analysis.
- We will apply the newton's second law of motion in component forms. And determine the components of force F, as ( Fx ) and ( Fy ) as follows:
[tex]F_x = \frac{m* ( v_x - u_x)}{dt} \\\\F_x = \frac{4* ( 8 - 3)}{8} \\\\F_x = 2.5 N\\\\F_y = \frac{m* ( v_y - u_y)}{dt} \\\\F_y = \frac{4* ( 10 - 0)}{8} \\\\F_y = 5 N\\\\[/tex]
- We will apply the Pythagorean theorem and determine the magnitude of the thrust force produced by the rocket with which the puck accelerated:
[tex]| F | = \sqrt{( F_x)^2 + ( F_y)^2} \\\\| F | = \sqrt{( 2.5)^2 + ( 5)^2} \\\\| F | = \sqrt{31.25} \\\\| F | = 5.590[/tex]
Answer: the magnitude of the thrust force is F = 5.59 N
Two Earth satellites, A and B, each of mass m = 980 kg , are launched into circular orbits around the Earth's center. Satellite A orbits at an altitude of 4100 km , and satellite B orbits at an altitude of 12100 km The radius of Earth RE is 6370 km.
(a) What is the ratio of the potential energy of satellite B to that of satellite A, in orbit?
(b) What is the ratio of the kinetic energy of satellite B to that of satellite A, in orbit?
(c) Which satellite has the greater total energy if each has a mass of 14.6 kg?
(d) By how much?
Answer:
Do u have a picture of the graph?
Explanation:
I can solve it with refraction
Please help! Which statements correctly describe the effect of distance in determining the gravitational force and the electrical force? Check all that apply.
There are six statements on the list.
The first 2 are true, and the last 2 are true.
The 2 in the middle aren't true. They are false.
In a circus act a 64.3 kg magician lies on a bed of nail. The bed consists of a large number of evenly spaced, relatively sharp nails mounted in a board so that the points extend vertically outward from the board/while the magician lying down, approximately 1900 nails make contact with hisbody.
1. What is the average force exerted by each nail on the magician's body?
2. If the area of contact at the head of each nail is1.26x10-6 m2 , what is the average pressure at each contact?
Answer:
(a) Fn = 0.33 N
(b) Pn = 263.22 x 10³ Pa = 263.22 KPa
Explanation:
(a)
First, we need to calculate the total force exerted by all nails on the magician. This force must be equal to the weight of magician:
F = W
where,
F = Total Force exerted by all nails = ?
W = Weight of magician = mg = (64.3 kg)(9.8 m/s²) = 630.14 N
Therefore,
F = 630.14 N
Now, we calculate the force exerted by each nail:
Fn = F/n
where,
Fn = force exerted by each nail = ?
n = Total no. of nails = 1900
Therefore,
Fn = 630.14 N/1900
Fn = 0.33 N
(b)
The pressure exerted by each nail is given as:
Pn = Fn/An
where,
Pn = Pressure exerted by each nail = ?
An = Area of contact for each nail = 1.26 x 10⁻⁶ m²
Therefore,
Pn = 0.33 N/1.26 x 10⁻⁶ m²
Pn = 263.22 x 10³ Pa = 263.22 KPa
Which of the following statements is accurate? A) Compressions and rarefactions occur throughout a transverse wave. B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave. C) Sound waves passing through the air will do so as transverse waves, which vibrate vertically and still retain their horizontal positions. D) Amplitude of longitudinal waves is measured at right angles to the direction of the travel of the wave and represents the maximum distance the molecule has moved from its normal position.
Answer:
B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave
Answer:
B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.
Explanation: hope this helps ;)
What does the vertical polarization axis of polarized sunglasses indicate about the direction of polarization of light bouncing off a horizontal surface, such as a wet road or lake surface
Answer:
it is desired that the lenses stop this ray, its polarization must be vertical
Explanation:
To answer this exercise, let's analyze the rays of light reflected on a horizontal surface, when the incident light that we consider non-polarized is reflected on a surface, the electric field of light moves the electrons on the surface horizontally and this re-emits the radiation same shape, that is horizontal.
The other vertical direction the atoms have a lot of movement restricted by the attraction on the surface, so for the reflected ray this polarization is attenuated, this does not stop the transmitted ray where the two polarizations are transmitted.
Total polarizations only for one angle, but in general as we approach dominant polarization it horizontal. Specifically the angle for full polarization is
n = tan teaP
Now we can analyze what polarization the lenses have, if the ray that comes is polarized horizontally and it is desired that the lenses stop this ray, its polarization must be vertical
An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
The new voltage between the plates of the capacitor is 18 V
Explanation:
The charge on parallel plate capacitor is calculated as;
q = CV
Where;
V is the battery voltage
C is the capacitance of the capacitor, calculated as;
[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]
[tex]q = \frac{\epsilon _0A V}{d}[/tex]
where;
ε₀ is permittivity of free space
A is the area of the capacitor
d is the space between the parallel plate capacitors
If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;
[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]
Therefore, the new voltage between the plates of the capacitor is 18 V
A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?
Answer:
The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]
Explanation:
The Diagram for this question is gotten from the first uploaded image
From the question we are told that
The mass of the rod is [tex]M = 3.41 \ kg[/tex]
The mass of each small bodies is [tex]m = 0.249 \ kg[/tex]
The moment of inertia of the three-body system with respect to the described axis is [tex]I = 0.929 \ kg \cdot m^2[/tex]
The length of the rod is L
Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as
[tex]I = I_r + 2 I_m[/tex]
Where [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as
[tex]I_r = \frac{ML^2 }{12}[/tex]
And [tex]I_m[/tex] the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as
[tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]
Thus [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]
Hence
[tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]
=> [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]
substituting vales we have
[tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]
[tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]
[tex]L = 1.5077 \ m[/tex]
A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the first charge.
Required:
a. What is the value of the unknown charge (magnitude and sign)?
b. What is the magnitude of the force that the unknown charge exerts on the -0.590 μC charge?
c. What is the direction of this force?
Answer:
a. q2 = 16.4μC, positive charge
b. F = 0.900N
c. downward
Explanation:
a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:
[tex]F_e=k\frac{q_1q_2}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
r: distance between the charges = 0.300m
q1: charge 1 = -0.550 μC = 0.550*10^-6C
q2: charge 2 = ?
Fe: electric force = 0.900N
The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.
You solve the equation (1) for the second charge ans replace the values of the other parameters:
[tex]q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C[/tex]
The values of the second charge is 1.64 μC
b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.
The force exerted on the first charge is 0.900N
c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.
A box with an initial speed of 15 m/s slides along a surface where the coefficient of sliding friction is 0.45. How long does it take for the block to come to rest
Answer:
t = 3.4 s
The box will come to rest in 3.4 s
Explanation:
For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:
Unbalanced Force = Frictional Force
but,
Unbalanced Force = ma
Frictional Force = μR = μW = μmg
Therefore,
ma = μmg
a = μg
where,
a = acceleration of box = ?
μ = coefficient of sliding friction = 0.45
g = 9.8 m/s²
Therefore,
a = (0.45)(9.8 m/s²)
a = -4.41 m/s² (negative sign due to deceleration)
Now, for the time to stop, we use first equation of motion:
Vf = Vi + at
where,
Vf = Final Speed = 0 m/s (since box stops at last)
Vi = Initial Speed = 15 m/s
t = time to stop = ?
Therefore,
0 m/s = 15 m/s + (-4.41 m/s²)t
(-15 m/s)/(-4.41 m/s²) = t
t = 3.4 s
The box will come to rest in 3.4 s
A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is
The lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.
What is diffraction?The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.
Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So, The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.
Learn more about diffraction here:
https://brainly.com/question/11176463
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A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s. What is her acceleration on the rough ice?
Answer:
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Explanation:
The distance travelled on the rough ice is equal to the width of the rough ice.
distance d = 5.0 m
Initial speed u = 9.2 m/s
Final speed v = 5.8 m/s
The time taken to move through the rough ice can be calculated using the equation of motion;
d = 0.5(u+v)t
time t = 2d/(u+v)
Substituting the given values;
t = 2(5)/(9.2+5.8)
t = 2/3 = 0.66667 second
The acceleration is the change in velocity per unit time;
acceleration a = ∆v/t
a = (v-u)/t
Substituting the values;
a = (5.8-9.2)/0.66667
a = -5.099974500127
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
A child has a toy car on a horizontal platform. The car starts from rest and reaches a maximum speed in 4 s. If the mass of the car is
0.1 kg and engine has an effective pull of 0.4 N Find the acceleration of the car.
Answer:
a=4m/s²
Explanation:
F=ma
0.4=0.1a
Answer:
a=4m/s
Explanation:
F=ma
0.4=0.1a
[tex] \frac{0.4}{0.1} = \frac{0.1}{0.1} [/tex]
a =4m/ s
A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction. At a particular instant, the velocity of the particle has components Vx=−1.68×10^4m/s, Vy=−2.61×10^4m/s, and Vz=5.85×10^4m/s. What is the z-component of the force on the particle at this time?
Answer:
The z-component of the force is [tex]\= F_z = 0.00141 \ N[/tex]
Explanation:
From the question we are told that
The charge on the particle is [tex]q = 7.76 *0^{-8} \ C[/tex]
The magnitude of the magnetic field is [tex]B = 0.700\r i \ T[/tex]
The velocity of the particle toward the x-direction is [tex]v_x = -1.68*10^{4}\r i \ m/s[/tex]
The velocity of the particle toward the y-direction is
[tex]v_y = -2.61*10^{4}\ \r j \ m/s[/tex]
The velocity of the particle toward the z-direction is
[tex]v_y = -5.85*10^{4}\ \r k \ m/s[/tex]
Generally the force on this particle is mathematically represented as
[tex]\= F = q (\= v X \= B )[/tex]
So we have
[tex]\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)[/tex]
[tex]\= F = q (v_y B(-\r k) + v_z B\r j)[/tex]
substituting values
[tex]\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)[/tex]
[tex]\= F= 0.00303\ \r j +0.00141\ \r k[/tex]
So the z-component of the force is [tex]\= F_z = 0.00141 \ N[/tex]
Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).
The voltage between the cathode and the screen of a television set is 30 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed (m/s) just before it hits the screen
Answer:
The speed is [tex]v =10.27 *10^{7} \ m/s[/tex]
Explanation:
From the question we are told that
The voltage is [tex]V = 30 kV = 30*10^{3} V[/tex]
The initial velocity of the electron is [tex]u = 0 \ m/s[/tex]
Generally according to the law of energy conservation
Electric potential Energy = Kinetic energy of the electron
So
[tex]PE = KE[/tex]
Where
[tex]KE = \frac{1}{2} * m* v^2[/tex]
Here m is the mass of the electron with a value of [tex]m = 9.11 *10^{-31} \ kg[/tex]
and
[tex]PE = e * V[/tex]
Here e is the charge on the electron with a value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]e * V = \frac{1}{2} * m * v^2[/tex]
=> [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]
[tex]v =10.27 *10^{7} \ m/s[/tex]
A 100 kg lead block is submerged in 2 meters of salt water, the density of which is 1096 kg / m3. Estimate the value of the hydrostatic pressure.
Answer:
21,920 Pascals
Explanation:
P = ρgh
P = (1096 kg/m³) (10 m/s²) (2 m)
P = 21,920 Pa
In a sinusoidally driven series RLC circuit, the inductive resistance is XL = 100 Ω, the capacitive reactance is XC = 200 Ω, and the resistance is R = 50 Ω. The current and applied emf would be in phase if
Answer:
The current and the applied emf can be in phase if either of the two changes are made.
1) The inductance of the inductor is doubled, with everything else remaining constant.
2) The capacitance of the capacitor is doubled, with everything else remaining constant.
Explanation:
The current and applied emf for this type of circuit would be in phase when there is no phase difference between the two quantities. That is, Φ = 0°.
The phase difference between current and applied emf is given as
Φ = tan⁻¹ [(XL - Xc)/R]
XL = Impedance due to the inductor
Xc = Impedance due to the capacitor
R = Resistance of the resistor.
For Φ to be 0°, tan⁻¹ [(XL - Xc)/R] = 0
But only tan⁻¹ 0 = 0 rad
So, for the phase difference to be 0,
[(XL - Xc)/R] = 0
Meaning
XL = Xc
But for this question,
XL = 100 Ω, Xc = 200 Ω
For them to be equal, we have to find a way to increase the impedance of the inductor or reduce the impedance of the capacitor.
The impedance are given as
XL = 2πfL
Xc = (1/2πfC)
f = Frequency
L = Inductance of the inductor
C = capacitance of the capacitor
The impedance of the inductor can be increased from 100 Ω to 200 Ω by doubling the inductance of the inductor.
And the impedance of the capacitor can be reduced from 200 Ω to 100 Ω by also doubling the capacitance of the capacitor.
So, these are either of the two ways to make the current and applied emf to be in phase.
Hope this Helps!!!
Julie is playing with a toy car and is pushing it around on the floor. The little car has a mass of 6.3 g. The car has a velocity of 2.5 m/s. What is the car's momentum?
Answer:
Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second
Explanation:
Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg
Velocity of the car is 2.5 meter per second
Since formula to calculate the momentum of an object is,
p = mv
Where, p = momentum of the object
m = mass of the object
v = velocity of the object
By substituting these values in the formula,
p = [tex](6.3\times 10^{-3})\times 2.5[/tex]
= [tex]1.575\times 10^{-2}[/tex] Kg meter per second
Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.
What is the maximum height the rock will reach?
Answer:
Explanation:
initial vertical velocity = 17.5 m/s
using g=-9.81 m/s^2
apply kinematics equation
v1^2-v0^2=2gS
solve for S with v1=0, v0=+17.5
S = (v1^2-v0^2)/2g
=(0-17.5^2)/(2*(-9.81))
= 15.61 m
The lower the value of the coefficient of friction, the____the resistance to sliding
Answer: lower
There are a number of factors that can affect the coefficient of friction, including surface conditions.
Values of the coefficient of sliding friction can be a good reference for specific combinations of materials. The frictional force and normal reaction are directly proportion but an increase or decrease in coefficient of friction will cause an increase or decrease in the resistance of sliding respectively
3. Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
Object 2 has the larger drag coefficient
Explanation:
The drag force, D, is given by the equation:
[tex]D = 0.5 c \rho A v^2[/tex]
Object 1 has twice the diameter of object 2.
If [tex]d_2 = d[/tex]
[tex]d_1 = 2d[/tex]
Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]
Area of object 1:
[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]
Since all other parameters are still the same except the drag coefficient:
For object 1:
[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]
For object 2:
[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]
Since the drag force for the two objects are the same:
[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]
Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient
Find acceleration. Will give brainliest!
Answer:
16200 km/s
270 km/min
4.5 km/h
Explanation:
Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)
Simply plug in our known variables and solve:
a = (45.0 - 0)/10
a = 45.0/10
a = 4.5 km/h
Answer:
[tex]\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}[/tex]
Explanation:
[tex]\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}[/tex]
[tex]\displaystyle \mathrm{a = \frac{v - u}{t}}[/tex]
[tex]\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}[/tex]
[tex]\displaystyle \mathrm{a = \frac{45- 0}{10}}[/tex]
[tex]\displaystyle \mathrm{a = \frac{45}{10}}[/tex]
[tex]\displaystyle \mathrm{a = 4.5}[/tex]
[tex]\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }[/tex]
If 62.9 cm of copper wire (diameter = 1.15 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.43 mT/s, at what rate is thermal energy generated in the loop?
Answer:
The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"
Explanation:
length of the copper wire:
L= 62.9 cm
r is the radius of the loop then:
[tex]r=\frac{L}{2 \pi}\\[/tex]
[tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]
area of the loop Is:
[tex]A_L= \pi r^2[/tex]
[tex]=100.2001\times 3.14\\\\=314.628[/tex]
change in magnetic field is:
[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]
then the induced emf is: [tex]e = A_L \times \frac{dB}{dt}[/tex]
[tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]
resistivity of the copper wire is: [tex]\rho =[/tex] 1.69 × 10-8Ω·m
diameter d = 1.15mm
radius (r) = 0.5mm
[tex]= 0.5 \times 10^{-3} \ m[/tex]
hence the resistance of the wire is:
[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]
[tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]
Power:
[tex]P=\frac{e^2}{R}[/tex]
[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]
The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]
a crate b of mass 40kg is raised by the rope of crane from the hold of a ship. mark and name forces on the crate . find acceleration if tension is 480N
Find the acceleration, a .
Formula used:-Force = Mass × Acceleration
Solution:-We know that ,
Force = Mass × Acceleration
★ Substituting the values in the above formula,we get:
⇒ 480 = 40 × Acceleration
⇒ Acceleration, a = 480/40
⇒ Acceleration,a = 12 m/s
Thus,the acceleration of a body is 12 metres per seconds.
The aorta pumps blood away from the heart at about 40 cm/s and has a radius of about 1.0 cm. It then branches into many capillaries, each with a radius of about 5 x 10−4 cm carrying blood at a speed of 0.10 cm/s.
How many capillaries are there?
Answer:
n = 1.6*10^9 capillaries
Explanation:
In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:
[tex]A_1v_1=nA_2v_2[/tex] (1)
A1: area of the aorta
v1: speed of the blood in the aorta = 40cm/s
n: number of capillaries = ?
A2: area of each capillary
v2: speed of the blood in each capillary
For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:
[tex]A=\pi r^2\\\\A_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\\\\A_2=\pi r_2^2=\pi (5*10^{-4}cm)^2=7.85*10^{-7}cm^2[/tex]
Where you have used the values of the radius for the aorta and the capillaries.
Next, you solve the equation (1) for n, and replace the values of all parameters:
[tex]n=\frac{A_1v_1}{A_2v_2}=\frac{(3.1415cm^2)(40cm/s)}{(7.85*10^{-7}cm^2)(0.10cm/s)}=1.6*10^9[/tex]
Then, the number of capillaries is 1.6*10^9
If an ant is starting from 0 radians, and travels all the way to 4.5 radians in 18 seconds, what is angular velocity? If the same ant then slows down back to 0 rad/s in 3 seconds what is its angular acceleration?
Answer:
Angular velocity is 0.25 rad/sAngular acceleration is 0.017 rad/s²Explanation:
Given;
initial angular displacement, θ₁ = 0 radians
final angular displacement, θ₂ = 4.5 radians
Angular velocity is calculated as;
[tex]\omega = \frac{\delta \theta}{\delta t}= \frac{4.5 -0}{18} \\\\\omega = 0.25 \ rad/s[/tex]
Angular acceleration is calculated as;
[tex]\alpha = \frac{\delta \omega}{\delta t} = \frac{\omega _f - \omega_i}{t_2 -t_1}[/tex]
where;
[tex]\omega_f[/tex] is the final angular velocity = 0 rad/s
[tex]\omega _i[/tex] is the initial angular velocity = 0.25 rad/s
t₂ is the final time of the motion = 3 seconds
t₁ is the initial time of the motion = 18 seconds
[tex]\alpha =\frac{\omega _f - \omega_i}{t_2 -t_1} \\\\\alpha = \frac{0 - 0.25}{3-18} \\\\\alpha = 0.017 \ rad/s^2[/tex]