Answer:
The complete question is
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26 W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.
a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.
For totally reflective, F = (2I/c)A ....1
for totally reflective, F = (I/c)A ....2
where I is the intensity of the light
c is the speed of light = 3 x 10^8 m/s
A is the area the sail
b) The intensity of the light from the sun = power/area
==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]
where r is the distance from the sun and the sail
The Force from the sail from equation 1 is therefore
[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]
gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]
where
G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.
m is the mass of the sail = 10000 kg
M is the mass of the sun = 1.99 x 10^30 kg.
==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
Equating the forces, we have
[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
the distance cancels out
A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2
==> 6428.2 km^2
c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation is inversely proportional to the square of the distance, so they both cancel out.
If you wanted to make your own lenses for a telescope, what features of a lens do you think would affect the images that you can see
Answer:
Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Explanation:
The length of the telescope is
L = f_ocular + f_objetive
the magnification of the telescope is
m = - f_objective / f_ocular
These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1.0 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).
Required:
a. Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.
b. Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm^2. How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?
Answer:
1. 6.99x 10^-6V/m
2. 18m
Explanation:
See attached file
Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.16 and μB = 0.23. Determine the incline angle θ for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2.1 lb/ft .
Answer:
[tex]\theta=10.20^{\circ}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Explanation:
First of all, we analyze the system of blocks before starting to move.
[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]
[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]
[tex]tan(\theta)=0.18[/tex]
[tex]\theta=arctan(0.18)[/tex]
[tex]\theta=10.20^{\circ}[/tex]
Hence, the incline angle θ for which both blocks begin to slide is 10.20°.
Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.
[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]
Where:
[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]
[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]
[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Therefore, the required stretch or compression in the connecting spring is 0.10 ft.
I hope it helps you!
(a) The inclined angle for which both blocks begin to slide is 10.3⁰.
(b) The compression of the spring is 0.22 ft.
The given parameters;
mass of block A, = 11 lbmass of block B, = 5 lbcoefficient of static friction for A, = 0.16coefficient of static friction for B, = 0.23 spring constant, k = 2.1 lb/ftThe normal force on block A and B:
[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]
The frictional force on block A and B:
[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]
The net force on the blocks when they starts sliding;
[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]
The change in the energy of the blocks is the work done in compressing the spring;
[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]
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If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First what are the mass units?
Answer:
The mas unit is the the 'Kilogram' written as 'kg'
Volume is 10 L
Explanation:
The complete question is
If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First, what are the mass units?
Second, what is the volume
mass units is the 'Kilogram', written as 'kg'
density = mass/volume = 100 kg/L
the mass = 1000 kg
volume = mass/density = 1000/100 = 10 L
Electrons are accelerated through a voltage difference of 270 kV inside a high voltage accelerator tube. What is the final kinetic energy of the electrons?
Each electron winds up with kinetic energy of
(270 keV)
plus
(whatever KE it had when it started accelerating).
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Answer:
[tex]I=2.71\times 10^{-5}\ A[/tex]
Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
[tex]C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}[/tex], r is radius
Let I is the displacement current. It is given by :
[tex]I=C\dfrac{dV}{dt}[/tex]
Here, [tex]\dfrac{dV}{dt}[/tex] is rate of increasing potential difference
So
[tex]I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A[/tex]
So, the value of displacement current is [tex]2.71\times 10^{-5}\ A[/tex].
An airplane flies 1,592 miles east from Phoenix, Arizona, to Atlanta, Georgia, in 3.68 hours.
What is the average velocity of the airplane? Round your answer to the nearest whole number.
Answer:
433
Explanation:
3. Which of the following accurately describes circuits?
O A. In a parallel circuit, the same amount of current flows through each part of the circuit
O B. In a series circuit, the amount of current passing through each part of the circuit may vary
O C. In a series circuit, the current can flow through only one path from start to finish
O D. In a parallel circuit, there's only one path for the current to travel.
Answer:
Option (c)
Explanation:
In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.
(C.) is the only correct statement in the list of choices.
In a series circuit, the current can flow through only one path from start to finish.
Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.
Answer:
E. None of these statements are true.
Explanation:
We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.
for A, the tension on the rope does not depend on if both teams pull are in equilibrium.
for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.
for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.
for D, just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.
we go with option E.
In a two-slit experiment, the slit separation is 3.34 ⋅ 10 − 5 m. The interference pattern is created on a screen that is 3.30 m away from the slits. If the 7th bright fringe on the screen is 29.0 cm away from the central fringe, what is the wavelength of the light?
Answer:
The wavelength is [tex]\lambda = 419 \ nm[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 3.34 *10^{-5} \ m[/tex]
The distance of the screen is [tex]D = 3.30 \ m[/tex]
The order of the fringe is n = 7
The distance of separation of fringes is y = 29.0 cm = 0.29 m
Generally the wavelength of the light is mathematically represented as
[tex]\lambda = \frac{y * d }{ n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}[/tex]
[tex]\lambda = 4.19*10^{-7}\ m[/tex]
[tex]\lambda = 419 \ nm[/tex]
A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exerts a resistive force with magnitude proportional to the square of the speed with k
Answer:
The velocity is 40 ft/sec.
Explanation:
Given that,
Force = 3200 lb
Angle = 30°
Speed = 64 ft/s
The resistive force with magnitude proportional to the square of the speed,
[tex]F_{r}=kv^2[/tex]
Where, k = 1 lb s²/ft²
We need to calculate the velocity
Using balance equation
[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]
Put the value into the formula
[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]
Put the value of k
[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]
[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]
At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]
So, [tex]1600-v^2=0[/tex]
[tex]v=\sqrt{1600}[/tex]
[tex]v=40\ ft/sec[/tex]
Hence, The velocity is 40 ft/sec.
Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations was 0.15 m, its wavelength was two meters, and the period was 2/15 s. If a point on the wave at a specific time has a displacement of 0.12 m, what is the transverse speed of that point?
Answer:
15m/sExplanation:
The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.
From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;
[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]
Before we can get the transverse speed, we need to get the frequency and the wavelength.
frequency = 1/period
Given period = 2/15 s
Frequency = [tex]\frac{1}{(2/15)}[/tex]
frequency = 1 * 15/2
frequency f = 15/2 Hertz
Given wavelength [tex]\lambda[/tex] = 2m
Transverse speed [tex]v = f \lambda[/tex]
[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]
Hence, the transverse speed at that point is 15m/s
The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 6.1 mm. Using light with a wavelength of 578 nm, how far could you be from this tile and still resolve these holes
Answer:
8.65x10^3m
Explanation:
See attached file
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled
Answer:
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
Explanation:
The diffraction of a slit is given by the expressions
a sin θ = m λ
where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.
sin θ = m λ / a
If this equation
a ’= 2 a
we substitute
2 a sin θ'= m λ
sin θ'= (m λ / a) 1/2
sin θ ’= sin θ / 2
We can use trigonometry to find the width
tan θ = y / L
as the angle is small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
y ’/ L = y/L 1/2
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 meters. The mass then slides without friction down the inner surface toward the bottom of the bowl. At a certain point of its path the mass achieves a speed v = 3.57 m/s. At this point, what angle \theta\:θ ( in degrees) does the mass make with the top of the bowl?
Answer:
The angle is [tex]\theta = 36.24 ^o[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.6 \ kg[/tex]
The radius is [tex]r = 1.1 \ m[/tex]
The speed is [tex]v = 3.57 \ m /s[/tex]
According to the law of energy conservation
The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e
[tex]m * g * h = \frac{1}{2} * m * v^2[/tex]
=> [tex]h = \frac{1}{2 g } * v^2[/tex]
Here h is the vertical distance traveled by the mass which is also mathematically represented as
[tex]h = r * sin (\theta )[/tex]
So
[tex]\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2][/tex]
substituting values
[tex]\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2][/tex]
[tex]\theta = 36.24 ^o[/tex]
In a physics lab, light with a wavelength of 490 nm travels in air from a laser to a photocell in a time of 17.5 ns . When a slab of glass with a thickness of 0.800 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 21.5 ns to travel from the laser to the photocell.What is the wavelength of the light in the glass? Use 3.00×108 m/s for the speed of light in a vacuum. Express your answer using two significant figures.
Answer:
196 nm
Explanation:
Given that
Value of wavelength, = 490 nm
Time spent in air, t(a) = 17.5 ns
Thickness of glass, th = 0.8 m
Time spent in glass, t(g) = 21.5 ns
Speed of light in a vacuum, c = 3*10^8 m/s
To start with, we find the difference between the two time spent
Time spent on glass - Time spent in air
21.5 - 17.5 = 4 ns
0.8/(c/n) - 0.8/c = 4 ns
Note, light travels with c/n speed in media that has index of refraction
(n - 1) * 0.8/c = 4 ns
n - 1 = (4 ns * c) / 0.8
n - 1 = (4*10^-9 * 3*10^8) / 0.8
n - 1 = 1.2/0.8
n - 1 = 1.5
n = 1.5 + 1
n = 2.5
As a result, the wavelength of light in a medium with index of refraction would then be
490 / 2.5 = 196 nm
Therefore, our answer is 196 nm
3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000 rpm, completando un total de 12 revoluciones.
d. Determina la aceleración tangencial, centrípeta y la total en m.s-2 cuando el tambor a alcanzado los 60000 rpm
e. Explica lo que ocurre con la magnitud y dirección de los vectores aceleración tangencial, aceleración centrípeta, aceleración total, aceleración angular, velocidad angular cuando la lavadora ha girado desde 3000 rpm hasta 6000 rpm.
Answer:
d) α = 1693.5 rad / s² , a = 392.7 m / s² , a_total = α √(R² +1) ,
e) tan θ = a / α
Explanation:
This is an exercise in linear and angular kinematics.
We initialize reduction of all the magnitudes to the SI system
w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s
w = 6000 rev / mi = 628.32 rad / s
θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad
d) ask for centripetal, tangential and total acceleration.
Let's start by looking for centripetal acceleration, let's use the formula
w² = w₀² + 2 α θ
α = (w²- w₀²) / 2θ
we calculate
α = (628.32²2 - 314.16²) / 2 75.398
α = 1693.5 rad / s²
the quantity is linear and angular are related
the linear or tangential acceleration is
a = α R
where R is the radius of the drum
a = 1693.5 R
Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m
a = 1693.5 0.20
a = 392.7 m / s²
the total acceleration is
a_total = √(a² + α²)
a_total = √ (α² R² + α²)
a_total = α √(R² +1)
e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant
Tangential acceleration is tangency to radius and its value varies proportionally radius
the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry
tan θ = a / α
the angular velocity increases linearly when with centripetal acceleration
what is the average flow rate in of gasoline to the engine of a plane flying at 700 km/h if it averages 100.0 km/l
Answer:
1.94cm³/s
Explanation:
1L = 1000cm³
Ihr = 3600s
So
Using
Average flow rate
Fr= 1L/100Km x 700Km/1hr x 1hr/3600s x 1000cm³/ 1L
= 1.94cm³/s
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?
Answer:
The current is [tex]I = 8.9 *10^{-5} \ A[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]
The current density is [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]
The distance we are considering is [tex]r = 0.5 R = 0.001585[/tex]
Generally current density is mathematically represented as
[tex]J = \frac{I}{A }[/tex]
Where A is the cross-sectional area represented as
[tex]A = \pi r^2[/tex]
=> [tex]J = \frac{I}{\pi r^2 }[/tex]
=> [tex]I = J * (\pi r^2 )[/tex]
Now the change in current per unit length is mathematically evaluated as
[tex]dI = 2 J * \pi r dr[/tex]
Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows
[tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]
[tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]
substituting values
[tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]
[tex]I = 8.9 *10^{-5} \ A[/tex]
A stationary coil is in a magnetic field that is changing with time. Does the emf induced in the coil depend
Answer:
Explanation:
The e.m.f induced in the coil depend on the following :
(a) No. of turns in the coil
(b) Cross-sectional Area of the coil
(c) Magnitude of Magnetic field
(d) Angular velocity of the coil
An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically grounded object. Positively charged paint droplets are repelled away from the paint sprayer's positively charged sphere and towards the grounded object. What charge must a 0.168-mg drop of paint have so that it will arrive at the object with a speed of 18.8 m/s
Answer:
The charge is [tex]Q = 2.177 *10^{-9} \ C[/tex]
Explanation:
From the question we are told that
The electric potential is [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]
The mass of the drop is [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]
The speed is [tex]v = 18.8 \ m/s[/tex]
Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question is mathematically represented as
[tex]Q = \frac{m v^2 }{ 2 * V }[/tex]
Substituting values
[tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]
[tex]Q = 2.177 *10^{-9} \ C[/tex]
A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire?
Answer:
The potential will be Va/b
Explanation:
So Let sphere A charged Q to potential V.
so, V= KQ/a. ....(1
Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.
therefore, potential will be ,
V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.11 mm and place your screen 8.63 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.71 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength lambda expressed in nanometers?
Answer:
λ = 605.80 nm
Explanation:
These double-slit experiments the equation for constructive interference is
d sin θ = m λ
where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.
In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm
Let's use trigonometry to find the angle
tan θ = y / L
as the angles are very small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in the first equation
d y / L = m λ
λ = d y / m L
let's calculate
λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)
λ = 6.05805 10⁻⁷ m
let's reduce to nm
λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)
λ = 605.80 nm
At what speed, as a fraction of c, will a moving rod have a length 65% that of an identical rod at rest
Answer:
v/c = 0.76
Explanation:
Formula for Length contraction is given by;
L = L_o(√(1 - (v²/c²))
Where;
L is the length of the object at a moving speed v
L_o is the length of the object at rest
v is the speed of the object
c is speed of light
Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.
Thus;
0.65L_o = L_o[√(1 - (v²/c²))]
Dividing both sides by L_o gives;
0.65 = √(1 - (v²/c²))
Squaring both sides, we have;
0.65² = (1 - (v²/c²))
v²/c² = 1 - 0.65²
v²/c² = 0.5775
Taking square root of both sides gives;
v/c = 0.76
You want the current amplitude through a 0.450 mH inductor (part of the circuitry for a radio receiver) to be 1.50 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor. What frequency is required?
Answer:
3.067MHzExplanation:
The formula for calculating the voltage across an inductor is expressed as
[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]
Given parameters
current amplitude I = 1.50mA = 1.5*10⁻³A
inductance L = 0.450mH = 0.450*10⁻³H
Voltage across the inductor [tex]V_l[/tex] = 13.0V
Required
frequency f
Substituting the given parametres into the formula, we have;
[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]
Hence, the frequency required is 3.067MHz
With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
How much heat is required to convert 5.0 kg of ice from a temperature of - 20 0C to water at a temperature of 205 0F
Answer:
Explanation:
To convert from °C to °F , the formula is
( F-32 ) / 9 = C / 5
F is reading fahrenheit scale and C is in centigrade scale .
F = 205 , C = ?
(205 - 32) / 9 = C / 5
C = 96°C approx .
Let us calculate the heat required .
Total heat required = heat required to heat up the ice at - 20 °C to 0°C + heat required to melt the ice + heat required to heat up the water at 0°C to
96°C.
= 5 x 2.04 x (20-0) + 5 x 336 + 5 x ( 96-0 ) x 4.2 kJ .
= 204 + 1680 + 2016
= 3900 kJ .
The accommodation limits for a nearsighted person's eyes are 20.0 cm and 82.0 cm. When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly
Answer;
26.45cm
See attached file for explanation
The metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. This simple
process is which kind of a change?
OA a physical change
OB. a chemical change
OC. a nuclear change
OD
an ionic change
B. A chemical change
Explanation:
I'm guessing ?
A bungee cord with a spring constant of 800 StartFraction N over m EndFraction stretches 6 meters at its greatest displacement. How much elastic potential energy does the bungee cord have? The bungee cord has J of elastic potential energy.
Explanation:
EE = ½ kx²
EE = ½ (800 N/m) (6 m)²
EE = 14,400 J
Answer:
14,400 J
Explanation:
Its the answer
