Answer:
The time rate of change of the electric field between the plates is [tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 2.8 \ cm = 0.028 \ m[/tex]
The distance of separation is [tex]d = 1.1 \ mm = 0.0011 \ m[/tex]
The current is [tex]I = 5 \ A[/tex]
Generally the electric field generated is mathematically represented as
[tex]E = \frac{q }{ \pi * r^2 \epsilon_o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value
[tex]\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
So the time rate of change of the electric field between the plates is mathematically represented as
[tex]\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }[/tex]
But [tex]\frac{q}{t } = I[/tex]
So
[tex]\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }[/tex]
substituting values
[tex]\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }[/tex]
[tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]
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.
Three identical resistors are connected in series to a battery. If the current of 12 A flows from the battery, how much current flows through any one of the resistors
Answer:
Current that flows through any one of the resistors has a value of 12 amperes.
Explanation:
When a group of resistors are connected in series, the same current flows through each resistor. According to the Ohm's Law, the circuit can be represented as follows:
[tex]V_{batt} = i\cdot (R_{1}+R_{2}+R_{3})[/tex]
[tex]i = \frac{V_{batt}}{R_{1}+R_{2}+R_{3}}[/tex]
Where:
[tex]V_{batt}[/tex] - Voltage of the battery, measured in volts.
[tex]R_{1}[/tex], [tex]R_{2}[/tex], [tex]R_{3}[/tex] - Electric resistance of the first, second and third resistors, measured in ohms.
[tex]i[/tex] - Current, measured in amperes.
If [tex]R_{1} = R_{2} = R_{3} = R[/tex], then:
[tex]i = \frac{V_{batt}}{3\cdot R}[/tex]
Current that flows through any one of the resistors has a value of 12 amperes.
The current flows via any of the resistors should have a value of 12 amperes.
Ohm law:At the time When a group of resistors are linked in series, so there is a similar current flow via each resistor.
Here the circuit should be
vbatt = i.(R1 + R2+ R3)
i = Vbatt/R1 + R2 + R3
here
Vbatt means the voltage of the battery
R1,R2, and R3 means the resistance of the first, second and third resistors
I means the current
So, in the case when
R1 = R2 = R3 = R
So,
i = Vbatt/3.R
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The index of refraction of a sugar solution in water is about 1.5, while the index of refraction of air is about 1. What is the critical angle for the total internal reflection of light traveling in a sugar solution surrounded by air
Answer:
The critical angle is [tex]i = 41.84 ^o[/tex]
Explanation:
From the question we are told that
The index of refraction of the sugar solution is [tex]n_s = 1.5[/tex]
The index of refraction of air is [tex]n_a = 1[/tex]
Generally from Snell's law
[tex]\frac{sin i }{sin r } = \frac{n_a }{n_s }[/tex]
Note that the angle of incidence in this case is equal to the critical angle
Now for total internal reflection the angle of reflection is [tex]r = 90^o[/tex]
So
[tex]\frac{sin i }{sin (90) } = \frac{1 }{1.5 }[/tex]
[tex]i = sin ^{-1} [\frac{ (sin (90)) * 1 }{1.5} ][/tex]
[tex]i = 41.84 ^o[/tex]
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 850 m . After this point, its acceleration is that of gravity, downward.
Answer:
v = 73.75 m/s
Explanation:
It is given that,
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 850 m.
Let us assume we need to find the velocity of the rocket when it runs out of fuel.
Let v is the final speed. Using the third equation of kinematics as :
[tex]v^2-u^2=2as[/tex]
u = 0
[tex]v=\sqrt{2as} \\\\v=\sqrt{2\times 3.2\times 850}\\\\v=73.75\ m/s[/tex]
So, the velocity of the rocket when it runs out of the fuel is 73.75 m/s
Two slits are separated by 0.370 mm. A beam of 545-nm light strikes the slits, producing an interference pattern. Determine the number of maxima observed in the angular range−26.0° ≤ θ ≤ 26.0°.
Answer:
There are 586maxima
Explanation:
Pls see attached file
A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time
Answer:
The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
Explanation:
The magnetic field at the center of the solenoid is given by;
B = μ(N/L)I
Where;
μ is permeability of free space
N is the number of turn
L is the length of the solenoid
I is the current in the solenoid
The rate of change of the field is given by;
[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]
The induced emf in the shorter coil is calculated as;
[tex]E = NA\frac{\delta B}{\delta t}[/tex]
where;
N is the number of turns in the shorter coil
A is the area of the shorter coil
Area of the shorter coil = πr²
The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m
Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²
[tex]E = NA\frac{\delta B}{\delta t}[/tex]
E = 14 x 0.000491 x 0.02514
E = 1.728 x 10⁻⁴ V
Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]
Induced EMF:The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:
[tex]E=-\frac{\delta\phi}{\delta t}[/tex]
where E is the induced emf,
[tex]\phi[/tex] is the magnetic flux through the coil.
The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:
[tex]\phi = \frac{\mu_oNIA}{L}[/tex]
so;
[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]
where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.
[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]
The emf produced in the coil at the center of the solenoid which has 14 turns will be:
[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]
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What do Equations 1 and 2 predict will happen to the single-slit diffraction pattern (intensity, fringe width, and fringe spacing) as the slit width is increased.
Equation 1:
Sinθ = mλ/ω
Equaiton 2:
I= Io [Sinθ (πωλ/πωλ/Rλ)
Answer:
the firtz agrees with the expression for the shape of the curve of diracion of a slit
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
where a is the width of the slit, t is the angle from the center of the slit, l is the wavelength and m is an integer that corresponds to the maximum diffraction.
the previous equation qualitatively describes the curve of the diffraction phenomenon the equation takes the form
I = I₀ [(sin ππ a y / R λ) / π a y / Rλ]²
I = I₀ ’[sin π a y /Rλ]²
I₀ ’= I₀ / (π a y /Rλ)²
By reviewing the two expressions given
equation 1
w sin θ = m λ
where w =a w is the slit width
we see that the firtz agrees with the expression for the shape of the curve of diracion of a slit
Equation 2
the squares are missing
A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev
Answer:
a
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
b
[tex]\theta = 9124.5 \ rev[/tex]
Explanation:
From the question we are told that
The maximum angular speed is [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]
The time taken is [tex]t = 2.8 \ s[/tex]
The minimum angular speed is [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that
[tex]w_{max} = w_{mini} + \alpha * t[/tex]
=> [tex]\alpha = \frac{ w_{max}}{t}[/tex]
substituting values
[tex]\alpha = \frac{40950.73}{2.8}[/tex]
[tex]\alpha = 14625 .3 \ rad/s^2[/tex]
converting to [tex]rev/s^2[/tex]
We have
[tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
According to the first equation of motion the angular displacement is mathematically represented as
[tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]
substituting values
[tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]
[tex]\theta = 57331.2 \ radian[/tex]
converting to revolutions
[tex]revolution = 57331.2 * 0.159155[/tex]
[tex]\theta = 9124.5 \ rev[/tex]
The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show your calculations, and explain.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The spring scale [tex]F_2[/tex] reads [tex]F_2 = 2.4225 \ N[/tex]
Explanation:
From the question we are told that
The first force is [tex]F_1 = 10.5 \ N[/tex]
The acceleration by which the cart moves to the right is [tex]a = 2.50 \ m/s^2[/tex]
The mass of the cart is m = 3.231 kg
Generally the net force on the cart is
[tex]F_{net} = F_1 - F_2[/tex]
This net force is mathematically represented as
[tex]F_{net} = m * a[/tex]
So
[tex]m* a = 10 - F_2[/tex]
[tex]F_2 = 10.5 - 2.5 (3.231)[/tex]
[tex]F_2 = 2.4225 \ N[/tex]
A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v= 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
Answer:
The wait time is [tex]t_w = 3.4723 \ s[/tex]
Explanation:
From the question we are told that
The distance of the hot air balloon above the ground is [tex]z = 50 \ m[/tex]
The distance of the hot air balloon from the target is [tex]k = 100 \ m[/tex]
The speed of the wind is [tex]v = 15 \ m/s[/tex]
Generally the time it will take the balloon to hit the ground is
[tex]t = \sqrt{ \frac{2 * z }{g} }[/tex]
where g is acceleration due to gravity with value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]t = \sqrt{ \frac{2 * 50 }{9.8} }[/tex]
[tex]t = 3.194 \ s[/tex]
Now at the velocity the distance it will travel before it hit the ground is mathematically represented as
[tex]d = v * t[/tex]
substituting values
[tex]d = 15 * 3.194[/tex]
[tex]d = 47.916 \ m[/tex]
Now in order for the balloon to hit the target on the ground it will need to travel b distance on air before the balloonist drops it and this b distance can be evaluated as
[tex]b = k - d[/tex]
substituting values
[tex]b =100 -47.916[/tex]
[tex]b = 52.084 \ m[/tex]
Hence the time which the balloonist need to wait before dropping the balloon is mathematically evaluated as
[tex]t_w = \frac{b}{v}[/tex]
substituting values
[tex]t_w = \frac{52.084}{15}[/tex]
[tex]t_w = 3.4723 \ s[/tex]
Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb. Assume that light is completely absorbed.
..........................................................
The radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb and the light is completely absorbed is 1.5707x10⁻⁶ N/m².
What is the Radiation pressure?Radiation pressure was defined as the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.
Radiation pressure always includes the Momentum of light or electromagnetic radiation of any wavelength that can be absorbed, reflected, or otherwise emitted by matter on any scale.
E.g: Black-body radiation
Given the values are,
Wattage of bulb = W = 25 W
distance = d = 6.5 cm = 0.065 m
To know the Radiation Pressure,
It can be given by
P = I/c
Where, c = 299792458 m/s is the speed of light,
I is the intensity of radiation and given by
I = W/4πd²
Where W is the Wattage of bulb and d is the distance
I = 25/4π*0.065²
I = 470.872 w/m²
so, the radiation pressure becomes
P = 470.872/299792458
P = 1.5707x10⁻⁶ N/m²
Therefore, the radiation pressure due to a 25 W bulb at a distance of 6.5 cm is 1.5707x10⁻⁶ N/m²
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What would be the correct value of m in the interference equation if the growth process were started again and the diamond layer were grown to three times the thickness t
Answer:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9
Explanation:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9
A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be
Answer:
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Explanation:
Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:
[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.
[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.
Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]
And final speed is now calculated after clearing it:
[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]
[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Isaac drop ball from height og 2.0 m, and it bounces to a height of 1.5 m what is the speed before and after the ball bounce?
Explanation:
It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.
We need to find the speed before and after the ball bounce.
Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,
[tex]\dfrac{1}{2}mu^2=mgh\\\\u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2} \\\\u=6.26\ m/s[/tex]
Let v is the final speed when it bounces to a height of 1.5 m. So,
[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 1.5} \\\\v=5.42\ m/s[/tex]
So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.
A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb
Answer:
931.00ft-lb
Explanation:
Pls see attached file
The work done in moving the object from x = 1 ft to x = 18 ft is 935 ft-lb.
What is work?
Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.
Given that force = 6x - 2 pounds.
So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]
= [ 3x² - 2x]¹⁸₁
= 3(18² - 1² ) - 2(18-1) ft-lb
= 935 ft-lb.
Hence, the work done is 935 ft-lb.
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Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnetic flux to estimate the self-inductance L of the coil.
Answer:
a
[tex]\phi = 1.78 *10^{-7} \ Weber[/tex]
b
[tex]L = 1.183 *10^{-7} \ H[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 6 \ cm = \frac{6}{100} = 0.06 \ m[/tex]
The current it carries is [tex]I = 1.50 \ A[/tex]
The magnetic flux of the coil is mathematically represented as
[tex]\phi = B * A[/tex]
Where B is the magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * I}{2 * r}[/tex]
Where [tex]\mu_o[/tex] is the magnetic field with a constant value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting value
[tex]B = \frac{4\pi * 10^{-7} * 1.50 }{2 * 0.06}[/tex]
[tex]B = 1.571 *10^{-5} \ T[/tex]
The area A is mathematically evaluated as
[tex]A = \pi r ^2[/tex]
substituting values
[tex]A = 3.142 * (0.06)^2[/tex]
[tex]A = 0.0113 m^2[/tex]
the magnetic flux is mathematically evaluated as
[tex]\phi = 1.571 *10^{-5} * 0.0113[/tex]
[tex]\phi = 1.78 *10^{-7} \ Weber[/tex]
The self-inductance is evaluated as
[tex]L = \frac{\phi }{I}[/tex]
substituting values
[tex]L = \frac{1.78 *10^{-7} }{1.50 }[/tex]
[tex]L = 1.183 *10^{-7} \ H[/tex]
A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g glass ball charged to 4.7 nC is shot straight up at 4.8 m/s from the floor level. How high does the ball go if the ceiling voltage is +3.0x10^6V?
Answer:
The ball traveled 0.827 m
Explanation:
Given;
distance between the metal plates of the room, d = 3.1 m
mass of the glass, m = 1.1g
charge on the glass, q = 4.7 nC
speed of the glass ball, v = 4.8 m/s
voltage of the ceiling, V = +3.0 x 10⁶ V
The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;
F = qV/d
|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)
|F| = 4.548 x 10⁻³ N
F = - 4.548 x 10⁻³ N
The net horizontal force experienced by this ball is;
[tex]F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N[/tex]
The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.
[tex]W = F_{net} *h\\\\W = 15.328 *10^{-3} * h[/tex]
W = K.E
[tex]15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m[/tex]
Therefore, the ball traveled 0.827 m
The height at which the ball goes for the given parameters is; 0.827 m
What is the height of the ball?We are given;
distance between the metal plates; d = 3.1 m
mass of glass; m = 1.1g = 0.0011 kg
charge on the glass; q = 4.7 nC = 4.7 × 10⁻⁹ C
speed of the glass ball; v = 4.8 m/s
voltage of the ceiling; V = +3.0 × 10⁶ V
The repulsive force experienced by the ball is gotten from the formula;
F = qV/d
|F| = (4.7 × 10⁻⁹ × 3 × 10⁶)/3.1
|F| = 4.548 × 10⁻³ N
F = -4.548 × 10⁻³ N (negative because it is repulsive force)
The net horizontal force experienced by the ball is;
F_net = F - mg
F_net = (-4.548 × 10⁻³) - (0.0011 × 9.8)
F_net = -15.328 × 10⁻³ N
To get the height of the ball, we will use the formula;
F_net * h = ¹/₂mv²
h = (¹/₂ * 0.0011 * 4.8²)/(15.328 × 10⁻³)
We took the absolute value of F_net, hence it is not negative
h = 0.827 m
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Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus
Answer:
2.55m
Explanation:
Using 1/do+1/di= 1/f
di= (1/f-1/do)^-1
( 1/0.0500-1/0.0510)^-1
= 2.55m
A spaceship is moving past Earth at 0.99c. The spaceship fires two lasers. Laser A is in the same direction it is traveling, and Laser B is in the opposite direction. How fast will the light from each laser be traveling according to an observer on Earth?
Answer:
Vx' = (Vx - u) / (1 - Vx *u / c^2) velocity transformation formula
In both cases we wish to measure the velocity in the frame of the earth which is moving at speed u = -.99 c relative to the spaceship
VA' = (c + .99c) / (1 - (-.99 c * c) / c^2) = 1.99c / 1.99 = c
VB' = (-c + .99c) / (1 - (-c * -.99c) / c^2) = .01 c / .01 = c
In both cases an observer on earth will observe the light traveling at speed c.
The temperature coefficient of resistivity for copper is 0.0068 (C°)-1. If a copper wire has a resistance of 104 Ω at 20°C, what is its resistance 80°C?
Answer:
R₈₀ = 146.43 Ω
Explanation:
The resistance of a resistor depends upon many factors. One of the main factors of the change in resistance of a resistor is the change in temperature. The formula for the resistance at a temperature other than 20°C is given as follows:
R₈₀ = R₀(1 + αΔT)
where,
R₈₀ = Resistance of wire at 80°C = ?
R₀ = Resistance of wire at 20° C = 104 Ω
α = Temperature coefficient of resistance for copper = 0.0068 °C⁻¹
ΔT = T₂ - T₁ = 80°C - 20°C = 60°C
Therefore,
R₈₀ = (104 Ω)[1 + (0.0068°C⁻¹)(60°C)]
R₈₀ = 146.43 Ω
An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. What is the ratio of the masses X/Y of the asteroids
Explanation:
It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.
As they are in equilibrium, the force of gravity due to each other is same. So,
[tex]\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847[/tex]
So, the ratio of masses X/Y is 0.0847
A sound wave of frequency 162 Hz has an intensity of 3.41 μW/m2. What is the amplitude of the air oscillations caused by this wave? (Take the speed of sound to be 343 m/s, and the density of air to be 1.21 kg/m3.)
Answer:
I believe it is 91
Explanation:
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite from the planet is 6600 N. What is the kinetic energy of the satellite
Answer:
The kinetic energy is [tex]KE = 7.59 *10^{10} \ J[/tex]
Explanation:
From the question we are told that
The radius of the orbit is [tex]r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m[/tex]
The gravitational force is [tex]F_g = 6600 \ N[/tex]
The kinetic energy of the satellite is mathematically represented as
[tex]KE = \frac{1}{2} * mv^2[/tex]
where v is the speed of the satellite which is mathematically represented as
[tex]v = \sqrt{\frac{G M}{r^2} }[/tex]
=> [tex]v^2 = \frac{GM }{r}[/tex]
substituting this into the equation
[tex]KE = \frac{ 1}{2} *\frac{GMm}{r}[/tex]
Now the gravitational force of the planet is mathematically represented as
[tex]F_g = \frac{GMm}{r^2}[/tex]
Where M is the mass of the planet and m is the mass of the satellite
Now looking at the formula for KE we see that we can represent it as
[tex]KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r[/tex]
=> [tex]KE = \frac{ 1}{2} *F_g * r[/tex]
substituting values
[tex]KE = \frac{ 1}{2} *6600 * 2.3*10^{7}[/tex]
[tex]KE = 7.59 *10^{10} \ J[/tex]
The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The amount of charge on the plates is now equal to:__________.
a) 4 Q.
b) 2 Q.
c) Q.
d) Q/2.
e) Q/4.
Answer:
D. Q/2
Explanation:
See attached file
Which has more mass electron or ion?
collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity
Answer:
metre per seconds
Explanation:
because velocity = distance ÷ time
why was the observation of the double-slit interference pattern more convincing evidence for the wave theory of light than the observation of diffraction
Answer:
The double slit experiment showed for the first time that light can be interfered, producing bands of light and dark fringes on a screen. This phenomenon was a unique and typical characteristic of waves.
Explanation:
Th double slit experiment by Thomas Young proved, and sealed for the first time the wave nature of light; showing that light just as any other wave can produce interference which was a unique, typical phenomenon of waves. The Interference of light was shown by allowing light to pass through narrow slits and superimpose on a wall or screen, at a distance away from the slit, producing a clear pattern of light and dark fringes. This was the first experiment to proof that darkness can be produced by the addition of light on light. Interference is accompanied by a spatial redistribution of the optical intensity without violation of power conservation. The phenomenon of interference proved the intuitive ideas of Huygens regarding the wave nature of light against Newton's particle nature of light theory.
A centrifuge's angular velocity is initially at 159.0 radians/second to test the stability
of a high speed drill component. It then increases its angular velocity to 999.0
radians/second. If this is achieved in 4,100.0 radians what is the angular acceleration
of the centrifuge?
Answer:
118.6 rad/s²
Explanation:
Δθ = 4100.0 rad
ω₀ = 159.0 rad/s
ω = 999.0 rad/s
Find: α
ω² = ω₀² + 2αΔθ
(999.0 rad/s)² = (159.0 rad/s)² + 2α (4100.0 rad)
α = 118.6 rad/s²
Two parallel plates have charges of equal magnitude but opposite sign. What change could be made to increase the strength of the electric field between the plates
Answer:
The electric field strength between the plates can be increased by decreasing the length of each side of the plates.
Explanation:
The electric field strength is given by;
[tex]E = \frac{V}{d}[/tex]
where;
V is the electric potential of the two opposite charges
d is the distance between the two parallel plates
[tex]E =\frac{V}{d} = \frac{\sigma}{\epsilon _o} \\\\(\sigma = \frac{Q}{A} )\\\\E = \frac{Q}{A\epsilon_o} \\\\E = \frac{Q}{L^2\epsilon_o}[/tex]
Where;
ε₀ is permittivity of free space
L is the length of each side of the plates
From the equation above, the electric field strength can be increased by decreasing the length of each side of the plates.
Therefore, decreasing the length of each side of the plates, could be made to increase the strength of the electric field between the plates
A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and the branch is 12.0 m long.
(a) If the mass of the branch is negligible, what force must be exerted on the free end to just barely lift the rock?
(b) What is the mechanical advantage of this lever system?
Answer:
a
[tex]F =326.7 \ N[/tex]
b
[tex]M = 6[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 200 \ kg[/tex]
The length of the small object from the rock is [tex]d = 2 \ m[/tex]
The length of the small object from the branch [tex]l = 12 \ m[/tex]
An image representing this lever set-up is shown on the first uploaded image
Here the small object acts as a fulcrum
The force exerted by the weight of the rock is mathematically evaluated as
[tex]W = m_r * g[/tex]
substituting values
[tex]W = 200 * 9.8[/tex]
[tex]W = 1960 \ N[/tex]
So at equilibrium the sum of the moment about the fulcrum is mathematically represented as
[tex]\sum M_f = F * cos \theta * l - W cos\theta * d = 0[/tex]
Here [tex]\theta[/tex] is very small so [tex]cos\theta * l = l[/tex]
and [tex]cos\theta * d = d[/tex]
Hence
[tex]F * l - W * d = 0[/tex]
=> [tex]F = \frac{W * d}{l}[/tex]
substituting values
[tex]F = \frac{1960 * 2}{12}[/tex]
[tex]F =326.7 \ N[/tex]
The mechanical advantage is mathematically evaluated as
[tex]M = \frac{W}{F}[/tex]
substituting values
[tex]M = \frac{1960}{326.7}[/tex]
[tex]M = 6[/tex]