Answer:
the water that remain in the tank in one hour will be 200 lb
Explanation:
Initial mass of water in the tank = 2000 lb
hot water is delivered through the first inlet pipe at a rate of = 0.8 lb/s
cold water is delivered through the second inlet pipe at a rate of = 1.2 lb/s
exit pipe flow rate = 2.5 lb/s
amount of water in the tank after one hour = ?
In one hour, there are 60 x 60 seconds = 3600 sec, therefore
the water through the first inlet pipe in one hour = 0.8 x 3600 = 2880 lb
the water through the second inlet pipe in one hour = 1.2 x 3600 = 4320 lb
the water through the exit in one hour = 2.5 x 3600 = 9000 lb
The total amount of water in the tank = 2000 + 2880 + 4320 = 9200 lb
The total amount of water that leaves the tank = 9000 lb
therefore, in one hour, the water that remain in the tank will be
==> 9200 lb - 9000 lb = 200 lb
15. The blank
of a sine wave is the time it takes to complete one cycle of the wave.
O A. maximum amplitude
O B. minimum amplitude
O C. average value
O D. wavelength
That time is the "period" of the wave.
(It's not one of the choices.)
The blank of a sine wave is the time it takes to complete one cycle of the wavelength, the correct answer is D.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
It is the total length of the wave for which it completes one cycle.
The wavelength is inversely proportional to the frequency of the wave as from the following relation.
C = νλ
where c is the speed of light
ν is the frequency of the wave
λ is the wavelength of the wave
The time taken by the sine wave to complete one cycle of the wavelength is called blank the correct answer is D.
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A vertically polarized light wave of intensity 1000 mW/m2 is coming toward you, out of the screen. After passing through this polarizing filter, the wave's intensity is
Answer:
The intensity is [tex]I = 500 mW/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]
Generally the intensity of the light emerging from the polarizer is mathematically represented as
[tex]I = \frac{I_o}{2}[/tex]
substituting values
[tex]I = \frac{1000 *10^{-3}}{2}[/tex]
[tex]I = 500 *10^{-3} W/m^2[/tex]
[tex]I = 500 mW/m^2[/tex]
What is the major cause of the muffled noise from a radio station?
Answer:
The major cause is "lack of high frequencies in a sound wave".
Explanation:
Muffling derives from either the absence of such a radio signal of a higher or specific frequency. This very same phenomenon has been observed whenever you overhear conversations through some kind of wall and perhaps door. The approach is equalization. This method helps them to raise those frequencies although these overprotective wavelengths decrease.So that the above would be the correct solution.
change in entropy of universe during 900g of ice at 0 degree celcus to water at 10 degree celcius at room temp=30 degree celcius
Answer:
4519.60 J/KExplanation:
Change in entropy is expressed as ΔS = ΔQ/T where;
ΔQ is the total heat change during conversion of ice to water.
T is the room temperature
First we need to calculate the total change in heat using the conversion formulae;
ΔQ = mL + mcΔθ (total heat energy absorbed during phase change)
m is the mass of ice/water = 900g = 0.9kg
L is the latent heat of fusion of ice = 3.33 x 10⁵J/kg
c is the specific heat capacity of water = 4200J/kgK
Δθ is the change in temperature of water = 10°C - 0C = 10°C = 283K
Substituting the given values into ΔQ;
ΔQ = 0.9(333000)+0.9(4200)(283)
ΔQ = 299700 + 1069740
ΔQ = 1,369,440 Joules
Since Change in entropy ΔS = ΔQ/T
ΔS = 1,369,440/30+273
ΔS = 1,369,440/303
ΔS = 4519.60 J/K
Hence, the change in entropy of the universe is 4519.60 J/K
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].
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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.
"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
Find the distance to a Sun-like star (L=3.8x1026 watts) whose apparent brightness at Earth is 1.0 x10-10 watt/m2.
Answer:
5.49 x 10^17 m is the distance between the sun-like star to the earth
Explanation:
Radiation intensity on Earth = 1.0 x 10^-10 W/m^2
Power of radiation of the star = 3.8 x 10^26 W
Recall that the intensity of radiation is given as
[tex]I[/tex] = [tex]\frac{P}{A}[/tex]
where
[tex]I[/tex] = intensity of radiation
P = power of radiation
A is the area through which the radiation spreads out in all three dimensional direction.
A = [tex]\frac{P}{I}[/tex] = [tex]\frac{3.8*10^{26} }{1.0*10^{-10} }[/tex] = 3.8 x 10^36 m^2
This area is spread out in the form of a sphere of area
A = [tex]4\pi r^{2}[/tex] = 4 x 3.142 x [tex]r^{2}[/tex]
3.8 x 10^36 = 12.568[tex]r^{2}[/tex]
[tex]r^{2}[/tex] = (3.8 x 10^36)/12.568 = 3.02 x 10^35
r = [tex]\sqrt{3.02*10^{35} }[/tex] = 5.49 x 10^17 m this is the distance of the star to the Earth
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.
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
The Goliath six flags Magic Mountain roller coaster ride starts at 71.6 m (235 feet) above the ground. Assuming the coaster starts from rest and ignoring any friction, what is the speed of the coaster when it reaches the ground level
Answer:
The velocity is [tex]v = 37 .46 \ m/s[/tex]
Explanation:
From the question we are told that
The start distance above the ground is [tex]h = 71.6 \ m[/tex]
Generally according to the law of energy conservation we have that
[tex]PE_{top} = KE_{bottom }[/tex]
Where [tex]PE_{top}[/tex] is potential energy at the top which is mathematically represented as
[tex]PE_{top} = m * g * h[/tex]
And [tex]KE_{bottom }[/tex] is the kinetic energy at the bottom which is mathematically represented as
[tex]KE_{bottom } = \frac{1}{2} * m * v^2[/tex]
Therefore
[tex]m * g * h = \frac{1}{2} * m * v^ 2[/tex]
=> [tex]v = \sqrt{2 * g * h }[/tex]
substituting value
[tex]v = \sqrt{2 * 9.8 * 71.6 }[/tex]
[tex]v = 37 .46 \ m/s[/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
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.
differences between
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and horse
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.
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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.
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.
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A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)
Answer:
[tex]3.1\times 10^{5}m/s[/tex]
Explanation:
The computation of the speed does the proton gain is shown below:
The potential difference is the difference that reflects the work done as per the unit charged
So, the work done should be
= Potential difference × Charge
Given that
Charge on a proton is
= 1.6 × 10^-19 C
Potential difference = 500 V
[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]
[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]
Simply we applied the above formulas
A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the switch is open and the capacitor is uncharged. What is the voltage across the resistor and the capacitor at the moment the switch is closed
Answer:
The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.
Explanation:
This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery
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
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
The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm
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 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.
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
A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40 MΩ. After a time of 4.00 s, the voltmeter reads 3.0 V.
A) What are the capacitance?
B) What is the time constant of the circuit?
Answer:
a. 0.849 micro farad
b. 2.89 s
Explanation:
a) V=V0 e^-t/RC
3=12*e^-4/3.4*10^6*C
3/12=e^-4/3.4*10^6*C
-1.3863 =-4/3.4*10^6*C
C=8.49*10^-7 F
=0.849 micro farad
B) time constant= R*C
=3.4*10^6*8.49*10^-7
=2.89 S
a. The capacitance is 0.849 micro farad
b. The time constant of the circuit is 2.89 s
Calculation of capacitance & time constant:a)
We know that
V=V0 e^-t/RC
3=12*e^-4/3.4*10^6*C
3/12=e^-4/3.4*10^6*C
-1.3863 =-4/3.4*10^6*C
C=8.49*10^-7 F
=0.849 micro farad
B)
Now
time constant= R*C
=3.4*10^6*8.49*10^-7
=2.89 S
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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
g The current in a series circuit is 15.0 A. When an additional 8.00-% resistor is inserted in series, the current drops to 12.0 A. What is the resistance in the original circuit
Answer:
Explanation:
Let the original resistance be R and voltage be V
Applying ohm's law
V / R = 15
V = 15 R
In second case
V / (R+8 ) = 12
V = 12 R + 96
15 R = 12 R + 96
3R = 96
R = 32 ohm .
1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?
Answer:
speed when it reaches y = 4.00cm is
v = 14.9 g.m/s
Explanation:
given
q₁=q₂ =2.00 ×10⁻⁶
distance along x = 3.00cm= 3×10⁻²
q₃= 4×10⁻⁶C
mass= 10×10 ⁻³g
distance along y = 4×10⁻²m
r₁ = [tex]\sqrt{3^{2} +2^{2} }[/tex] = [tex]\sqrt{13}[/tex] = 3.61cm = 0.036m
r₂ = [tex]\sqrt{4^{2} + 3^{2} }[/tex] = [tex]\sqrt{25}[/tex] = 5cm = 0.05m
electric potential V = [tex]\frac{kq}{r}[/tex]
change in potential ΔV = [tex]V_{1} - V_{2}[/tex]
ΔV = [tex]\frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} }[/tex] , where [tex]q_{1} = q_{2}=[/tex]2.00μC
ΔV = [tex]2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })[/tex]
ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × [tex](\frac{1}{0.036} - \frac{1}{0.05} )[/tex]
ΔV= 2.789×10⁵
[tex]\frac{1}{2}mv^{2}[/tex] = ΔV × q₃
[tex]\frac{1}{2}[/tex] ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶
v² = 223.12 g.m/s
v = 14.9 g.m/s
The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s
We are given;
Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C
Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C
Distance of charge 1 along x = 3 cm = 3 × 10⁻² m
Distance of charge 2 along x = -3 cm = -3 × 10⁻² m
Charge 3; q₃ = +4.00 μC = 4 × 10⁻⁶ C
mass; m = 0.01 g
distance of charge 3 along y = 4 cm = 4 × 10⁻² m
q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.
Thus;
Distance of charge 1 from the initial position of q₃;
r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)
r₁ = 0.0361 m
Distance of charge 2 from the final position of q₃;
r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)
r₂ = 0.05 m
Now, formula for electric potential is;
V = kq/r
Where k = 9 × 10⁹ N.m²/s²
Thus,change in potential is;
ΔV = V₁ - V₂
Now, Net V₁ = 2kq₁/r₁
Net V₂ = 2kq₂/r₂
Thus;
ΔV = 2kq₁/r₁ - 2kq₂/r₂
ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]
ΔV = 277229.92 V
Now, from conservation of energy;
½mv² = q₃ΔV
Thus;
½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92
v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01
v = √(221.783936)
v = 14.89 m/s
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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/s