Answer:
the answer should be b) 3.8 x 10-5 v
Consider a piston filled with 3 mols of an ideal gas, kept at a constant temperature 290 K. We slowly compress the gas starting at 2 m3 and ending at 1 m3. How much work do we need to do on the gas to perform this operation
Answer: [tex]-5013.65\ J[/tex]
Explanation:
Given
No of moles [tex]n=3[/tex]
Temperature [tex]T=290\ K[/tex]
Initial volume [tex]V_1=2\ m^3[/tex]
Final volume [tex]V_2=1\ m^3[/tex]
Work done in constant temperature process is
[tex]W=nRT\ln \left(\dfrac{V_2}{V_1}\right)[/tex]
Insert the values
[tex]\Rightarrow W=3\times 8.314\times 290\ln \left (\dfrac{1}{2}\right)\\\\\Rightarrow W=-870\times 8.314\times \ln (2)\\\Rightarrow W=-5013.65\ J[/tex]
what is the distance time how can we find the speed of an object from its distance time graph
Answer:
speed is the gradient of the graph
Answer:
Speed is the slope of a distance time graph.
Explanation:
Speed= d/t
Slope is equal to rise/run
If the rise of the graph is the distance and the run is the time, calculating slope is the equivalent of calculating average speed.
A lightning bolt has a current of 56,000 A and lasts for 80 x 10-6 seconds (80 μs). How much charge (in Coulombs) has flowed in this bolt?
Answer:
A cloud can discharge as much as 20 coulombs in a lightning bolt.
The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars
Answer:
the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Explanation:
Given the data in the question;
Time period = 6 months = 1.577 × 10⁷ s
orbital speed v = 64000 m/s
since its a circular orbit,
v = 2πr / T
we solve for r
r = vT/ 2π
r = ( 64000 × 1.577 × 10⁷ ) / 2π
r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU
Now, from Kepler's law
T² = r³ / ( m₁ + m₂ )
T = 6 months = 0.5 years
we substitute
(0.5)² = (1.0737)³ / ( m₁ + m₂ )
0.25 = 1.2378 / ( m₁ + m₂ )
( m₁ + m₂ ) = 1.2378 / 0.25
( m₁ + m₂ ) = 4.9512
m₁ = m₂ = 4.9512 / 2 = 2.4756 solar mass
we know that solar mass = 1.989 × 10³⁰ kg
so
m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg
m₁ = m₂ = 4.92 × 10³⁰ kg
Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
why meter cube is called derived unit
Answer:
Because it is the result of two more fundamental units, a derived unit is termed that. For volume, the cubic meter (m³) is the fundamental unit of area. Any number that cannot be measured directly with any equipment is referred to as a derived unit. For example, we can't quantify a substance's density using a rule, scale, or bucket.
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Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
The current through each resistor in the two-resistor circuit is _________ the current through the resistor in the one-resistor circuit (the circuit in Part A). The voltage across each resistor in the two-resistor circuit is ___________ the voltage across the resistor in the one-resistor circuit.
Answer:
Serial circuit. the current is constant.
the voltage across a given resistor is half the rating in a one-resistor circuit.
Parallel circuit the voltage is constant
the current is half the value of the current with a single resistor
Explanation:
To answer exactly this exercise, you need the diagram or the indication of the type of circuit being used, since there are two possibilities, let's consider the results of each one.
Serial circuit.
The two resistance are one after the other.
In this case the current in the resistor sides is the same, that is, the current is constant in the circuit.
The voltage is proportional to the value of each resistor and if the two resistors are equal, the voltage across a given resistor is half the rating in a one-resistor circuit.
Parallel circuit
the two resistance is next to each other.
In this case the voltage is constant, that is, the voltage across the two resistors is the same as in the case of a single resistor.
The current is inversely proportional to the value of the resistance
i₁ = V / R₁
i₂ = V / R₂
for a single resistance
I = V / R
these currents are related
i = i₁ + i₂
if the two resistors have the same value the current is half the value of the current with a single resistor
Define hydropower or hydroelectric power ?
No Spam..
[tex]\:[/tex]
Hydroelectric power, also called hydropower is the electricity produced from generators driven by turbines that convert the potential energy of falling or fast-flowing water into mechanical energy.
Answer:
Hydroelectric power/hydropower - electricity produced by a hydraulic source, specifically energy generated falling or flowing water
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to 109 rad/sec in 1.87 seconds. The blades of the fan have a diameter of 6.7 meters and their deceleration rate is 4.7 rad/sec2.
What was the initial angular speed of the fan in rev/sec?
ωi =
Answer:
wo = 18.75 rev / s
Explanation:
This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²
w_f = w₀ - a t
w₀ = w_f + a t
w₀ = 109 + 4.7 1.87
w₀ = 117.8 rad / s
let's reduce to revolutions / s
w₀ = 117.8 rad / s (1 rev / 2pi rad)
w₀ = 18.75 rev / s
calculate the pressure of water having density 1000 kilo per metre square at a depth of 20 m inside the water
Answer:
the pressure of the water at the given depth is 196,200 N/m².
Explanation:
Given;
density of the water, ρ = 1000 kg/m³
depth of the water, h = 20 m
acceleration due to gravity, g = 9.81 m/s²
The pressure at the given depth of the water is calculated as;
P = ρgh
P = 1000 x 9.81 x 20
P = 196,200 N/m²
Therefore, the pressure of the water at the given depth is 196,200 N/m².
Joule is a SI unit of power
Measuring cylinder is used to measure the volume of a liquid
Answer:
The SI unit of power is watt
Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 655 nm and a peak electric field magnitude of 1.5 V/m. 0.002984 W/m2 (b) an electromagnetic wave with an angular frequency of 6.5 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T. 1.19366E-6 W/m2
The intensity of the electromagnetic wave in terms of the electric field is 0.00298 W/m² and the intensity of the electromagnetic wave in terms of the magnetic field is 1.193x10⁻⁶ W/m².
The intensity of the electromagnetic wave is related to the electric field as well as to the magnetic field.
a) Intensity of the electromagnetic wave for the electromagnetic field.
The intensity of the electromagnetic wave (I) in terms of the electromagnetic field is given by:
[tex] I = \frac{E^{2}*c*\epsilon_{0}}{2} [/tex] (1)
Where:
c: is the speed of light = 3.00*10⁸ m/s
E: is the magnitude of the electric field = 1.5 V/m
ε₀: is the permittivity of free space = 8.85*10⁻¹² C²/Nm²
Hence, the intensity of the electromagnetic wave (eq 1) is:
[tex] I = \frac{(1.5 V/m)^{2}*3.00 \cdot 10^{8} m/s*8.85 \cdot 10^{-12} C^{2}/(N*m^{2})}{2} = 0.00298 W/m^{2} [/tex]
b) Intensity of the electromagnetic wave for the magnetic field
We can calculate the intensity of the electromagnetic wave (I) in terms of the magnetic field with the following equation:
[tex] I = \frac{cB^{2}}{2\mu_{0}} [/tex] (2)
Where:
B: is the magnitude of the magnetic field = 10⁻¹⁰ T
μ₀: is the vacuum permeability = 4π*10⁻⁷ m*T/A
Therefore, the intensity of the electromagnetic wave (eq 2) is:
[tex] I = \frac{3.00 \cdot 10^{8} m/s*(1\cdot 10^{-10} m*T/A)^{2}}{2*4\pi \cdot 10^{-7} T/A} = 1.193 \cdot 10^{-6} W/m^{2} [/tex]
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A wire, 0.60 m in length, is carrying a current of 2.0 A and is placed at a certain angle with respect to the magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field
Answer:
[tex]\theta=30 \textdegree[/tex]
Explanation:
From the question we are told that:
Current [tex]I=2.0A[/tex]
Length [tex]L=0.60m[/tex]
Magnetic field [tex]B=0.30T[/tex]
Force [tex]F=0.18N[/tex]
Generally the equation for Force is mathematically given by
[tex]F = BIL sin\theta[/tex]
[tex]sin\theta=\frac{F}{BIL}[/tex]
[tex]\theta=sin^{-1}\frac{0.18}{0.3*2*0.6}[/tex]
[tex]\theta=30 \textdegree[/tex]
Suppose a teenager on her bicycle. The rear wheel is spinning at an angular velocity of 281.133 rpm. She stops it in 3.686 s. How many revolution did it take to stop it?
Answer:
Explanation:
The formula for angular velocity is
[tex]\omega=\frac{\theta}{t}[/tex] where omega is the angular velocity, theta is the change in the angular rotation, and t is the time in seconds. First and foremost, we have the angular rotation in minutes and the time in seconds, so that's a problem we have to amend. Let's change the angular rotation to rotations per second:
[tex]281.133\frac{r}{min}*\frac{1min}{60s}=4.68555\frac{r}{s}[/tex]
Now we're ready to set up the problem:
[tex]4.68555=\frac{\theta}{3.686}[/tex] and we multiply both sides by 3.686 to get the rotations per seconds:
θ = 17.27 rotations
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C. where, d= 11 cm Q= 12.5 C
Answer:
The electric field is 9.3 x 10^12 N/C and the direction is away from the charge.
Explanation:
charge, Q = 12.5 C
distance, d = 11 cm = 0.11 m
Let the electric field is E.
[tex]E =\frac{K Q}{d^2}\\\\E = \frac{9\times 10^9\times 12.5}{0.11\times 0.11}\\\\E = 9.3\times 10^{12} N/C[/tex]
The direction of electric filed is away from the charge.
You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)
Answer:
The current is more in the parallel combination than in the series combination.
Explanation:
two resistances, R1 and R2 are connected to a battery of voltage V.
When they are in series,
R = R1 + R2
In series combination, the current is same in both the resistors, and it is given by Ohm's law.
V = I (R1 + R2)
[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)
When they are connected in parallel.
the voltage is same in each resistor.
The effective resistance is R.
[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]
So, the current is
[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)
So, the current is more is the parallel combination.
Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]
Answer:
Gap left = Change in length on heating
Gap=Initial length×Coefficient of linear expansion×change in temperature
Gap=10×0.000012×15m
⟹Gap=0.0018 m
this is an example u have to put your equation in it
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 24 ft/s2. What is the distance (in ft) traveled before the car comes to a stop? (Round your answer to one decimal place.)
The car has initial speed 50 mi/h ≈ 73 ft/s, so it covers a distance x such that
0² - (73 ft/s)² = 2 (-24 ft/s²) x
==> x ≈ 111.0 ft
Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to view distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?
Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
Two friends, Al and Jo, have a combined mass of 194 kg. At the ice skating rink, they stand close together on skates, at rest and facing each other. Using their arms, they push on each other for 1 second and move off in opposite directions. Al moves off with a speed of 7.9 m/sec in one direction and Jo moves off with a speed of 6.7 m/sec in the other. You can assume friction is negligible.
What is Al's mass? 110.58 What is Jo's mass? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Al on Jo? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Jo on Al?
Answer:
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
Explanation:
Given the absence of external forces, this situation can be described will by Principle of Linear Momentum Conservation and Impact Theorem on each skater:
Al:
[tex]m_{1}\cdot (v_{1, f}-v_{1, o}) = -F \cdot \Delta t[/tex] (1)
Jo:
[tex]m_{2}\cdot (v_{2,f}-v_{2,o}) = F\cdot \Delta t[/tex] (2)
Total mass:
[tex]m_{1} + m_{2} = 194\,kg[/tex]
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the skaters, in kilograms.
[tex]v_{1,o}[/tex], [tex]v_{1,f}[/tex] - Initial and final velocities of Al, in meters per second.
[tex]v_{2,o}[/tex], [tex]v_{2,f}[/tex] - Initial and final velocities of Jo, in meters per second.
[tex]F[/tex] - Impact force between skaters, in newtons.
[tex]\Delta t[/tex] - Impact time, in seconds.
If we know that [tex]v_{1,o} = 0\,\frac{m}{s}[/tex], [tex]v_{1,f} = -7.9\,\frac{m}{s}[/tex], [tex]\Delta t = 1\,s[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex] and [tex]v_{2,f} = 6.7\,\frac{m}{s}[/tex], then the masses of the skaters are, respectively:
[tex](194-m_{2})\cdot (-7.9) = -F[/tex] (1b)
[tex]m_{2} \cdot 6.7 = F[/tex] (2b)
(2b) in (1b):
[tex](194-m_{2})\cdot (-7.9) = -m_{2}\cdot 6.7[/tex]
[tex]-1532.6 +7.9\cdot m_{2} = -6.7\cdot m_{2}[/tex]
[tex]14.6\cdot m_{2} = 1532.6[/tex]
[tex]m_{2} = 104.973\,kg[/tex]
[tex]m_{1} = 194\,kg - 104.973\,kg[/tex]
[tex]m_{1} = 89.027\,kg[/tex]
And the magnitude of the force is:
[tex]F = 6.7\cdot m_{2}[/tex]
[tex]F = 596.481\,N[/tex]
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
A 3.1-mole sample of an ideal gas is gently heated at constant temperature 320 K. It expands from initial volume 23 L to final volume V2. A total of 1.7 kJ of heat is added during the expansion process. What is V2? Let the ideal-gas constant R = 8.314 J/(mol • K).
From the ideal gas law,
PV = nRT ==> P = nRT/V
where P is the pressure exerted by the gas on the container. The work W done by this pressure as the volume of the gas changes from V₁ to V₂ is given by the integral,
[tex]W = \displaystyle \int_{V_1}^{V_2}P\,\mathrm dV \implies W = nRT \ln\left(\dfrac{V_2}{V_1}\right)[/tex]
and solving for V₂ gives
[tex]V_2 = V_1\exp\left(\dfrac{W}{nRT}\right)[/tex]
If you add 1.7 kJ of heat to the system, which does the aforementioned work, the gas will expand to a volume of
[tex]V_2 = (23\,\mathrm L)\exp\left(\dfrac{1.7\,\mathrm{kJ}}{(3.1\,\mathrm{mol})\left(8.314\frac{\rm J}{\mathrm{mol}\cdot\mathrm K}\right)(320\,\mathrm K)}\right) \approx \boxed{28 \,\mathrm L}[/tex]
A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.
Answer:
B = 0.013(-j) T
Explanation:
Given that,
The speed of a proton, [tex]v=1.75\times 10^7\ m/s[/tex]
Acceleration experienced by the proton,[tex]a=2.25\times 10^3\ m/s[/tex]
We need to find the magnitude and the direction of the magnetic field. At equilibrium,
[tex]ma=qvB\\\\B=\dfrac{ma}{qv}\\\\B=\dfrac{1.67\times 10^{-27}\times 2.25\times 10^{13}}{1.6\times 10^{-19}\times 1.75\times 10^{7}}\\\\B=0.013\ T[/tex]
The velocity is in +z direction, force in +x direction, then the field must be in -y direction.
If the accuracy in measuring the velocity of a particle increases, the accuracy in measuring its position will:__________.
a. It is impossible to say since the two measurements are independent and do not affect each other.
b. remain the same.
c. increase.
d. decrease.
When the accuracy in measuring the velocity of a particle increases, that of its position decreases. The correct option is d. decrease.
The Heisenberg's uncertainty principle states that there is always an uncertainty in any attempt to measure accurately the value of any two complementary variables simultaneously. This implies that measuring of position and momentum of a particle simultaneously would lead to uncertainty in their values.
So that;
Δx.ΔP ≥ [tex]\frac{h}{2\pi }[/tex]
Where Δx is the uncertainty in the value of its position, ΔP is the uncertainty in the value of momentum and h is the Planck's constant.
This principle simply explains and provides the required answer to the given question. So that according to Heisenberg's uncertainty principle, the accuracy in measuring its position will definitely decrease.
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Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.
Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Solution :
We know,
Distance,
[tex]$S=ut+\frac{1}{2}at^2$[/tex]
[tex]$S=ut+0.5(a)(t)^2$[/tex]
For the first 20 ms,
[tex]$S=0+0.5(220)(0.020)^2$[/tex]
S = 0.044 m
In the remaining 30 ms, it has constant velocity.
[tex]$v=u+at$[/tex]
[tex]$v=0+(220)(0.020)[/tex]
v = 4.4 m/s
Therefore,
[tex]$S=ut+0.5(a)(t)^2$[/tex]
[tex]$S'=4.4 \times 0.030[/tex]
S' = 0.132 m
So, the required distance is = S + S'
= 0.044 + 0.132
= 0.176 m
Therefore, the tongue can reach = 0.176 m or 17.6 cm
Answer:
The total distance is 0.176 m.
Explanation:
For t = 0 s to t = 20 ms
initial velocity, u = 0
acceleration, a = 220 m/s^2
time, t = 20 ms
Let the final speed is v.
Use first equation of motion
v = u + at
v = 0 + 220 x 0.02 = 4.4 m/s
Let the distance is s.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]
Now the distance is
s' = v x t
s' = 4.4 x 0.03 = 0.132 m
The total distance is
S = s + s' = 0.044 + 0.132 = 0.176 m
You and your friends find a rope that hangs down 11 m from a high tree branch right at the edge of a river. You find that you can run, grab the rope, swing out over the river, and drop into the water. You run at 2.0 m/s and grab the rope, launching yourself out over the river.
Required:
How long must you hang on if you want to drop into the water at the greatest possible distance from the edge?
Answer:
if you want to drop into the water at the greatest possible distance from the edge, you must hang for 1.662s.
Explanation:
The time period of the oscillation is,
[tex]T = 2\pi \sqrt{ \frac{I} {g }[/tex]
[tex]T = 2\pi \sqrt{\frac{11}{9.8} } \\\\T= 6.65 s[/tex]
This would be the time taken for the person to move from.
The duration of time he hangs over the river be one-fourth of the time period.
Here,
[tex]t= \frac{T}{4} \\\\t=\frac{6.65}{4}\\\\t = 1.662 s[/tex]
What computer measures physical quantities?
Answer:
Three Types of Computer The Computer are classified into three main types:: • Analog Computers • Digital Computers • Hybrid Computers (Analog + Digital) Analog Computers:: Analog Computer measures “Physical Quantities” for example Temperature, Voltage, Pressure, and Electric Current.
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is
The question is incomplete. The complete question is :
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s) and the dissolution rate (kg/s).
Solution :
From flow over sphere, the mass transfer equation can be written as :
[tex]$Sh = 2 + 0.6 Re^{1/2} Sc^{1/3}$[/tex]
where, Sherood number, [tex]$Sh = \frac{K_L d}{D_{eff}}$[/tex]
Reynolds number, [tex]$Re=\frac{Vd\rho}{\mu}$[/tex]
Schmid number, [tex]$Sc= \frac{\mu}{\rho D_{eff}}$[/tex]
So,
[tex]$\frac{K_L d}{D_{eff}}=2+0.6 \left( \frac{V d \rho}{\mu} \right)^{1/2} \ \left( \frac{\mu}{\rho D_{eff}} \right)^{1/3}$[/tex]
Diameter, d = 1 cm = [tex]$1 \times 10^{-2}$[/tex] m
V = 1 m/s
[tex]$\rho = 1000 \ kg/m^3$[/tex]
[tex]$\mu = 10^{-3} \ kg/m/s$[/tex]
[tex]$D_{eff} = 2 \times 10^{-9} \ m^2/s$[/tex]
[tex]$\frac{K_L \times 10^{-2}}{2 \times 10^{-9}}=2+0.6 \left( \frac{1 \times 10^{-2} \times 10^3}{10^{-3}} \right)^{1/2} \ \left( \frac{10^{-3}}{10^3 \times 2 \times 10^{-9}} \right)^{1/3}$[/tex]
[tex]$K_L \times 5 \times 10^6=478.22$[/tex]
[tex]$K_L=9.5644 \times 10^{-5}$[/tex] m/s
So the mass transfer coefficient is 9.5644 [tex]$\times 10^{-5}$[/tex] m/s. It is given solubility,
[tex]$\Delta C = 2 \ kg/m^3$[/tex]
[tex]$N = Md^2 \times \Delta C \times K_L$[/tex]
[tex]$N= M \times (10^{-2})^2 \times 2 \times 9.5644 \times 10^{-5}$[/tex]
[tex]$N= 6 \times 10^{-8}$[/tex] kg/s (dissolution rate)
I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above
Answer:
b = frequency
point charges q1=50 uc and q2=-25 uc are placed 1 m apart. what is the force on a third chare q3=2 uc placed midway between q1 and q2? where must q3 of the preceding problem be placed so that the net force on it is zero?
Answer:
d = -1 m
The negative sign indicates that the charge from that force of the space of the two spheres.
Explanation:
That is a problem of electric forces, given by Coulomb's law
F = [tex]k \frac{ q1q2}{r^2}[/tex]
We use that charges of the same sign repel and charges of different signs do not attract, so the net force is
∑ = F₁₃ + F₂₃
F_ {net} = [tex]k \frac{q_1q_3}{r_{13}^2} + k \frac{q_2q_3}{ r_{23}^}[/tex]
a) the charge is placed at the midpoint between the other two
r₁₃ = r₁₂ = R = ½ m = 0.5mF_ {net} =[tex]\frac{k}{R^2 } \ q3 ( q1+q2)[/tex]
calculate us
F_ {net} = 9 10⁹ / 0.5² 2 10⁻⁶ (50 -25) 10⁻⁶
F_ {net} = 1,800 N
b) where must be placed q3 so that the force is zero
for this case the charge q3 is outside the spheres
∑ F = 0
F₁₁₃ = F₂₃
k q_1 / r_{13}² = k q₂ q₃ / r₂₃²
q₁/ r₁₂² = q₂ / r₂₃²
suppose the distance
r₁₂ = d
the he other sphere is
r₂₃ = d + 1
we substitute
q₃ / d² = q₂ / (d + 1) ²
(d + 1) ² = q₂ / q₃ d²
d² (1 - q₂/ q₃) + 2d + 1 = 0
we solve the equation of a second
d = [-2 + [tex]\sqrt{2^2 - 4 1 ( 1+25/50}[/tex] ] / 2
d = -2 /2
d = -1 m
The negative sign indicates that the charge from that force of the space of the two spheres.
The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?
Answer:
[tex]v_1=87.40m/s[/tex]
Explanation:
From the question we are told that:
Mass of arrow [tex]m=52g[/tex]
Mass of rock [tex]m_r=1.50kg[/tex]
Height [tex]h=0.47m[/tex]
Generally the equation for Velocity is mathematically given by
[tex]v = \sqrt{(2gh)}[/tex]
[tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]
[tex]v= 3.035m/s[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_1v_1=m_2v_2[/tex]
[tex]0.052kg * v = 1.5 * 3.03m/s[/tex]
[tex]v_1=87.40m/s[/tex]
