Answer:
(a) W = 3293 J = 3.293 KJ
(b) W = 0 KJ
(c) W = -506.6 J = -0.507 KJ
(d) W = 0 KJ
Net Work = 2.786 KJ = 2786 J
Explanation:
(a)
The work done is given as:
[tex]W = P\Delta V\\W = P(V_2-V_1)[/tex]
where,
P = Constant Pressure = (6.5 atm)(101325 Pa/1 atm) = 6.59 x 10⁶ Pa
V₁ = initial volume = (1 L)(0.001 m³/1 L) = 0.001 m³
V₂ = final volume = (6 L)(0.001 m³/1 L) = 0.006 m³
Therefore,
[tex]W = (6.59\ x\ 10^6\ Pa)(0.006\ m^3-0.001\ m^3)[/tex]
W = 3293 J = 3.293 KJ
(b)
Since the volume is constant in this stage. Therefore,
ΔV = 0
As a result:
W = 0 KJ
(c)
The work done is given as:
[tex]W = P\Delta V\\W = P(V_2-V_1)[/tex]
where,
P = Constant Pressure = (1 atm)(101325 Pa/1 atm) = 1.01 x 10⁵ Pa
V₁ = initial volume = (6 L)(0.001 m³/1 L) = 0.006 m³
V₂ = final volume = (1 L)(0.001 m³/1 L) = 0.001 m³
Therefore,
[tex]W = (1.01\ x\ 10^5\ Pa)(0.001\ m^3-0.006\ m^3)[/tex]
W = -506.6 J = -0.507 KJ
negative sign show that the work is done on the gas by the environment.
(d)
Since the volume is constant in this stage. Therefore,
ΔV = 0
As a result:
W = 0 KJ
Net work will be the sum of all the works:
Net Work = 3.293 KJ + 0 KJ - 0.507 KJ + 0 KJ
Net Work = 2.786 KJ = 2786 J
A current 2a flows in a circuit fir 2 minutes. Calculate the charge generated in a circuit
Answer:
the charge generated in the circuit is 240 C.
Explanation:
Given;
current flowing in the circuit, I = 2A
time of current flow, t = 2 minutes = 2 x 60s = 120 s
The current flowing through a given circuit is defined as the quantity of charge flowing through the circuit in a given time.
[tex]I = \frac{Q}{t} \\\\Q = I t[/tex]
where;
Q is the charge flowing in the circuit
Q = 2 x 120
Q = 240 C
Therefore, the charge generated in the circuit is 240 C.
The picture shows an investigation conducted by Galileo many years ago. He learned that the speed of each ball increased as it fell, and that the balls fell at the same average speed. image Which of these statements is best supported by Galileo’s investigation?
Answer:
Italian astronomer Galileo Galilei provided a number of scientific insights that laid the foundation for future scientists. His investigation of the laws of motion and improvements on the telescope helped further the understanding of the world and universe around him.Nov 14, 2017
Explanation:
You know that for any object to escape the gravitational pull of a planet, star, and so on, the object must first achieve escape velocity, or vesc. What velocity must the Veracious achieve to escape Earth
Answer:
11.186 * 10^3 m/s
Explanation:
For a Veracious to escape the gravitational pull of the earth it has to a attain
Vesc = [tex]\sqrt{\frac{2GM}{R} }[/tex] -------- ( 1 )
where: G = ( gravitational constant ) = 6.674 * 10^-11 m^3 kg^-1 s^-2 )
M = ( mass of earth) = 5.972 * 10^24 kg
R = ( Radius of earth ) = 6.371 * 10^6 m
next we will Input values into equation ( 1 )
The required Vesc that the Veracious can attain to escape the gravitational pull of the earth = 11.186 * 10^3 m/s
Justin is riding his bike up Dunmore Hill. The effort force from his feet on his pedals is 1160N.
The resulting load force to move the bike forward is 675N. What is the mechanical advantage of
the bike?
Explanation:
the answer is 1835N that how best I can help
1. A bus starting from rest moves
with a uniform acceleration of
0.1 m s2 for 2 minutes. Find (a)
the speed acquired. (b) the
distance travelled.
Answer:
Speed acquired (Final velocity) = 12 m/s
Distance travelled = 720 meter
Explanation:
Given:
Uniform acceleration = 0.1 m/s²
Initial velocity = 0 m/s
Time taken = 2 minutes = 2 x 60 = 120 second
Find:
(a) Speed acquired (Final velocity)
(b) Distance travelled
Computation:
v = u + at
v = 0 + 0.1(120)
Speed acquired (Final velocity) = 12 m/s
Distance travelled = ut + (1/2)(a)(t²)
Distance travelled = (0)(120) + (1/2)(0.1)(120²)
Distance travelled = (1/2)(0.1)(14400)
Distance travelled = 720 meter
7. Tumutukoy ito sa lakas ng pagbigkas sa isang pantig ng salitang binibigkas.
Answer:
ang sagot po dyan ay diin
With the temperature held constant, the piston of a cylinder containing a gas is pulled out so that the volume increases from 0.3 m3 to 0.9 m3. If the final pressure of the gas is 80 kPa, what was the initial pressure
Answer:
P1 = 240 kPa.
Explanation:
Given the following data;
Initial volume = 0.3 m³
Final volume, V2 = 0.9 m³
Final pressure, P2 = 80 kPa
To find the initial pressure, we would use Boyle's law;
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by the formula;
[tex] PV = K[/tex]
[tex] P_{1}V_{1} = P_{2}V_{2} [/tex]
Substituting into the formula, we have;
[tex] P_{1} * 0.3 = 80 * 0.9 [/tex]
[tex] 0.3P_{1} = 72 [/tex]
[tex] P_{1} = \frac {72}{0.3} [/tex]
[tex] P_{1} = 240 [/tex]
Therefore, the initial pressure of the gas is 240 kPa
A 1.2-kg mass is projected from ground level with a velocity of 31.3 m/s at some unknown angle above the horizontal. A short time after being projected, the mass barely clears a 16-m tall fence. Disregard air resistance and assume the ground is level. What is the kinetic energy of the mass as it clears the fence
Answer:
The kinetic energy will be "399.65 J".
Explanation:
Given:
Mass,
m = 1.2 kg
Velocity,
v = 31.3 m/s
The total energy of mass will be:
⇒ [tex]E=K+U[/tex]
or,
⇒ [tex]E=.5mv^2+mgh[/tex]
By putting the values, we get
[tex]=.5(1.2)(31.3)^2+0[/tex]
[tex]=0.6\times 979.69+0[/tex]
[tex]=587.81 \ J[/tex]
Since,
The system's total energy is unchanged, then
⇒ [tex]E=K+U[/tex]
or,
⇒ [tex]E=K+mgh[/tex]
[tex]587.81=K+1.2(9.8)(16)[/tex]
[tex]587.81=K+188.16[/tex]
[tex]K=587.81-188.16[/tex]
[tex]=399.65 \ J[/tex]
a car drives 24 meters to the left in three seconds what is the velocity of the car?
Answer:
8 m/s to the left.
Explanation:
Applying,
V = d/t...................... Equation 1
Where V = Velocity of the car, d = distance, t = time
From the question,
Given: d = 24 meters, t = 3 seconds
Substitute these values into equation 1
V = 24/3
V = 8 m/s to the left.
Hence the velocity of the car is 8 m/s to the left.
Please please help!!!
The half-life of uranium-238 is about 4.5 billion years. How does this isotope's long half-life make it useful for finding the
age of rocks on Earth and in space?
Answer:
Because the half-life is as long as Earth has existed, virtually all rocks on Earth still have most of their uranium-238. So, measurements can easily be obtained.
Explanation:
exact answer from pluto/ edmentum
The significant part of the chemical composition of rocks is uranium. The half -life of uranium is close to the age of earth. Hence, by measuring the uranium content in rocks we can predict its age.
What is uranium?Uranium is 92nd element in periodic table. It is an actinide element. All the actinides are radioactive in which the most common is uranium. They all undergo radioactive decay process. Radioactive decay is a first order reaction. The rate constant K can be written as:
k = 1/t ln (W0/Wt)
where t is the tome of decay, W0 be the initial amount and Wt is the amount after time t.
here k = 0.693/t1/2
By this relation we can find t the age of a substance, if know the amounts and half life t1/2. Uranium is present in rocks and minerals. Uranium undergo nuclear decay and will forms its isotopes or new products.
Thus, by measuring the current amount of uranium in rocks by knowing the half life of uranium we can predict the age of rocks.
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Which exercise type will improve a person's range of motion?
Answer:
stretching prior to an exercise is a great way to improve a range of motion, along with yoga and Pilates!
If you scored 2 times in tennis and your opponent didn't score, how would the score read on the scoreboard?
1) 2-0
2) 15-Love
3) 0-0
4) 40-0
Answer:
None of them. It would be 30-Love.
Explanation:
0 points is Love
1 point is 15
2 points is 30
3 points is 40
If you see lightning strike and then hear the thunder 4.9 sec later, how far away did the lightning strike?
A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them, and they appear concentric when viewed from either end). The lenses are identical, each with a positive (converging) focal length of 14.8 cm. They are separated by a distance of 39.4 cm. Lens 1 is to the left of Lens 2.
Required:
a. What is the final image's distance (in cm) from Lens 2?
b. Where is the final image located?
c. What is the overall magnification of the lens pair, considered as a single optical instrument?
Answer:
A) q₂ = 75.98 cm, B) q₂' = 115.38 cm, C)
Explanation:
A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.
Lens 1. More to the left
let's use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively,
We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.
[tex]\frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}[/tex]
[tex]\frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}[/tex]
1 / q₁ = 0.04756
q₁ = 21.0227 cm
this image is the object for the second lens that has f₂ = 14.8 cm
the distance must be measured from the second lens
p₂ = 39.4 -q₁
p₂ = 39.4 -21.0227
p₂ = 18.38 cm
let's use the constructor equation
1 / q₂ = 1 / f - 1 / p2
[tex]\frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}[/tex]
[tex]\frac{1}{q_2}[/tex] = 0.01316
q₂ = 75.98 cm
measured from the second lens
B) the position of the final image with respect to the first lens is
q₂’= q₂ + 39.4
q₂'= 75.98 +39.4
q₂' = 115.38 cm
C) the magnification of a lens is
m = - q / p
in this case the image measured from lens 2 is q2 = 75.98 cm
the distance to the object from the first lens is p1 = 50cm
m = - 75.98 / 50
m = -1.5 X
the negative sign indicates that the image is inverted
What is the period of a wave with a speed of 10 m/s and frequency of 23 Hz?
The frequency calculator will let you find a wave's frequency given the wavelength and its velocity or period in no ... velocity equals 320 m/s and its wavelength is 8 m. Find its frequency. f = 320 m/s / 8 m. f = 40/s. f = 40 Hz.
A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the tires rolled without slipping, what was the magnitude of the average angular acceleration of the tires during the time the car slowed to a rest
Answer: deceleration of [tex]1.904\ rad/s^2[/tex]
Explanation:
Given
Car is traveling at a speed of u=20 m/s
The diameter of the car is d=70 cm
It slows down to rest in 300 m
If the car rolls without slipping, then it must be experiencing pure rolling i.e. [tex]a=\alpha \cdot r[/tex]
Using the equation of motion
[tex]v^2-u^2=2as\\[/tex]
Insert [tex]v=0,u=20,s=300[/tex]
[tex]0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2[/tex]
Write acceleration as [tex]a=\alpha \cdot r[/tex]
[tex]-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2[/tex]
So, the car must be experiencing the deceleration of [tex]1.904\ rad/s^2[/tex].
The angular accelaration of a car traveling at 20.0 m/s on tires with a diameter of 70.0 cm, when car slows down to rest after traveling 300.0 m is 1.91 m/s
Angular acceleration: This can be defined as the rate of change of angular velocity. The s.i unit of angular acceleration is rad/s²
In other to solve the problem above, we will use the formula for calculating angular acceleration
a = αr.............. Equation 1
Where a = acceleration of the car's tires, α = angular acceleration of the car's tires, r = radius of the car's tires
Therefore,
α = a/r................. Equation 2
But, we need to calculate a using the equations of motion
v² = u²+2as................ Equation 3
Where v = final velocity = 0 m/s (rest), u = initial velocity= 20m/s, s = distance = 300 m
Substitute these values into equation 3
0² = 20²+2(300²)a
-400 = 600a
a = -400/600
a = -0.67 m/s²
going back to equation 2,
Given: r = 70/2 = 35 cm = 0.35 m
Substitute this values into equation 2,
α = -0.67/0.35
α = -1.91 rad/s²
Hence, the magnitude of the angular acceleration is 1.91 rad/s²
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What are the biotic factors in this image?
Someone help thank you!!
Choose True or False. If False, choose the statement that makes it True.
James Clerk Maxwell laid the foundation for the discovery of electromagnetism.
one deterrent to burglary is to leave your front porch light on all the time if your fixture contains the 23 watt compact fluorescent bulb at 120 volts and your local power utility cell's energy at ten cents per kilowatt-hour how much will it cost to leave the bulb on for the entire month
Answer:
Cost = 165.6 cents = $1.656
Explanation:
First, we will calculate the total energy consumed by the bulb in a month:
[tex]E=Pt\\[/tex]
where,
E = Energy Consumed = ?
P = power = 23 W
t = time = (1 month)(30 days/1 month)(24 h/1 day) = 720 h
Therefore,
[tex]E = (23\ W)(720\ h)\\E = 16560\ Wh = 16.56\ Kilowatt-hour[/tex]
Now, the cost will be:
[tex]Cost = P(Unit\ Cost)\\Cost = (16.56\ kilowatt-hour)(10\ cents/kilowatt-hour)[/tex]
Cost = 165.6 cents = $1.656
A wave of frequency 13000Hz travel 1300m in 4.0 seconds, what is the wavelength of the wave
Explanation:
v = 1300/ 4
v = 325m/s
wavelength = 325/13000
=0.025 m
A spinning disc with a mass of 2.5kg and a radius of 0.80m is rotating with an angular velocity of 1.5 rad/s. A ball of clay with unknown mass is dropped onto the disk and sticks to the very edge causing the angular velocity of the disk to slow to 1.13 rad/s. What is the mass of the ball of clay
Answer:
M = 1.90 Kg
Explanation:
Given data: mass = 2.5 Kg
radius R = 0.8 m
angular velocity ω = 1.5 rad/s
Angular momentum L =0.5×Iω^2
Where, I is the moment of inertia of the spinning disc.
I = 0.5MR^2
I = 0.5×2.5×0.8^2
I = 0.8 Kg/m^2
Then L = 0.5×0.8×1.5^2 = 0.8×2.25 = 0.9 Kg-m^2/sec
Let unknown mass be M
New mass of disc = (2.5+M) Kg, R = 0.8 m
New I = 0.5(2.5+M)(0.8)^2
Since, angular momentum is conserved
Angular momentum before = angular momentum after
0.5×0.5(2.5+M)(0.8)^2×(1.13)^2 = 0.9
Solving for M we get
0.204304(2.5+M)=0.9
M = 1.90 Kg
A sheet of glass having an index of refraction of 1.52 is to be coated with a film of material having a refractiveindex of 1.75 such that green light with a wavelength of 525 nm (measured in vacuum) is preferentially transmitted when white light is incident normally on the film.What is the minimum thickness of the film that will achieve this result
Answer:
The correct answer is "164.06 nm".
Explanation:
Refractive index,
n = 1.75
Wavelength,
λ = 525 nm
Now,
The maximum thickness of the film will be:
⇒ [tex]t=\frac{\lambda}{2n}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{525}{2\times 1.75}[/tex]
⇒ [tex]=\frac{525}{3.5}[/tex]
⇒ [tex]=164.06 \ nm[/tex]
A certain CD has a playing time of 74.0 minutes. When the music starts, the CD is rotating at an angular speed of 480 revolutions per minute (rpm). At the end of the music, the CD is rotating at 210 rpm. Find the magnitude of the average angular acceleration of the CD. Express your answer in rad/s2.
Answer: [tex]0.00636\ rad/s^2[/tex]
Explanation:
Given
CD has a playing time of [tex]t=74\ min\ or\ 74\times 60\ s[/tex]
Initial angular speed of CD is [tex]480\ rpm[/tex]
Final angular speed of DC is [tex]210\ rpm[/tex]
Angular speed, when rpm is given
[tex]\omega =\dfrac{2\pi N}{60}[/tex]
[tex]\omega_i=\dfrac{2\pi \times 480}{60}\\\\\Rightarrow \omega_i=16\pi \ rad/s[/tex]
Final speed
[tex]\Rightarrow \omega_f=\dfrac{2\pi \times 210}{60}\\\\\Rightarrow \omega_f=7\pi \ rad/s[/tex]
Using equation of angular motion
[tex]\Rightarrow \omega_f=\omega_i+\alpha t[/tex]
Insert the values
[tex]\Rightarrow 7\pi =16\pi +\alpha \times 74\times 60\\\Rightarrow -9\pi =\alpha \cdot (4440)\\\\\Rightarrow \alpha=-\dfrac{9\pi}{4440}\\\\\Rightarrow \alpha=-0.00636\ rad/s^2[/tex]
Magnitude of angular acceleration is [tex]0.00636\ rad/s^2[/tex]
A 3kg cat falls out of a window that is 10m above the ground and lands
safely in the grass. Calculate the velocity at which the cat hit the ground.
Answer:
p = 30 kg·m/s
Explanation:
Consider a loop of wire with a resistor. In the same plane, (with switch S closed) a long wire has a current I flowing from left to right. R I S If switch S is closed for a long period of time, and then suddenly opened, what is the direction of the induced current through R as the current I rapidly decreases
Solution :
The given figure is a loop of a wire with a resistor.
When the switch S is closed for long time and is suddenly opened, the direction of the induced current can be find out by using the rule of right hand screw. According to the right hand screw rule, the direction of the magnetic field at the loop is in the direction that points outwards. The strength of the current rapidly decreases as it is switch off and the magnetic flux that is linked with the loop wire will also decrease.
According to the Lenz's law, the direction of the induced current must be such [tex]$\text{that it opposes}$[/tex] the decrease in the magnetic flux. It means the direction of the magnetic field must be outwards and also normal to the plane of the screen. The direction of the induced anti clockwise or from right to left in the resistance.
The load across a 12 V battery consists of a series combination of three resistors 34 Ω, 42Ω, and 30Ω. What is the total resistance of the load? Answer in units of Ω. What is the current in the circuit? Answer in units of A.
Answer:
The total resistance is 106 Ω and the current in the circuit is 0.11 A.
Explanation:
Given that,
Voltage of the battery, V = 12 V
Resistors 34 Ω, 42Ω, and 30Ω are connected in series.
The total resistance is given by :}
R = 34 + 42 +30
= 106Ω
Let I is the total current in the circuit. Using ohm's law to find it such that,
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{12}{106}\\\\I=0.11\ A[/tex]
Hence, the total resistance is 106 Ω and the current in the circuit is 0.11 A.
Which type of energy is demonstrated by a person jogging?
A. electrical energy
B. nuclear energy
C. electromagnetic energy
D. kinetic energy
A car of mass 2 320 kg is descending a sloping road, inclined at 10.0° to the horizontal. The driver sees a hazard and brakes to make an emergency stop. The distance travelled while braking is 22.5 m.
The initial speed of the car was 13.4 m s−1. Calculate the average braking force. Assume air resistance is negligible.
f = 1.32×10^4 N
Explanation:
We can use the work-energy theorem to find the work done by the braking force f:
W = ∆KE + ∆PE
= (KEf - KEi) + (PEf - PEi)
= [(1/2)mvf^2 - (1/2)mvi^2] + (mhf - mghi)
At the bottom of the slope, vf = 0 and hf = 0 and hi = dsin10° (d = braking distance) so work W becomes
W = -[(1/2)mvi^2 + mgdsin10°]
= -m[(1/2)vi^2 + gdsin10°]
= -(2320kg)[(1/2)(13.4m/s)^2 + (9.8 m/s^2)(22.5m)sin10]
= -2.97×10^5 J
Since W = fd, where f is the braking force, we can now solve for f:
f = W/d = (-2.97×10^5 J)/(22.5 m)
= -1.32×10^4 N
Note: the negative sign means that it is a dissipative force.
Average braking force is 1.32×10^4 N.
What is force?A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
We can use the work-energy theorem to find the work done by the braking force f:
W = ∆KE + ∆PE
= (KEf - KEi) + (PEf - PEi)
= [(1/2)mvf^2 - (1/2)mvi^2] + (mhf - mghi)
At the bottom of the slope, vf = 0 and hf = 0 and hi = dsin10° (d = braking distance) so work W becomes
W = -[(1/2)mvi^2 + mgdsin10°]
= -m[(1/2)vi^2 + gdsin10°]
= -(2320kg)[(1/2)(13.4m/s)^2 + (9.8 m/s^2)(22.5m)sin10]
= -2.97×10^5 J
Since W = fd, where f is the braking force, we can now solve for f:
f = W/d = (-2.97×10^5 J)/(22.5 m)
= -1.32×10^4 N, the negative sign means that it is a dissipative force.
Average braking force is 1.32×10^4 N.
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What are the functionssss of non-geostationary satellite?
Please list oust at least 5 function. Thanks a lot!
Answer:
A communication system comprising a set of non-geostationary satellites provides communication links between ground stations and terminals, a communication link from a particular station via a satellite terminating at associated terminals in a limited geographical area including the station and providing a.Explanation:
Hopes this helps. Mark as brainlest plz!g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in series do the current is maximized
Answer: [tex]71.16\ Hz[/tex]
Explanation:
Given
Capacitance [tex]C=100\ \mu F[/tex]
Resistance [tex]R=500\ \Omega[/tex]
Inductance [tex]L=50\ mH[/tex]
In LCR circuit, current is maximum at resonance frequency i.e.
[tex]X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}[/tex]
Insert the values
[tex]\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}[/tex]
Also, frequency is given by
[tex]\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}[/tex]
[tex]\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz[/tex]
