Answer:
C. The two cart system has a kinetic energy of 2.4 J.
Explanation:
Hi there!
The momentum of the two cart system is conserved. That means that the momentum of the system before the collision is equal to the momentum of the system after the collision. The momentum of the system is calculated by adding the momenta of the two carts:
initial momentum of the system = final momentum of the system
pA + pB = p (A + B)
mA · vA + mB · vB = (mA + mB) · v
Where:
pA and pB = initial momentum of carts A and B respectively.
p (A +B) = momentum of the two cart system after the collision.
mA and mB = mass of carts A and B respectively.
vA and vB = initial velocity of carts A and B.
v = velocity of the two cart system.
We have the following data:
mA = 0.4 kg
mB = 0.8 kg
vA = 6 m/s
vB = 0 m/s
Solving the equation for v:
mA · vA + mB · vB = (mA + mB) · v
0.4 kg · 6 m/s + 0.8 kg · 0 m/s = (0.4 kg + 0.8 kg) · v
2.4 kg m/s = 1.2 kg · v
v = 2.4 kg m/s / 1.2 kg
v = 2 m/s
The equation of kinetic energy (KE) is the following:
KE = 1/2 · m · v²
Where m is the mass of the object and v its speed.
Replacing with the data we have obtained:
KE = 1/2 · 1.2 kg · (2 m/s)²
KE = 2.4 J
The two cart system has a kinetic energy of 2.4 J.
After the collision, the two carts stick together and move with a common velocity. The kinetic energy of the two carts after the collision is 2.4 J.
Explanation:First, we need to calculate the initial momentum of cart A and the final momentum of the two carts combined.
The initial momentum of cart A is given by the formula: momentum = mass * velocity.
So, momentum of cart A = 0.4 kg * 6 m/s = 2.4 kg*m/s.
After the collision, the two carts stick together and move with a common velocity.
Using the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Hence, 2.4 kg*m/s = (0.4 kg + 0.8 kg) * final velocity.
Solving for the final velocity, we get: final velocity = 2.4 kg*m/s / 1.2 kg = 2 m/s.
Finally, we can calculate the kinetic energy of the two carts after the collision using the formula: kinetic energy = (1/2) * mass * velocity^2.
Kinetic energy = (1/2) * (0.4 kg + 0.8 kg) * (2 m/s)^2 = 2.4 J.
Therefore, the correct answer is option C. 2.4 J.
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If the net work done on an object is positive, what can you conclude about the object's motion?
- The object is moving with a constant velocity.
- The object is speeding up.
- The object is slowing down.
- The object is at rest; its position is constant.
Since there is a positive work on an object, this directly indicates that there is an increase in Energy on the body.
Work can be expressed in terms of Force and distance as
W = Fd
Where,
F = Force
d = Distance
The force in turn is a description of how mass accelerates according to Newton's second law, that is
F = ma
The mass (m) remains constant, but the acceleration (a) is constant, which implies directly that there is a gradual change in the velocity of the object.
Therefore the correct answer is: The object is speeding up.
The work is positive so the energy of the object is increasing so the object is speeding up
What can you conclude about objects' motion?As we know that the work is the product of the force and the displacement
[tex]W=F\times D[/tex]
Where,
F = Force
D= Distance
And from newtons second law we can see that
[tex]F=m\times a[/tex]
Since here mass will be constant to there will be a change in the velocity that is acceleration in the body so the energy of the body will change
Thus work is positive so the energy of the object is increasing so the object is speeding up
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Selena uses a garden hose to fill a bucket of water. The water enters the hose through a faucet with a 6.0 cm diameter.
The speed of the water at the faucet is 5/ms.
If the faucet and nozzlw are at the same height, and the water leaves the nozzle with a speed of 20 m/s, what is the diameter of the nozzle?
Answer:
3 cm
Explanation:
To answer this question the equation of continuity can be used
For us to use the equaition of continuity, we will make a few assumptions:
That the temperature of the water does not change, therefore there is not expansion or contraction (change in volume)That the flow is non-viscousThat there is a single entry and a single exitThat the water is incompressible[tex]A_{1}v_{1} = A_{2} v_{2} \\\pi (3cm)^{2} * 5m/s = \pi r^{2} *20m/s\\\\r = 1.5 cm\\D = 2r = 2*1.5cm=3 cm[/tex]
A solid brass (bulk modulus 6.7 x 1010 N/m2) sphere is subjected to a pressure of 1.0 x 105 Pa due to the earth's atmosphere. On Venus the pressure due to the atmosphere is 9.0 x 106 Pa. By what fraction r/r0 (including the algebraic sign) does the radius of the sphere change when it is exposed to the Venusian atmosphere? Assume that the change in radius is very small relative to the initial radius.
Answer:
[tex]\frac{r}{r_0}=-4.4x10^{-5}[/tex]
Explanation:
Using the equation to find the fraction does the radius of the sphere change when it is exposed to Venusian atmosphere.
[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{V}{V_0}[/tex]
Replacing to find the relation between the radius
[tex]\frac{V}{V_0}=-\frac{P}{B}[/tex]
Replacing numeric to find the relation
[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{8.9x10^6pa}{6.7x10^{10}pa}[/tex]
[tex]\frac{r}{r_0}=\frac{1}{3}*-1.33x10^{-4}[/tex]
So the relation is
[tex]\frac{r}{r_0}=-4.4x10^{-5}[/tex]
Identify the procedure to determine a formula for self-inductance, or inductance for short. Using the formula derived in the text, find the inductance in henries for a solenoid with 900 loops of wire wound on a rod 9 cm long with radius 3 cm.
Answer:
L = 0.0319 H
Explanation:
Given that,
Number of loops in the solenoid, N = 900
Radius of the wire, r = 3 cm = 0.03 m
Length of the rod, l = 9 cm = 0.09 m
To find,
Self inductance in the solenoid
Solution,
The expression for the self inductance of the solenoid is given by :
[tex]L=\dfrac{\mu_o N^2 A}{l}[/tex]
[tex]L=\dfrac{4\pi\times10^{-7}\times(900)^{2}\times\pi(0.03)^{2}}{0.09}[/tex]
L = 0.0319 H
So, the self inductance of the solenoid is 0.0319 henries.
Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% 232Th, with a half-life of 1.405 ✕ 1010 y. If an average lantern mantle contains 200 mg of thorium, what is its activity (in Bq)?
Answer:
The activity is 811.77 Bq
Solution:
As per the question:
Half life of Thorium, [tex]t_{\frac{1}{2}} = 1.405\times 10^{10}\ yrs[/tex]
Mass of Thorium, m = 200 mg = 0.2 g
M = 232 g/mol
Now,
No. of nuclei of Thorium in 200 mg of Thorium:
[tex]N = \frac{N_{o}m}{M}[/tex]
where
[tex]N_{o }[/tex] = Avagadro's number
Thus
[tex]N = \frac{6.02\times 10^{23}\times 0.2}{232} = 5.19\times 10^{20}[/tex]
Also,
Activity is given by:
[tex]\frac{0.693}{t_{\frac{1}{2}}}\times N[/tex]
[tex]A= \frac{0.693}{1.405\times 10^{10}}\times 5.19\times 10^{20} = 2.56\times 10^{10}\ \yr[/tex]
[tex]A = \frac{2.56\times 10^{10}}{365\times 24\times 60\times 60} = 811.77\ Bq[/tex]
The earth has a radius of 6.38 × 106 m and turns on its axis once every 23.9 h.
(a) What is the tangential speed (in m/s) of a person living in Ecuador, a country that lies on the equator?
(b) At what latitude (i.e., the angle theta in the drawing) is the tangential speed one-third that of a person living in Ecuador?
Answer:
a) V = 465.9 m/s
b) θ = 70.529°
Explanation:
Let's first calculate angular velocity of earth:
[tex]\omega=\frac{2\pi}{23.9h}*1h/3600s[/tex]
Velocity of a person on Ecuador will be:
[tex]V_E = \omega*R[/tex]
[tex]V_E = 465.9 m/s[/tex]
For part b, since angular velocity is the same:
[tex]\frac{\omega*R}{3}=\omega*(R*cos\theta )[/tex]
Solving for θ:
[tex]\theta=acos(1/3)[/tex]
[tex]\theta=70.529\°[/tex]
(a) speed at the equator: 465.9 m/s
(b) Latitude for 1/3 speed: 50.6°
(a) The tangential speed of a person living in Ecuador is equal to the circumference of the Earth at the equator divided by the period of rotation. The circumference of the Earth at the equator is 2πr, where r is the radius of the Earth. The period of rotation is 23.9 hours, which is equal to 23.9 × 3600 seconds. Therefore, the tangential speed of a person living in Ecuador is:
v = 2πr / T = 2π(6.38 ×[tex]10^6[/tex] m) / (23.9 × 3600 s) ≈ 465.9 m/s
(b) The tangential speed of a person living at a latitude of θ is equal to vr * cos(θ), where vr is the tangential speed of a person living in Ecuador and r is the radius of the Earth. Therefore, we can solve for θ:
cos(θ) = vr / v = 465.9 m/s / v
Solving for θ, we get:
θ = arccos(vr / v) ≈ 50.6°
Therefore, the latitude at which the tangential speed is one-third that of a person living in Ecuador is approximately 50.6°.
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A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the poorly secured load falls off! Immediately after losing the load, the truck begins to accelerate at 1.5 m/s2. Part A What was the mass of the load? Ignore rolling friction.
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
The mass of the load is 2636.8 kg
given information:Mass of the truck m = 7100 kg
Angle θ = 15°
velocity v = 15m/s
Acceleration a = 1.5 m/s²
Now let us assume that the mass of the truck is m₁ and the mass of the load is m₂. The thrust from the engine be T.
resolving forces:Initially, the thrust (T) balances the total weight:
[tex]T = (m_1+m_2) g sin\theta[/tex]
When the load falls off, the thrust (T) remains the same
so the net force on the truck
[tex]F = T -m_1gsin\theta[/tex]
[tex]T-m_1gsin\theta = m_1a \\\\ T = m_1(a + gsin\theta)[/tex]
comparing both the equation for T we get:
[tex](m_1+m_2)gsin\theta = m_1(a + gsin\theta)\\\\m_1gsin\theta + m_2gsin\theta =m_1a + m_1gsin\theta\\\\m_2gsin\theta = m_1a[/tex]
Now, m₁+m₂ = 7100kg
m₁= 7100 – m₂
[tex]m_2gsin\theta = (7100-m_2)a\\\\m_2gsin\theta = 7100a-m_2a\\\\m_2gsin\theta + m_2a = 7100a\\\\m_2 (gsin\theta + a) = 7100a\\\\m_2 = 7100a/(gsin\theta + a)\\\\m_2 = (7100 \times1.5) / (9.8sin(15) + 1.5)\\\\m_2 = 2636.8 kg[/tex]
The load has a mass of 2636.8 kg
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⦁ A speed skater increases her speed form 10 m/s to 12.5 m/s over a period of 3 seconds while coming out of a curve of 20 m radius. What are the magnitudes of her radial, tangential and total accelerations as she leaves the curve?
Answer:
7.85 m/s^2
Explanation:
linear or tangential acceleration= dv/dt
⇒[tex]a_t= \frac{12.5-10}{3}[/tex]
=0.83 m/s^2
radial acceleration is given by = [tex]\frac{v^2}{r}[/tex]
⇒[tex]a_r =\frac{12.5^2}{20}[/tex]
= 7.81 m/s^2
total acceleration
[tex]a_T= \sqrt{a_t^2+a_r^2}[/tex]
putting values we get
[tex]a_T= \sqrt{0.83^2+7.81^2}[/tex]
= 7.85 m/s^2
Answer:
Explanation:
Tangential acceleration = ( 12.5 - 10 )/ 3
a_t= .833 m /s²
radial acceleration
= v² / R
12.5² / 20 ( 12.5 m/s is the velocity when it leaves the curve )
a_r= 7.81 ms⁻²
Total acceleration √( .833² + 7.81²)
= √( 61.6939)
= 7.85 m/s⁻
Ethanol has a heat of vaporization of 38.56 kJ/mol and a vapor pressure of 760 torr at 78.4 oC. What is the vapor pressure of ethanol at 38.8 oC?
To develop this problem it is necessary to apply the concepts developed by Clausius - Claperyron.
This duet found the relationship between temperature and pressure expressed as,
[tex]ln P = constant - \frac{\Delta H}{RT}[/tex]
For the two states that we have then we could define the pressure and temperature in each of them as
[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
Where,
[tex]P_{1,2}[/tex]= Pressure at state 1 and 2
[tex]T_{1,2}[/tex]= Temperature at state 1 and 2
[tex]\Delta H[/tex]= Enthalpy of Vaporization of a substance
R = Gas constant (8.134J/mol.K)
Our values are given by,
[tex]P_1 = 1atm \\\Delta H = 38.56*10^{-3} J/mol \\R = 8.134J/mol.K\\T_1 = 78.4\°C = 351.55K\\T_2 =38.8\°C = 311.95K[/tex]
Therefore replacing we have that,
[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
[tex]ln(\frac{P_2}{1atm}) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]
[tex]ln(P_2) - Ln(1atm) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]
[tex]P_2 = e^{\frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})}[/tex]
[tex]P_2 = 0.187355atm[/tex]
Therefore the pressure of the Ethanol at 38.8°C is 0.187355atm
Two generators use the same magnetic field and operate at the same frequency. Each has a single-turn circular coil. One generator has a coil radius of 5.1 cm and a peak emf of 1.8 V. The other generator has a peak emf of 3.9 V. Find the coil radius r of this other generator.
Answer:
The radius of the second coil will be 7.5 cm
Explanation:
We have given that radius of the first coil [tex]r_1=5.1cm=0.051m[/tex]
Peak emf of the first generator [tex]e_1=1.8Volt[/tex]
Peak emf of the second generator [tex]e_2=3.9Volt[/tex]
We know that induced emf is rate of change of magnetic flux , that is equal to magnetic field multiply by change in area
So emf will be proportional to the change in area
So [tex]\frac{e_1}{e_2}=\frac{A_1}{A_2}[/tex]
[tex]\frac{1.8}{3.9}=\frac{0.051^2}{r_2^2}[/tex]
[tex]r_2=0.075m = 7.5cm[/tex]
So the radius of the second coil will be 7.5 cm
A car with tires of radius 25.0 cm accelerates from rest to 30.0 m/s in 10.0 s. When the car's speed is 2.00 m/s find the linear acceleration of the top of the wheel relative to (a) the center of the wheel, (b) the road. (Note that there will be a vertical and horizontal component to the acceleration.
Answer
given,
radius of tire = 25 cm = 0.25 m
initial velocity = 0 m/s
final velocity = 30 m/s
time = 10 s
a) acceleration with respect to center
v = u + a t
30 = 0 + 10 a
a = 3 m/s²
radial acceleration
[tex]a_r = \dfrac{v^2}{r}[/tex]
[tex]a_r = \dfrac{2^2}{0.25}[/tex]
[tex]a_r =16\ m/s^2[/tex]
[tex]a=\sqrt{a_r^2 + a_t^2}[/tex]
[tex]a=\sqrt{16^2 + 3^2}[/tex]
[tex]a=16.28\ m/s[/tex]
acceleration w.r.t center of tire = [tex]a=16.28\ m/s[/tex]
b) acceleration with respect to road will be equal to 3 m/s²
A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction is mu_k. During time interval t, the sled moves a distance s, starting from rest.Find the average velocity vavg of the sled during that time interval.Express your answer in terms of the given quantities and, if necessary, appropriate constants. You may or may not use all of the given quantities.
Answer:
The average velocity of the sled is vavg = s/t.
Explanation:
Hi there!
The average velocity is calculated as the traveled distance over time:
vavg = Δx/Δt
Where:
vavg = average velocity.
Δx = traveled distance.
Δt = elapsed time.
We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:
vavg = s/t
Have a nice day!
The average velocity of the sled, given by the formula v_avg = s / t, evaluates the distance covered over a certain duration of time. The sled has moved a distance 's' during a time interval 't', and so its average velocity over this interval is simply 's' divided by 't'. The presence of friction does not affect the calculation of average velocity.
Explanation:The average velocity of the sled is simply the distance traveled over a certain time interval. In this case, the sled starts from rest and is being pulled by a horizontal force F against a frictional force. Average Velocity, v_avg, is given by the formula:
v_avg = s / t
In physics, velocity is a measure of the rate of change of position concerning time, so the average velocity is simply the total distance (or displacement) divided by the total time. Given the fact that the sled has moved a distance, s, during the time interval t, the average velocity is simply s divided by t.
With this definition, the sled's average velocity can be determined without needing to tabulate its speed or direction throughout the time interval even in the presence of friction.
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An object of mass m = 8.0 kg is attached to an ideal spring and allowed to hang in the earth's gravitational field. The spring stretches 2.2 cm before it reaches its equilibrium position.
If it were now allowed to oscillate by this spring, what would be its frequency?
Answer:
Frequency, f = 3.35 Hz
Explanation:
It is given that,
Mass of the object, m = 8 kg
Stretching in the spring, x = 2.2 cm
When the spring is hanged up in the Earth's gravitational field, its weight is balanced by the force in the spring. So,
[tex]mg=kx[/tex]
k is the spring constant
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{8\times 9.8}{2.2\times 10^{-2}}[/tex]
k = 3563.63 N/m
Let f is the frequency of oscillation. Its expression is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{3563.63}{8}}[/tex]
f = 3.35 Hz
So, the frequency of oscillation by the spring is 3.35 Hz. Hence, this is the required solution.
An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal resistance.
(a) What is the current to the motor?
(b) What voltage is applied to it?
(c) What power is supplied to the motor?
(d) Repeat these calculations for when the battery connections are corroded and add 0.0900Ω to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)
Answer
given,
resistance = 0.05 Ω
internal resistance of battery = 0.01 Ω
electromotive force = 12 V
a) ohm's law
V = IR
and volage
[tex]V = \epsilon - Ir[/tex]
now,
[tex]IR = \epsilon - Ir[/tex]
[tex]I(R+r) = \epsilon[/tex]
[tex]I= \dfrac{\epsilon}{R+r}[/tex]
inserting the values
[tex]I= \dfrac{12}{0.05+0.01}[/tex]
I = 200 A
b) Voltage
V = I R
V = 200 x 0.05
V = 10 V
c) Power
P = I V
P = 200 x 10 = 2000 W
d) total resistance = 0.05 + 0.09 = 0.14 Ω
[tex]I= \dfrac{\epsilon}{R+r}[/tex]
[tex]I= \dfrac{12}{0.14+0.01}[/tex]
I = 80 A
V = 80 x 0.05 = 4 V
P = 4 x 80 = 320 W
Answer:
Explanation:
Resistance of motor, R = 0.05 ohm
internal resistance of battery, r = 0.01 ohm
Voltage of battery, V = 12 V
(a) Total resistance, R' = R + r = 0.05 + 0.01 = 0.06 ohm
Let the current be i.
use Ohm's law
i = V / R'
i = 12 / 0.06 = 200 A
(b) Voltage across motor, V' = i x R = 200 x 0.05 = 10 V
(c) Power, P = i²R = 200 x 200 x 0.05 = 2000 Watt.
(d) Total resistance, R' = 0.05 + 0.1 + 0.09 = 0.15 ohm
i = V / R' = 12 / 0.15 = 80 A
V' = i x R = 80 x 0.05 = 4 V
P' = i²R = 80 x 80 x 0.05 = 320 Watt
Vadim bought his wife a "gold" bracelet from a guy on the street for $100. If he drops it in a beaker of water and the volume raises 400 cm3 and the mass is determined to be 0.5 kg, then is the jewelry gold? If it is not gold then what do you think it is made of?
Answer: It is not gold. The density is 1.25 g/cm3, very little compared with gold.
Explanation:
According to Archimedes’ principle, any body submerged in a liquid, receives an upward force, called buoyant force, which magnitude is equal to the weight of the liquid that displaces.
This weight can be calcultated as follows:
Fb= δH2O . Vbody g
We are told that the volume of the liquid displaced (equal to the volume of the body as it is completely submerged in water) is 400 cm3.
If we know that the mass of the bracelet is 0.5 Kg, we can find out the density of the material, just applying the same relationship between mass, density and volume:
m = δbracelet. V
δbracelet = m / V = 500 g / 400 cm3= 1.25 g/cm3
As the gold density is approximately 19.3 g/cm3, it is clear that the bracelet is not made from gold, but from something less dense.
The density of Vadim's bracelet is calculated to be 1.25 g/cm³, which is significantly lower than the density of pure gold (19.3 g/cm³). Thus, the bracelet is not made of gold and could be made from a lighter metal or alloy.
To determine if Vadim's bracelet is gold, we must calculate its density and compare it to the density of pure gold. The density of a substance is the mass-to-volume ratio, and for gold, it is 19.3 g/cm³. Given that the bracelet has a mass of 0.5 kg (or 500 g) and a volume displacement of 400 cm³ when submerged in water, we can calculate its density using the formula:
Density = Mass / Volume
density = 500 g / 400 cm³ = 1.25 g/cm³
The calculated density of 1.25 g/cm³ is significantly lower than the density of pure gold, which suggests that the bracelet is not made of gold. Since the density is lower than that of gold, the bracelet could be made of a metal or alloy with a lower density, such as aluminium or a mixture that includes materials with lighter densities.
When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of the calorimeter increases by 10.0°C. The heat capacity of the calorimeter is 6.97 kJ/°C. Determine the energy flow, q (reaction).
Answer:
Q (reaction) = -69.7 kJ
Explanation:
Octane reacts with oxygen to give carbon dioxide and water.
C₈H₁₈ + 25 O₂ ---> 16 CO₂ +18 H₂O
This reaction is exothermic in nature. Therefore, the energy is released into the atmosphere. This reaction took place in a calorimeter, there the temperature (T) increases by 10 C. The heat capacity of the calorimeter is 6.97 kJ/C
The heat (q) of the reaction is calculated as follows:
Q= -cT, where c is the heat capacity of the calorimeter and T is the increase in temperature
q = -(6.97) x (10) = -69.7kJ
Since the heat capacity is given in kilo -joule per degree Celsius, therefore, the mass of octane is not required
A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object so that it strikes the stationary smaller object which then slides follows the curved track so that it rises vertically to a maximum height, h_{max}h max . The masses are equipped with elastic bumpers so that the impact between them is an elastic collision. If the larger object has a mass , M = 5.41 ~\text{kg}M=5.41 kg and the smaller object has a mass of m = 1.68~\text{kg}m=1.68 kg, then with what velocity, v_0v 0 should the player release the larger object so that the smaller object just reaches the target maximum height of h_{max} = 3.0 mh max =3.0m above the horizontal portion of the track?
The required velocity of the larger object for the smaller object to reach a maximum height of 3m in an elastic collision can be calculated using the equation v0 = sqrt((2*m*g*h_max)/M). This concept relies on the conservation of energy and momentum in an elastic collision.
Explanation:The central physics concept in this question is the conservation of energy and momentum in an elastic collision. In an elastic collision, both momentum and kinetic energy are conserved. According to the conservation of energy, the kinetic energy of the larger mass before the collision is converted into the potential energy of the smaller mass at its maximum height.
Expressing the conservation of energy mathematically, we have:
1/2*M*v0² = m*g*h_max
Solving for v0, we have:
v0 = sqrt((2*m*g*h_max)/M)
where v0 is the velocity with which the player should release the larger object, m is the mass of the smaller object (1.68 kg), g is the acceleration due to gravity (9.8 m/s²), h_max is the maximum height the smaller object will reach (3.0 m), and M is the mass of the larger object (5.41 kg).
Please do remember to plug in the correct values and units when solving the equation.
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The player must release the larger object with a velocity of 1.95 m/s so that the smaller object just reaches the target maximum height of 3.0 m.
Since the collision of the two masses is elastic, the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision.
We can use the principle of conservation of mechanical energy to solve this problem. The mechanical energy of a system is the sum of its potential energy and kinetic energy. At the beginning of the collision, the larger object has only potential energy due to its height, while the smaller object is at rest on the ground and has no energy. After the collision, the larger object has some kinetic energy and the smaller object has both potential energy and kinetic energy.
Let's denote the velocity of the larger object before the collision as v_0 and the velocity of the smaller object after the collision as v.
The potential energy of the larger object before the collision is:
PE_initial = Mgh
where:
M is the mass of the larger object (kg)
g is the acceleration due to gravity (m/s^2)
h is the height of the larger object above the ground (m)
The kinetic energy of the larger object before the collision is:
KE_initial = (1/2)Mv_0^2
The total mechanical energy of the system before the collision is:
E_initial = PE_initial + KE_initial = Mgh + (1/2)Mv_0^2
After the collision, the potential energy of the smaller object is:
PE_final = mgh_{max}
where:
m is the mass of the smaller object (kg)
g is the acceleration due to gravity (m/s^2)
h_{max} is the maximum height reached by the smaller object (m)
The kinetic energy of the smaller object after the collision is:
KE_final = (1/2)mv^2
The total mechanical energy of the system after the collision is:
E_final = PE_final + KE_final = mgh_{max} + (1/2)mv^2
Since the collision is elastic, we know that E_initial = E_final. Therefore, we can set the two equations equal to each other and solve for v_0.
Mgh + (1/2)Mv_0^2 = mgh_{max} + (1/2)mv^2
Solving for v_0, we get:
v_0 = sqrt((2mg(h - h_{max}))/(M + m))
Plugging in the values, we get:
v_0 = sqrt((2(5.41 kg)(9.81 m/s^2)(0.3 m))/(5.41 kg + 1.68 kg)) = 1.95 m/s
Therefore, the player must release the larger object with a velocity of 1.95 m/s so that the smaller object just reaches the target maximum height of 3.0 m.
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A wheel with a weight of 395 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at an angular velocity of 23.1 rad/s . The radius of the wheel is 0.652 m and its moment of inertia about its rotation axis is 0.800 MR2. Friction does work on the wheel as it rolls up the hill to a stop, at a height of h above the bottom of the hill; this work has a magnitude of 3470 J .
Answer:
h=12.04m
Explanation:
1) Notation and some important concepts
[tex]\omega_i[/tex] represent the initial angular velocity
[tex]v_i[/tex] represent the initial velocity
[tex]h[/tex] represent the final height reached by the mass
[tex]M[/tex] represent the mass of the object
[tex]W_f[/tex] represent the work due the friction force (variable of interest)
[tex]KE_{rot}[/tex] represent the rotational energy
[tex]KE_{tra}[/tex] represent the transitional kinetic energy
[tex]PE=mgh[/tex] represent the potential energy
[tex]I=0.8MR^2[/tex] represent the rotational inertia
W= 395 N is the weight of the object
For this problem we can use the principle of energy conservation, this principle states that "the total energy of an isolated system remains constant; it is said to be conserved over time".
At the begin the wheel had rotational energy defined as "The kinetic energy due to rotational motion. Is a scalar quantity and have units of energy usually Joules". And this energy is represented by the following formula: [tex]KE_{rot}=\frac{1}{2}I\omega^2_i[/tex]
At the starting point the wheel also had kinetic energy defined as "The energy of mass in motion" and is given by the formula : [tex]KE_{tran}=\frac{1}{2}mv_i^2_i[/tex]
At the end of the movement we have potential energy since the mass is at height h the potential energy is defined as "The energy stored within an object, due to the object's position, arrangement or state" and is given by the formula [tex]PE=mgh[/tex].
Since we have friction acting we have a work related to the force of friction and we need to subtract this from the formula of conservation of energy
2) Formulas to use
The figure attached is an schematic draw for the problem
Using the principle of energy conservation we have:
[tex]KE_{rot}+KE_{tran}-W_f =PE[/tex]
Replacing the formulas for each energy w ehave:
[tex]\frac{1}{2}I\omega^2_i+\frac{1}{2}mv_i^2_i-W_f =Mgh[/tex] (1)
We also know that [tex]v_i =\omega_i R[/tex] and [tex]I=0.8MR^2[/tex] so if we replace this into equation (1) we got:
[tex]0.8(\frac{1}{2})M(R\omega_i)^2 +\frac{1}{2}M(\omega_i R)^2-W_f=Mgh[/tex] (2)
We also know that the weight is defined as [tex]W=mg[/tex] so then [tex]M=\frac{W}{g}[/tex], so if we replace this into equation (2) we have:
[tex]0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f=Wh[/tex] (3)
So then if we solve for h we got:
[tex]h=\frac{0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f}{W}[/tex] (4)
3) Solution for the problem
Now we can replace the values given into equation (4):
[tex]h=\frac{0.8(\frac{1}{2})\frac{395N}{9.8\frac{m}{s^2}}(0.652m(23.1\frac{rad}{s}))^2 +\frac{1}{2}\frac{395}{9.8\frac{m}{s^2}}(23.1\frac{rad}{s}(0.652m))^2-3470J}{395N}=12.04m[/tex] (4)
So then our final answer would be h=12.04m
A machine part has the shape of a solid uniform sphere of mass 220 g and diameter 4.50 cm . It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point.
The sphere's angular acceleration is [tex]2.064 rad/s^2[/tex], and it will take approximately 10.17 seconds to decrease its rotational speed by 21.0 rad/s when experiencing a constant friction force.
Part A: Calculating Angular Acceleration
To find the angular acceleration, we must first calculate the torque. The torque ( au) caused by the friction force (F) is the product of the force and the radius (r) at which the force is applied, and torque is given by [tex]\tau = r \times F[/tex]. Using the diameter (d) to find the radius, we have r = d / 2 = 4.50 cm / 2 = 2.25 cm = 0.0225 m. The given friction force is 0.0200 N. Therefore, [tex]\tau = 0.0225 m \times 0.0200 N = 0.00045 Nm[/tex].
Next, we use the moment of inertia (I) for a sphere, [tex]I=\frac{2}{5} mr^2[/tex], to calculate the angular acceleration ([tex]\alpha[/tex]). The mass (m) is 220 g which is 0.220 kg. So, [tex]I = \frac{2}{5} \times 0.220 kg \times (0.0225 m)^2 = 0.000218025 kg m^2[/tex]. Angular acceleration, [tex]\alpha = \frac{\tau}{I} = \frac{0.00045 Nm}{0.000218025 kgm^2} \longrightarrow \alpha= 2.064 rad/s^2[/tex].
Part B: Time to Decrease Rotational Speed
To find the time (t) to decrease the rotational speed by 21.0 rad/s, we can use the equation [tex]\omega = \omega_0 + \alpha \times t[/tex], where [tex]\omega_0[/tex] is the initial angular velocity and [tex]\omega[/tex] is the final angular velocity. Since the sphere is slowing down due to friction, the angular acceleration will be negative, [tex]\alpha = -2.064 rad/s^2[/tex]. We are looking for the time it takes for the speed to decrease by 21.0 rad/s, so if the initial speed is [tex]\omega_0[/tex] and the final is [tex]\omega_0 - 21.0 rad/s[/tex], then [tex]0 = \omega_0 - 21.0 rad/s + \alpha \times t[/tex]. Solving for t gives [tex]t = \frac{21.0 rad/s}{2.064 rad/s^2} \approx t = 10.17 s[/tex].
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.19 mm apart and position your screen 3.53 m from the slits. While Young had to struggle to achieve a monochromatic light beam of sufficient intensity, you simply turn on a laser with a wavelength of 635 nm. How far on the screen are the first bright fringe and the second dark fringe from the central bright fringe? Express your answers in millimeters.
Answer:
First bright fringe is at 1.82 mm
First dark fringe is at 2.83 mm
Solution:
As per the question:
Slit width, d = 1.19 mm = [tex]1.19\times 10^{- 3}\ m[/tex]
Distance from the screen, x = 3.53 m
Wavelength of the light, [tex]\lambda = 635\ nm = 635\times 10^{- 9}\ m[/tex]
Now,
We know that the 1st bright fringe from the central fringe is given by:
[tex]y = \frac{n\lambda x}{d}[/tex]
where
n = 1
[tex]y = \frac{1\times 635\times 10^{- 9}\times 3.53}{1.19\times 10^{- 3}} = 1.88\ mm[/tex]
Also, we know that the 1st dark fringe from the central fringe is given by:
[tex]y = \frac{(n + \frac{1}{2})\lambda x}{d}[/tex]
[tex]y = \frac{(1 + \frac{1}{2})\times 635\times 10^{- 9}\times 3.53}{1.19\times 10^{- 3}} = 2.83\ mm[/tex]
A rectangular loop of wire with width w = 5 cm, length L = 10cm, mass m = 40 g, and resistance R = 20 mΩ has an initial velocity v0 = 1 m/s to the right. It crosses from a region with zero magnetic field to a region with B = 2T pointing out of the page. How far does the loop penetrate into the magnetic field?
Answer:
The loop penetrate 4 cm into the magnetic field.
Explanation:
Given that,
Width w= 5 cm
Length L= 10 cm
mass m = 40 g
Resistance R = 20 mΩ
Initial velocity = 1 m/s
Magnetic field = 2 T
We need to calculate the induced emf
Using formula of emf
[tex]\epsilon=v_{0}Bw[/tex]
Put the value into the formula
[tex]\epsilon =1\times2\times5\times10^{-2}[/tex]
[tex]\epsilon =10\times10^{-2}\ volt[/tex]
We need to calculate the current
Using Lenz's formula
[tex]i=\dfrac{\epsilon}{R}[/tex]
[tex]i=\dfrac{10\times10^{-2}}{20\times10^{-3}}[/tex]
[tex]i=5\ A[/tex]
We need to calculate the force
Using formula of force
[tex]F=i(\vec{w}\times\vec{B})[/tex]
[tex]F=iwB[/tex]
Put the value into the formula
[tex]F=5\times5\times10^{-2}\times2[/tex]
[tex]F=0.5\ N[/tex]
We need to calculate the acceleration
Using formula of acceleration
[tex]a=\dfrac{F}{m}[/tex]
Put the value in to the formula
[tex]a=\dfrac{0.5}{40\times10^{-3}}[/tex]
[tex]a=12.5\ m/s^2[/tex]
We need to calculate the distance
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{0-1^2}{2\times(-12.5)}[/tex]
[tex]s=0.04\ m[/tex]
[tex]s=4\ cm[/tex]
Hence, The loop penetrate 4 cm into the magnetic field.
A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi).
(a) Compute the maximum and minimum stress levels.
(b) Compute the stress ratio.
(c) Compute the magnitude of the stress range.
Answer:
275 MPa, -175 MPa
-0.63636
450 MPa
Explanation:
[tex]\sigma_{max}[/tex] = Maximum stress
[tex]\sigma_{min}[/tex] = Minimum stress
[tex]\sigma_m[/tex] = Mean stress = 50 MPa
[tex]\sigma_a[/tex] = Stress amplitude = 225 MPa
Mean stress is given by
[tex]\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}[/tex]
Stress amplitude is given by
[tex]\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa[/tex]
[tex]\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa[/tex]
Maximum stress level is 275 MPa
Minimum stress level is -175 MPa
Stress ratio is given by
[tex]R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636[/tex]
The stress ratio is -0.63636
Stress range is given by
[tex]\sigma_{max}-\sigma_{min}=450\ MPa[/tex]
Magnitude of the stress range is 450 MPa
The maximum and minimum stress levels could be calculated as 275 MPa and -175 MPa respectively. The stress ratio comes out to be -0.64, and the magnitude of the stress range is 450 MPa.
Explanation:In material fatigue studies, the stress levels in a material are often evaluated. In your case:
The mean stress (σm) is given as 50 MPa, and the stress amplitude (σa) is 225 MPa.(a) The maximum stress (σmax) can be calculated as σmax = σm + σa = 50 + 225 = 275 MPa, and the minimum stress (σmin) can be calculated as σmin = σm - σa = 50 - 225 = -175 MPa.(b) The stress ratio (R) is derived as the ratio of the minimum to the maximum stress. So R = σmin / σmax = -175 / 275 = -0.64(rounded to two decimal places).(c) The stress range (Δσ) is defined as the difference between the maximum and minimum stresses in a cycle. So Δσ = σmax - σmin = 275 - (-175) = 450 MPa.Learn more about Material Fatigue here:https://brainly.com/question/33293027
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A 86 g particle undergoes SHM with an amplitude of 1.3 mm, a maximum acceleration of magnitude 6.8 x 103 m/s2, and an unknown phase constant φ. What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?
Answer:
Explanation:
mass, m = 86 g
Amplitude, A = 1.3 mm
Maximum acceleration, a = 6.8 x 10^3 m/s^2
(a) Let T be the period of motion and ω be the angular frequency.
maximum acceleration, a = ω²A
6.8 x 10^3 = ω² x 1.3 x 10^-3
ω = 2287.0875 rad/s
T = 2π/ω
T = ( 2 x 3.14) / 2287.0875
T = 2.75 x 10^-3 second
T = 2.75 milli second
(b) maximum speed = ωA
v = 2287.0875 x 1.3 x 10^-3
v = 2.97 m/s
(c) Total mechanical energy
T = 1/2 mω²A²
T = 0.5 x 86 x 10^-3 x 2287.0875 x 2287.0875 x 1.3 x 1.3 x 10^-6
T = 0.38 J
(d) At maximum displacement, the acceleration is maximum
F = m x a = 86 x 10^-3 x 6.8 x 1000
F = 541.8 N
(e) acceleration a = ω² y = ω² x A / 2
a = 2287.0875 x 2287.0875 x 1.3 x 10^-3 / 2
a = 3400 m/s^2
Force, F = 86 x 10^-3 x 3400
F = 292.4 N
** In Problem 4.26, the coefficient of static friction between the book and the vertical back of the wagon is μs. Determine an expression for the minimum acceleration of the wagon in terms of μs so that the book does not slide down. Does the mass of the book matter? Explain.
The minimum acceleration of the wagon required for the book not to slide down can be determined using the coefficient of static friction (μs) between the book and the vertical back of the wagon. The expression for the minimum acceleration is given by a = μs * g.
Explanation:The minimum acceleration of the wagon required for the book not to slide down can be determined using the coefficient of static friction (μs) between the book and the vertical back of the wagon. The expression for the minimum acceleration is given by a = μs * g, where g is the acceleration due to gravity.
The mass of the book does not matter in this scenario. The coefficient of static friction is a property that depends on the nature of the surfaces in contact, and it determines the maximum force of static friction that can act between the book and the wagon before the book starts to slide down. The acceleration required to prevent sliding is independent of the mass of the book.
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An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in time t, through what angle did it rotate in the time ½t?
a. 4θ
b. ¼θ
c. ½θ
d. 2θ
e. θ
Answer:
Angular displacement will be [tex]\frac{1}{4}\Theta[/tex]
So option (b) will be the correct option
Explanation:
We have given that firstly object is at rest
So [tex]\omega _i=0rad/sec[/tex]
From law of motion we know that angular displacement is given by
[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times t+\frac{1}{2}\alpha t^2=\frac{1}{2}\alpha t^2[/tex]
Now angular displacement by the object in [tex]\frac{t}{2}sec[/tex]
[tex]\Theta =0\times t+\frac{1}{2}\alpha (\frac{t}{2})^2=\frac{1}{4}(\frac{1}{2}\alpha t^2)=\frac{1}{4}\Theta[/tex]
So option (b) will be the correct option
The angle the object rotate through in the time [tex]\frac{1}{2} t[/tex] is [tex]\frac{1}{4} (\theta)[/tex]
Given the following data:
Initial angular speed = 0 m/s (since it starts from rest).Angle = [tex]\theta[/tex]Time = tTo determine the angle the object rotate through in the time [tex]\frac{1}{2} t[/tex]:
How to calculate angular displacement.Mathematically, angular displacement is given by this formula:
[tex]\theta = \omega_i t +\frac{1}{2} \alpha t^2[/tex]
Where:
[tex]\theta[/tex] is the angular displacement.[tex]\omega[/tex] is the initial angular speed.[tex]\alpha[/tex] is the angular acceleration.t is the time.Substituting the given parameters into the formula, we have;
[tex]\theta = 0( t )+\frac{1}{2} \alpha t^2\\\\\theta = \frac{1}{2} \alpha t^2[/tex]
when t = [tex]\frac{1}{2} t[/tex]:
[tex]\theta = \frac{1}{2} \alpha (\frac{t}{2} )^2\\\\\theta = \frac{1}{2} \alpha (\frac{t^2}{4} )\\\\\theta =\frac{1}{4} (\frac{1}{2} \alpha t^2)\\\\\theta =\frac{1}{4} (\theta)[/tex]
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A 25 - kg television set rests on four rubber pads , each having a height of 1 . 0 cm and a radius of 0 . 60 cm . A 200 - N horizontal force is applied to the television set . How far does it move sideways ? The shear modulus of rubber is 2 . 6 x 106 N / m2
The television set moves sideways approximately 0.15 µm.
Explanation:
To calculate the distance the television set moves sideways, we need to use the equation for shear deformation. The equation is Ax = (F * L0 * L2) / (S * A), where Ax is the displacement, F is the force applied, L0 is the original length, L2 is the height of the rubber pads, S is the shear modulus, and A is the cross-sectional area. Substituting the known values, we have Ax = (200 N * 1.0 cm) / (2.6 x 10^6 N/m^2 * (3.14 x (0.0060 m)^2)). Solving for Ax, we find Ax ≈ 1.50 x 10^-7 m, or 0.15 µm.
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A conducting loop of radius 1.50 cm and resistance 8 × 10−6Ω is perpendicular to a uniform magnetic field of magnitude 23.0 × 10−6T. The field magnitude drops to zero in 7 ms. How much thermal energy is produced in the loop by the change in the magnetic field?
To solve this problem it is necessary to apply the concepts related to electromotive force or induced voltage.
By definition we know that the induced emf in the loop is equal to the negative of the change in the magnetic field, that is,
[tex]\epsilon = -A \times \frac{\Delta B}{\Delta t}[/tex]
[tex]\epsilon = -A \times (\frac{B_f-B_i}{t_f-t_i})[/tex]
Where A is the area of the loop, B the magnetic field and t the time.
Replacing with our values we have that
[tex]\epsilon = -(\pi (1.5*10^{-2})^2)(\frac{0-23*10^{-6}}{7*10^{-3}-0})[/tex]
[tex]\epsilon = 2.3225*10^{-6}V[/tex]
Therefore the thermal energy produced is given by
[tex]E = P*t = \frac{\epsilon^2}{R}t[/tex]
[tex]E = \frac{(2.3225*10^{-6})^2}{8*10^{-6}}*(7*10^{-3})[/tex]
[tex]E = 4.719*10^{-9}J[/tex]
The thermal energy produced in the loop is [tex]4.719*10^{-9}J[/tex]
Halogen lightbulbs allow their filaments to operate at a higher temperature than the filaments in standard incandescent bulbs. For comparison, the filament in a standard lightbulb operates at about 2900K, whereas the filament in a halogen bulb may operate at 3400K. Which bulb has the higher peak frequency? Calculate the ratio of the peak frequencies. The human eye is most sensitive to a frequency around 5.5x10^14 Hz. Which bulb produces a peak frequency close to this value?
Answer:
Halogen
0.85294
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
b = Wien's displacement constant = [tex]2.897\times 10^{-3}\ mK[/tex]
T = Temperature
From Wien's law we have
[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{2900}\\\Rightarrow \lambda_m=9.98966\times 10^{-7}\ m[/tex]
Frequency is given by
[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{9.98966\times 10^{-7}}\\\Rightarrow \nu=3.00311\times 10^{14}\ Hz[/tex]
For Halogen
[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{3400}\\\Rightarrow \lambda_m=8.52059\times 10^{-7}\ m[/tex]
Frequency is given by
[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{8.52059\times 10^{-7}}\\\Rightarrow \nu=3.52088\times 10^{14}\ Hz[/tex]
The maximum frequency is produced by Halogen bulbs which is closest to the value of [tex]5.5\times 10^{14}\ Hz[/tex]
Ratio
[tex]\frac{3.00311\times 10^{14}}{3.52088\times 10^{14}}=0.85294[/tex]
The ratio of Incandescent to halogen peak frequency is 0.85294
Nitrogen (N2) undergoes an internally reversible process from 6 bar, 247°C during which pν1.2 = constant. The initial volume is 0.1 m3 and the work for the process is 50 kJ. Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in kJ, and the entropy change, in kJ/K.
The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.
Explanation:The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. In order to determine the heat transfer and entropy change, we need to use the first and second laws of thermodynamics. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.
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8. A solid wooden door, 90 cm wide by 2.0 m tall, has a mass of 35 kg. It is open and at rest. A small 500-g ball is thrown perpendicular to the door with a speed of 20 m/s and hits the door 60 cm from the hinged side, causing it to begin turning. The ball rebounds along the same line with a speed of 16.0 m/s relative to the ground. If the momentum of inertia of the door around the hinge is I=1/3 Ma 2 , where a is the width of the door, how much energy was lost during this collision?
Answer:
Explanation:
Kinetic energy of ball
= .5 x .5 x 20²
= 100 J
Original kinetic energy of door = 0
Total kinetic energy before ball hitting the door
= 100 J
We shall apply law of conservation of momentum to calculate angular velocity of the door after ball hitting it.
change in angular momentum of ball
= mvr - mur , u is initial velocity and v is final velocity of ball
= .5 ( 20 + 16 ) x .06
= 1.08
Change in angular momentum of door
= I x ω - 0
1/3 x 35 x .09² x ω
= .0945 x ω
so
.0945 x ω = 1.08
ω = 11.43
rotational K E of door after collision
= 1/2 I ω²
= .5 x .0945 x 11.43 ²
= 6.17 J
Kinetic energy of ball after collision
= 1/2 x .5 x 16²
= 64
Total KE of door and ball
= 64 + 6.17
= 70.17 J
LOSS OF ENERGY
= 100 - 70.17 J
= 29.83 J