Final answer:
The argon temperature at the heater exit is determined to be 348.08 ℃, while the argon volume flow rate at the heater exit is calculated to be 3769.23 m³/s.
Explanation:
To determine the argon temperature and volume flow rate at the heater exit, we can use the principles of thermodynamics. The problem states that heat transfer in the rate of 150 kW is supplied to the argon. Since there is no information given about the heater efficiency, we can assume it to be 100%, meaning all the heat is transferred to the argon. Therefore, the heat transfer can be equated to the change in internal energy of the argon, which is given by:
Q = m * Cp * ΔT
Where Q is the heat transfer, m is the mass flow rate, Cp is the specific heat capacity of argon, and ΔT is the change in temperature.
Given that Q = 150 kW, m = 6.24 kg/s, and Cp = 0.52 kJ/kg·℃ (specific heat capacity of argon), we can rearrange the equation to solve for ΔT:
ΔT = Q / (m * Cp)
Substituting the values, we have ΔT = (150 * 10³ J/s) / (6.24 kg/s * 0.52 kJ/kg·℃), which equals 48.08 ℃.
Therefore, the argon temperature at the heater exit is 300 ℃ + 48.08 ℃, which equals 348.08 ℃.
To calculate the argon volume flow rate at the heater exit, we can use the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation to solve for V, we have:
V = (nRT) / P
Given that the initial conditions are 300 K and 100 kPa, and a mass flow rate of 6.24 kg/s, we can calculate the number of moles of argon:
n = m / M
Where M is the molar mass of argon. The molar mass of argon is 39.948 g/mol, so n = 6.24 / (39.948 * 10⁻³) = 156.172 mol.
Substituting the values into the ideal gas law equation, we have V = (156.172 mol * 8.314 J/mol·K * 300 K) / 100 kPa. Converting kPa to Pa, we have V = (156.172 mol * 8.314 J/mol·K * 300 K) / 100,000 Pa, which equals 3769.23 m³.
Therefore, the argon volume flow rate at the heater exit is 3769.23 m³/s.
A velocity selector has an electric field of magnitude 2170 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 5.45 × 103 m/s, enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of +4.10 × 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.54 × 10-9 N, pointing directly upward. What is the speed of this particle?
Answer:
[tex]v = 4.51 \times 10^3 m/s[/tex]
Explanation:
electric field = 2170 N/C
now the speed of the charge particle is given as
[tex]v = 5.45 \times 10^3 m/s[/tex]
here we know that charge particle moves without any deviation
so we will have
[tex]qvB = qE[/tex]
now magnetic field in this region is given as
[tex]B = \frac{E}{v}[/tex]
[tex]B = \frac{2170}{5.45 \times 10^3}[/tex]
[tex]B = 0.398 T[/tex]
Now another charge particle enters the region with different speed and experience the force upwards
[tex]F = qE - qvB[/tex]
[tex]1.54 \times 10^{-9} = (4.10\times 10^{-12})[2170 - v(0.398)][/tex]
[tex]375.6 = 2170 - v(0.398)[/tex]
[tex]v = 4.51 \times 10^3 m/s[/tex]
The answer is: [tex]9.43 \times 10^5 \, \text{m/s}[/tex].
To determine the speed of the different particle that enters the velocity selector, we need to consider the forces acting on it due to the electric and magnetic fields. The net force acting on a charged particle in an electric field [tex]\( E \)[/tex] and a magnetic field [tex]\( B \)[/tex] is given by the Lorentz force equation:
[tex]\[ \vec{F} = q\vec{E} + q(\vec{v} \times \vec{B}) \][/tex]
where [tex]\( q \)[/tex] is the charge of the particle, [tex]\( \vec{v} \)[/tex] is the velocity of the particle, [tex]\( \vec{E} \)[/tex] is the electric field, and [tex]\( \vec{B} \)[/tex] is the magnetic field.
For the particles that pass through the velocity selector without being deflected, the forces due to the electric and magnetic fields must cancel each other out. This means that:
[tex]\[ qE = qvB \][/tex]
where [tex]\( v \)[/tex] is the speed of the particles that pass through undeflected.
Given that the electric field [tex]\( E \)[/tex] is 2170 N/C directed vertically upward, and the particles are traveling east, the magnetic field [tex]\( B \)[/tex] must be directed south to produce a force that cancels the electric force. The speed [tex]\( v \)[/tex] of the undeflected particles is given as [tex]\( 5.45 \times 10^3 \)[/tex] m/s.
Now, for the different particle with a charge of [tex]\( +4.10 \times 10^{-12} \)[/tex] C, the net force [tex]\( F \)[/tex] acting on it is [tex]\( 1.54 \times 10^{-9} \)[/tex] N, pointing directly upward. This means that the magnetic force is not sufficient to cancel the electric force, and the net force is due to the electric force only:
[tex]\[ F = qE \][/tex]
We can use this equation to find the speed [tex]\( v' \)[/tex] of the different particle. Since the net force is equal to the electric force, the magnetic force must be zero. This implies that the velocity of the particle is such that the magnetic force component is equal and opposite to the electric force component, but since the net force is not zero, the particle is not moving at the correct speed to pass through undeflected.
Let's solve for the speed [tex]\( v' \)[/tex] of the different particle:
[tex]\[ F = qE \][/tex]
[tex]\[ 1.54 \times 10^{-9} \, \text{N} = (4.10 \times 10^{-12} \, \text{C})(2170 \, \text{N/C}) \][/tex]
[tex]\[ v' = \frac{F}{qB} \][/tex]
We know [tex]\( F \), \( q \), and \( E \)[/tex], but we need to find [tex]\( B \)[/tex] from the information given for the undeflected particles:
[tex]\[ qvB = qE \][/tex]
[tex]\[ vB = E \][/tex]
[tex]\[ B = \frac{E}{v} \][/tex]
[tex]\[ B = \frac{2170 \, \text{N/C}}{5.45 \times 10^3 \, \text{m/s}} \][/tex]
[tex]\[ B = 3.98 \times 10^{-4} \, \text{T} \][/tex]
Now we can find [tex]\( v' \)[/tex]:
[tex]\[ v' = \frac{1.54 \times 10^{-9} \, \text{N}}{(4.10 \times 10^{-12} \, \text{C})(3.98 \times 10^{-4} \, \text{T})} \][/tex]
[tex]\[ v' = \frac{1.54 \times 10^{-9}}{1.6322 \times 10^{-15}} \][/tex]
[tex]\[ v' = 9.43 \times 10^5 \, \text{m/s} \][/tex]
Therefore, the speed of the different particle is [tex]\( 9.43 \times 10^5 \) m/s[/tex].
Which of the following is the kinetic rate equation for the addition-elimination mechanism of nucleophilic aromatic substitution? rate = k[aryl halide] rate = k[nucleophile] rate = k[aryl halide][nucleophile] rate = k[aryl halide][nucleophile]2
Answer:
Rate = k[aryl halide][nucleophile]
Explanation:
The simple aryl halides are almost inert to usual nucleophilic reagents but considerable activation on the ring can be produced by the addition of strongly electron-attracting substituents on either the ortho or para positions, or both. These groups deactivate the ring to allow the attack of the nucleophille on the ring.
Thus, these reactions can occur by following addition-elimination mechanism in which the nucleophille first attacks the aryl halide and then the elimination of the leaving group takes place.
Kinetic studies of this type of mechanism demonstrate that the reactions are of second-order kinetics– first order w.r.t. nucleophile and also, first-order w.r.t. aromatic substrate. The rate determining step (r.d.s.) is the formation of the addition intermediate.
Thus,
Rate = k[aryl halide][nucleophile]
Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B carries a charge of -q. Sphere C carries no net charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Lastly, sphere C is touched to sphere B and separated from it. (a) What is the ratio of the final charge on sphere C to q? What is the ratio of the final total charge on the three spheres to q (b) before they are allowed to touch each other and (c) after they have touched?
Answer:
Part a)
Final charge on C : q = 1.875
Part b)
Ratio for A = 6 : 1.25
Ratio for B = -1 : 1.875
Ratio for C = 0
Explanation:
When two identical metal sphere are connected together then the charge on them will get equally divided on both after connecting them by conducting wire
So here we have
[tex]q_A = + 6q[/tex]
[tex]q_B = -q[/tex]
[tex]q_c = 0[/tex]
Step 1: We connected A and B and then separate them
so we have
[tex]q_A' = q_B' = 2.5q[/tex]
Step 2: We connected A and C and then separate them
so we have
[tex]q_A'' = q_c' = 1.25q[/tex]
Step 3: We connected B and C and then separate them
so we have
[tex]q_c'' = q_b'' = 1.875q[/tex]
The final charge on sphere C is 2.5q. Before touching, the total charge is +5q, which remains the same after all interactions, demonstrating conservation of charge.
Explanation:When identical conductive spheres come into contact, the charges redistribute evenly across both spheres. If Sphere A is initially charged with +6q and Sphere B has a -q charge, touching them together allows their total charge to be shared, resulting in each sphere having (6q - q)/2 = 2.5q. After separation, both spheres A and B would have a charge of 2.5q. By touching Sphere C, which is uncharged, to A and then B in sequence, C gains a fraction of the charge from each, ending up with (2.5q)/2 from A and (2.5q)/2 from B, which totals 2.5q, since touching B does not change the charge obtained from A.
Before contact, the total charge is +5q (+6q from A and -q from B). After all the interactions, the total charge remains the same, +5q, but redistributed: A and B with 2.5q each and C with 2.5q.
In general, the total charge before and after remains constant, demonstrating conservation of charge.
A 15,000 N truck starts from rest and moves down a 15∘ hill with the engine providing a 8,000 N force in the direction of the motion. Assume the rolling friction force between the truck and the road is very small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill?
Answer:
[tex]v_f = 27.9 m/s[/tex]
Explanation:
Component of the weight of the truck along the inclined plane is given as
[tex]F_1 = W sin\theta[/tex]
[tex]F_1 = 15000 sin15[/tex]
[tex]F_1 = 3882.3 N[/tex]
also the engine is providing the constant force to it as
[tex]F_2 = 8000 N[/tex]
now the net force along the the plane is given as
[tex]F_{net} = 8000 + 3882.3[/tex]
[tex]F = 11882.3 N[/tex]
mass of the truck is given as
[tex]m = \frac{w}{g} = 1529 kg[/tex]
now the acceleration is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = 7.77 m/s^2[/tex]
now the speed of the truck after travelling distance of d = 50 m is given as
[tex]v_f^2 = v_i^2 + 2 a d[/tex]
[tex]v_f^2 = 0 + 2(7.77)(50)[/tex]
[tex]v_f = 27.9 m/s[/tex]
A baseball player wants to hit a home run over the wall of a stadium. He hits the ball 0.751 m above the ground with at an angle of 30 degrees above the horizontal and a speed of 37.5 m/s. What is the tallest wall the player can clear if the wall is 150 m away?
Answer:
The baseball does not reach the wall, because the ball falls at 125.33 meters and the wall is at 150 meters
Explanation:
V= 37.5 m/s
α= 30º
g= 9.8 m/s²
Vx= V * cos(30º) = 32.47 m/s
Vy= V * sin(30º) = 18.75 m/s
flytime of the baseball:
t= 2 * Vy/g
t= 3.86 sec
distance of baseball fall:
d= Vx * t
d= 125.33 m
Answer: 18.67 m wall
Explanation:this is projectile.
For Max height,
H = v²sin²©/2g
H= (37.5)²sin²(30)/(2*9.81)
H = 17.919 m
But he is standing 0.751 m above ground, so total height = 17.919+0.751= 18.67m
A pipe has a diameter of 20cm. What is the cross-sectional area of the pipe with units m^2.
Answer:
[tex]Area = 0.126 m^2[/tex]
Explanation:
Since the crossection of cylinder is of circular shape
so here the crossectional area will be given as
[tex]A = \pi r^2[/tex]
here we know that radius of the cylinder will be
[tex]R = 20 cm[/tex]
so in SI units it is given as
[tex]R = 0.20 m[/tex]
now we have
[tex]Area = \pi (0.20)^2[/tex]
[tex]Area = 0.126 m^2[/tex]
A negative charge -0.550 HC exerts an upward 0.700 N force on an unknown charge 0.220 m directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the -0.550 HC charge? downward upward
Answer:
Part a)
[tex]q_2 = -6.8 \mu C[/tex]
Part b)
[tex]F = 0.700 N[/tex]
direction = downwards
Explanation:
As we know that the negative charge will experience the force due to some other charge below it
the force is given as
[tex]F = 0.700 N[/tex]
now we know that
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now plug in all data
[tex]0.700 = \frac{(9 \times 10^9)(0.550\mu C)q_2}{0.220^2}[/tex]
[tex]0.700 = 1.022\times 10^5 q_2[/tex]
[tex]q_2 = -6.8 \mu C[/tex]
since this is a repulsion force so it must be a negative charge
Part b)
As per Newton's III law it will exert equal and opposite force on it
So here the force on the charge below it will be same in magnitude but opposite in direction
so here we have
[tex]F = 0.700 N[/tex]
direction = downwards
A 13.7 N force is applied to a cord wrapped around a disk of radius 0.43 m. The disk accelerates uniformly from rest to an angular speed of 30.3 rad/s in 3.43 s. Determine the angular acceleration of the disk. Hint: the average acceleration is the change in angular speed over time.
Answer:
8.83 rad/s²
Explanation:
w₀ = initial angular speed of the disk = 0 rad/s
w = final angular speed of the disk = 30.3 rad/s
t = time period of rotation of the disk = 3.43 sec
[tex]\alpha[/tex] = Angular acceleration of the disk
Angular acceleration of the disk is given as
[tex]\alpha = \frac{w-w_{o}}{t}[/tex]
inserting the values
[tex]\alpha[/tex] = (30.3 - 0)/3.43
[tex]\alpha[/tex] = 8.83 rad/s²
Find the speed of light in carbon tetrachlorideethyl alcohol. The refraction index is 1.461 using 3 x 10^8 m/s as the speed of light in vacuum. Answer in units of m/s.
Answer:
2.05 x 10^8 m /s
Explanation:
c = 3 x 10^8 m/s
μ = c / v
where, μ is the refractive index, c be the velocity of light in air and v be the velocity of light in the medium.
μ = 1.461
1.461 = 3 x 10^8 / v
v = 3 x 10^8 / 1.461
v = 2.05 x 10^8 m /s
An energy efficient light bulb uses 15 W of power for an equivalent light output of a 60 W incandescent light bulb. How much energy is saved each month by using the energy efficient light bulb instead of the incandescent light bulb for 4 hours a day? Assume that there are 30 days in one month.
Answer:
5400 W-hr
Explanation:
Given:
Power used by energy efficient bulb = 15 W
Power used by incandescent bulb = 60 W
Total time of bulb used = 4 hours/day × 30 days = 120 hours
Now,
Energy used by the individual bulb by using them for 120 hours
we know,
Energy = Power × Time
thus,
Energy consumed by energy efficient bulb = 15 W × 120 hour = 1800 W-hr
Energy consumed by incandescent bulb = 60 W × 120 hour = 7200 W-hr
hence, the energy saved will be = 7200 - 1800 = 5400 W-hr
Final answer:
Calculating the monthly energy savings involves finding the difference in power consumption between a 60 W incandescent bulb and a 15 W energy efficient bulb, multiplying by the daily usage, and the number of days in a month. The savings are 45 W per hour, equating to 180 Wh/day and 5.4 kWh/month.
Explanation:
The question asks how much energy is saved each month by using an energy efficient light bulb instead of an incandescent light bulb for 4 hours a day. To calculate the energy saved, we need to determine the difference in power consumption between the two types of bulbs, multiply that difference by the number of hours used per day, and then by the number of days in a month.
Firstly, we find the power saved per hour:
60 W (incandescent) - 15 W (efficient) = 45 W saved per hour
Next, we calculate the daily savings:
45 W x 4 hours/day = 180 Wh/day
Finally, we calculate the monthly savings:
180 Wh/day x 30 days/month = 5400 Wh/month
Since 1 kWh = 1000 Wh, the saving is 5.4 kWh/month
This is the amount of energy saved by using the energy efficient bulb instead of the incandescent bulb for 4 hours each day over a month.
What is the effective resistance of a car’s starter motor when 150 A flows through it as the car battery applies 12.0 V to the motor?
Answer:
From ohms law,
V=IR
R=V/I =12.0/150 =0.08 ohm.
What is the tangential velocity of a 15 kg mass tied to a string and moving in a circle of radious 10 m, if the period of the orbit is 0.2 seconds?
Answer:
Tangential speed, v = 314.15 m/s
Explanation:
It is given that,
Mass of the object, m = 15 kg
It is moving in a circle of radius, r = 10 m
Time period, t = 0.2 seconds
We need to find the tangential velocity of the object. It is given by :
[tex]v=\dfrac{2\pi r}{t}[/tex]
Where
v = tangential speed
[tex]v=\dfrac{2\pi \times 10\ m}{0.2\ s}[/tex]
v = 314.15 m/s
So, the tangential speed of the object is 314.15 m/s. Hence, this is the required solution.
A wire placed on the plane of this screen carries a current toward the top of the screen. The wire feels a magnetic force toward the right. The direction of the magnetic field causing this force is which of the following? (a) outward out of the screen (b) inward into the screen (c) in the plane of the screen and toward the left edge (d) in the plane of the screen and toward the bottom edge.
Answer:
option (a)
Explanation:
The direction of magnetic field is given by Fleming's left hand rule.
If we spread the fore finger, middle finger and the thumb of our right hand such that they are mutually perpendicular to each other, then the direction of force is in the direction of thumb, direction of magnetic field is given by the direction of fore finger and the middle finger indicates the direction of current .
Here, current is upwards, magnetic force is rightwards, so the direction of magnetic field is given in outwards from the screen.
An object moves on a trajectory given by Bold r left parenthesis t right parenthesis equals left angle 10 cosine 6 t comma 10 sine 6 t right angle for 0 less than or equals t less than or equals pi. How far does it travel?
The object is moving along a circular path. Given the time restriction, it completes a half revolution, which equates to half the circumference of the circle. Therefore, the object travels approximately 31.4 units.
Explanation:The object is moving on a circular trajectory defined by r(t) = <10 cos 6t, 10 sin 6t>. This equation describes a circular path with a radius of 10 units. To find out how far the object travels, we can use the formula for the circumference of a circle, which is 2πr. However, since the time t varies from 0 to π, the object only makes a half revolution along the circle. So, the total distance the object travels would be equal to half the circumference of the circle, which is πr.
Substituting r = 10, we get, distance traveled = π*10 = 31.4 units approx.
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Upper A 16-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the wall?
Answer:
The ladder is moving at the rate of 0.65 ft/s
Explanation:
A 16-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second. We need to find the rate at which the top of the ladder moving down when the foot of the ladder is 5 feet from the wall.
The attached figure shows whole description such that,
[tex]x^2+y^2=256[/tex].........(1)
[tex]\dfrac{dx}{dt}=2\ ft/s[/tex]
We need to find, [tex]\dfrac{dy}{dt}[/tex] at x = 5 ft
Differentiating equation (1) wrt t as :
[tex]2x.\dfrac{dx}{dt}+2y.\dfrac{dy}{dt}=0[/tex]
[tex]2x+y\dfrac{dy}{dt}=0[/tex]
[tex]\dfrac{dy}{dt}=-\dfrac{2x}{y}[/tex]
Since, [tex]y=\sqrt{256-x^2}[/tex]
[tex]\dfrac{dy}{dt}=-\dfrac{2x}{\sqrt{256-x^2}}[/tex]
At x = 5 ft,
[tex]\dfrac{dy}{dt}=-\dfrac{2\times 5}{\sqrt{256-5^2}}[/tex]
[tex]\dfrac{dy}{dt}=0.65[/tex]
So, the ladder is moving down at the rate of 0.65 ft/s. Hence, this is the required solution.
This is a calculation of a related rates problem in calculus, using the Pythagorean Theorem. The calculation finds the rate of descent (db/dt) for the top of a ladder sliding down a wall. The final answer found is -1.5 feet/second, meaning the top of the ladder is moving downward at a rate of 1.5 feet per second when the ladder is 5 feet away from the building.
Explanation:The question involves a concept in calculus known as related rates problem. It's a practical extension of the Pythagorean theorem where the sides of the right triangle formed by the ladder, building, and the ground are changing over time. In this case, a 16-feet ladder is sliding down a wall and we are to find the rate at which the top of the ladder is moving down when the foot of the ladder is 5 feet from the wall. First, we label the bottom (horizontal) side of our triangle 'a', the side along the wall (vertical) 'b', and the hypotenuse (ladder) 'c'. The Pythagorean theorem gives us a² + b² = c². Differentiating both sides with respect to time (t) gives us 2a(da/dt) + 2b(db/dt) = 2c(dc/dt). Here, we know that da/dt = 2 feet/sec (rate at which the ladder is moving away from the wall), c = 16 feet (length of the ladder), and a (distance from the wall) = 5 feet. Also, the ladder's length is not changing, so dc/dt = 0. Substituting these values, we can solve for db/dt (rate of descent of the ladder). The resultant rate of descent of the top of the ladder is -1.5 feet/second.
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A plank 2.65 m long is supported by a cable, and a small ledge. The cable attaches 79.7 cm from the ledge, and makes an angle of 40.3 with the plank. It can hold 2648 N. How close (in meters) to the end of the plank can a 76.4 kg person walk before the cable breaks? Ignore the mass of the plank itself.
Answer:
0.51 m
Explanation:
Given :
Plank length = 2.65 m
Distance from the ledge from which the cable is attached = 79.7 cm = 0.797 m
Angle = 40.3 degree
Mass of the man, m = 76.4 kg
Now the net torque acting on the plank is zero.
Therefore, maximum torque applied by cable is
= 2648 x 0.797 x cos (40.3)
= 1609.5 N-m
We know that the man should apply the same torque,
Therefore, x [tex]\times[/tex]76.4 [tex]\times[/tex] g = 1609.5
x =2.14 m
Therefore, length to the end of the plank is = 2.65-2.14
=0.51 m
A ball is released at a velocity of 400 m/s at an angle of 85 degrees from the horizontal. How far does it travel horizontally before the ball lands on the ground? (use a-10m/s2 if needed, round answer to a whole number-no decimals). Do not put the units in the answer. QUESTION 5 angle of 85 degrees from the horizontal. How long will it take for the ball to get if needed, round answer to a whole number-no decimals). Do not put the units in the answer. A ball is released at a velocity of 400 m/s at an to its apex? (use a-10m/s QUESTION 6 A ball is released at a velocity of 400 m/s at an angle of 45 degrees from the horizontal. What is the vertical component of the velocity? (use a--10m/s if needed, round answer to a whole number-no decimals), Do not put the units in the answer.
Answer:
8000
Explanation:
V²=U²+2as
V=400m/s
U=0
a=10
s=?
(400)²=(0)²+2*10*a
160,000=20s
s=160,000/20
s=8000
The pressure of a monatomic ideal gas is doubled, while the volume is cut in half. By what factor is the internal energy of the gas multiplied? (a) 1/4 (b) 1/2 (c) 1 (d) 2 (e) 4
Answer:
option C
multiplication factor n = 1 when volume change to half and pressure become double
Explanation:
we know by Ideal Gas law:
P1V1 = nRT1
P2V2 = nRT2
according to the question
pressure is doubled and volume is reduced to half
so we have
new pressure = 2*P1
new volume = V1/2
hence,
(2P1) * (V1/2) = nRT2
P1V1 = nRT2
we have now
nRT1 = nRT2
we get
T1 = T2
thus no change in temperature
we know that internal energy is given as
internal energy = nCvT,
since temperature is directly proportional to internal energy and since temperature remains constant therefore internal energy remains constant
So there is no change in internal energy
thus, multiplication factor n = 1 when volume change to half and pressure change to double
Suppose that you determine the density of a mineral by measuring its mass (m) (4.635±0.002) g and its volume (1.13±0.05) mL. d = m/V What is the uncertainty in the computed density?
Answer:
[tex]\Delta \rho =0.18/mL326gm[/tex]
Explanation:
we have error in division of 2 quantities is related as
[tex]z=\frac{a}{b}\\\\\frac{\Delta z}{z}=\frac{\Delta a}{a}+\frac{\Delta b}{b}[/tex]
where [tex]\Delta a,\Delta b[/tex] are errors in quantities a,b
Thus error in density becomes
[tex]\Delta \rho =\rho _{0}\times(\frac{\Delta m}{m}+\frac{\Delta V}{V})[/tex]
Applying values we get
[tex]\Delta \rho =\frac{4.635}{1.13} \times(\frac{0.002}{4.635}+\frac{0.05}{1.13})[/tex]
thus [tex]\Delta \rho =0.18326gm/mL[/tex]
A 310-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,190 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable
Answer:
t = 141.55 years
Explanation:
As we know that the radius of the wire is
r = 2.00 cm
so crossectional area of the wire is given as
[tex]A = \pi r^2[/tex]
[tex]A = \pi(0.02)^2[/tex]
[tex]A = 1.26 \times 10^{-3} m^2[/tex]
now we know the free charge density of wire as
[tex]n = 8.50 \times 10^{28}[/tex]
so drift speed of the charge in wire is given as
[tex]v_d = \frac{i}{neA}[/tex]
[tex]v_d = \frac{1190}{(8.50 \times 10^{28})(1.6 \times 10^{-19})(1.26\times 10^{-3})}[/tex]
[tex]v_d = 6.96 \times 10^{-5} m/s[/tex]
now the time taken to cover whole length of wire is given as
[tex]t = \frac{L}{v_d}[/tex]
[tex]t = \frac{310 \times 10^3}{6.96 \times 10^{-5}}[/tex]
[tex]t = 4.46 \times 10^9 s[/tex]
[tex]t = 141.55 years[/tex]
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15 cm from the wire, what is the maximum current the wire can carry?
Answer:
37.725 A
Explanation:
B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T
r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m
i = magnitude of current flowing through the wire = ?
Magnetic field by a long wire is given as
[tex]B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}[/tex]
Inserting the values
[tex]0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}[/tex]
i = 37.725 A
What is the length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters? A. 0.00180 meter B. 1.201386 meters C. 1.214855 meters D. 0.001386 meter
Answer:
Option B is the correct answer.
Explanation:
Thermal expansion
[tex]\Delta L=L\alpha \Delta T[/tex]
L = 1.2 meter
ΔT = 65 - 15 = 50°C
Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C
We have change in length
[tex]\Delta L=L\alpha \Delta T=1.2\times 24\times 10^{-6}\times 50=1.44\times 10^{-3}m[/tex]
New length = 1.2 + 1.44 x 10⁻³ = 1.2014 m
Option B is the correct answer.
Answer:
B. 1.201386 meters
Explanation:
The length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters is 1.201386 meters.
A curve ball is a type of pitch in which the baseball spins on its axis as it heads for home plate. If a curve ball is thrown at 34.5 m/s (77 mph ) with a spin rate of 26 rev/s , how many revolutions does it complete before reaching home plate? Assume that home plate is 18.3 m (60 ft) from the pitching mound and that the baseball travels at a constant velocity.
Answer:
13.79 revolutions
Explanation:
Velocity of ball = v = 34.5 m/s
Spin rate of ball = 26 revolutions/s
Distance between home plate and pitching mound = s = 18.3 m
Time taken by the ball to reach the home plate = t
[tex]t=\frac{18.3}{34.5}\ s[/tex]
Number of revolutions the ball completes is
[tex]26\times \frac{18.3}{34.5}=13.79[/tex]
∴Number of revolutions the ball completes before reaching home plate is 13.79
Answer:
The number of revolutions is 13.78
Explanation:
Given data:
v = speed = 34.5 m/s
d = distance = 18.3 m
spin rate = 26 rev/s
The time taken is equal to:
[tex]t=\frac{d}{v} =\frac{18.3}{34.5} =0.53s[/tex]
The number of revolutions is equal:
N = 0.53 * 26 = 13.78 rev
A 19 kg rock slides on a rough horizontal surface at 9.25 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.63. What average thermal power (in Watts) is produced as the rock stops?
Answer:
541.89 Watt
Explanation:
Consider the motion of rock on rough surface
μ = Coefficient of kinetic friction = 0.63
acceleration caused to kinetic friction is given as
a = - μg
a = - (0.63) (9.8)
a = - 6.2 m/s²
v₀ = initial velocity of the rock = 9.25 m/s
t = time taken to stop
v = final velocity = 0 m/s
Using the equation
v = v₀ + a t
0 = 9.25 + (- 6.2) t
t = 1.5 sec
m = mass of the rock = 19 kg
Energy lost due to friction is given as
E = (0.5) m (v₀² - v²)
E = (0.5) (19) (9.25² - 0²)
E = 812.84 J
Average thermal power is given as
[tex]P = \frac{E}{t}[/tex]
[tex]P = \frac{812.84}{1.5}[/tex]
P = 541.89 Watt
A nonconducting sphere contains positive charge distributed uniformly throughout its volume. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity. (There may be more than one correct choice)
a) The potential is highest at the center of the sphere. b) The potential at the center of the sphere is the same as the potential at the surface. c) The potential at the surface is higher than the potential at the center. d) The potential at the center is the same as the potential at infinity. e) The potential at the center of the sphere is zero.
Answer:
a). The potential is highest at the center of the sphere
Explanation:
We k ow the potential of a non conducting charged sphre of radius R at a point r < R is given by
[tex]E=\left [ \frac{K.Q}{2R} \right ]\left [ 3-(\frac{r}{R})^{2} \right ][/tex]
Therefore at the center of the sphere where r = 0
[tex]E=\left [ \frac{K.Q}{2R} \right ]\left [ 3-0 \right ][/tex]
[tex]E=\left [ \frac{3K.Q}{2R} \right ][/tex]
Now at the surface of the sphere where r = R
[tex]E=\left [ \frac{K.Q}{2R} \right ]\left ( 3-1 \right )[/tex]
[tex]E=\left [ \frac{2K.Q}{2R} \right ][/tex]
[tex]E=\left [ \frac{K.Q}{R} \right ][/tex]
Now outside the sphere where r > R, the potential is
[tex]E=\left [ \frac{K.Q}{r} \right ][/tex]
This gives the same result as the previous one.
As [tex]r\rightarrow \infty , E\rightarrow 0[/tex]
Thus, the potential of the sphere is highest at the center.
The electric field everywhere on the surface of a thin, spherical shell of radius 0.750 m is of magnitude 890 N/C and points radially toward the center of the sphere. What is the net charge in nano coulomb within the sphere’s surface?
Answer:
55.6 nC
Explanation:
The electric field at the surface of a charged sphere has the same expression of the electric field produced by a single point charge located at the centre of the sphere and having the same charge of the sphere, so it is given by
[tex]E=k\frac{Q}{r^2}[/tex]
where
[tex]k=9\cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb's constant
Q is the charge on the sphere
r is the radius of the sphere
In this problem we know
E = 890 N/C is the magnitude of the electric field on the sphere
r = 0.750 m is the radius of the sphere
So by re-arranging the equation we can find the net charge on the sphere:
[tex]Q=\frac{Er^2}{k}=\frac{(890)(0.750)^2}{9\cdot 10^9}=5.56\cdot 10^{-8} C=55.6 nC[/tex]
To find the net charge within the surface of the sphere, we can use Gauss's Law. The net electric flux through a closed surface enclosing a charge is equal to the charge divided by the electric constant, ε₀. Given the electric field and the radius of the sphere, we can calculate the net charge using the equation derived from Gauss's Law.
Explanation:The electric field everywhere on the surface of a thin, spherical shell is radially inward and has a magnitude of 890 N/C. To find the net charge within the surface of the sphere, we can use Gauss's Law. The net electric flux through a closed surface enclosing a charge is equal to the charge divided by the electric constant, ε₀. In this case, the closed surface is the spherical shell and the electric field is constant on the surface, so the net electric flux is equal to the electric field multiplied by the surface area of the sphere. The surface area of the sphere is given by 4πr², where r is the radius of the sphere. Setting the net electric flux equal to the charge divided by ε₀, we can solve for the charge.
Given that the electric field is 890 N/C, the radius of the sphere is 0.750 m, and the electric constant, ε₀, is 8.99 x 10⁹ Nm²/C², we can plug in these values to solve for the charge.
Let Q be the net charge within the sphere's surface:
Net electric flux = Electric field * Surface area
Q / ε₀ = Electric field * Surface area
Q / (8.99 x 10⁹ Nm²/C²) = (890 N/C) * (4π(0.750 m)²)
Q = (890 N/C) * (4π(0.750 m)²) * (8.99 x 10⁹ Nm²/C²)
Solving this equation will give us the net charge within the sphere's surface.
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Water is flowing in a straight horizontal pipe of variable cross section. Where the cross-sectional area of the pipe is 3.70·10-2 m2, the pressure is 6.10·105 Pa and the velocity is 0.260 m/s. In a constricted region where the area is 9.50·10-4 m2, what is the velocity?
Answer:
v = 10.1 m/s
Explanation:
As we know that by the law of conservation of volume the rate of volume flowing through the pipe will remain conserved
so here we have flow rate given as
[tex]Q = Area\times velocity[/tex]
now we have
[tex]A_1 v_1 = A_2 v_2[/tex]
now we have
[tex]A_1 = 3.70 \times 10^{-2} m^2[/tex]
[tex]v_1 = 0.260 m/s[/tex]
[tex]A_2 = 9.50 \times 10^{-4} m^2[/tex]
now from above equation we have
[tex]v_2 = \frac{A_1}{A_2} v_1[/tex]
[tex]v_2 = \frac{3.70\times 10^{-2}}{9.50\times 10^{-4}}(0.260)[/tex]
[tex]v_2 = 10.1 m/s[/tex]
A talented PHY210 student is tuning her car. She increase the engine speed from 660 rpm very (69.1 rad/s) to 4500 rpm (471 rad/s). If the flywheel has a moment of inertia of 0.525 kg-m2 calculate the work done by the engine on the flywheel during this change in angular velocity. Select one: o a. 105 b, 1.43 x 104 J c. 1.44 x 103 O d. 5.70 x 104 J e. 1.14 x 105J
Answer:
d. 5.7 x 10⁴ J
Explanation:
I = moment of inertia of the flywheel = 0.525 kg-m²
w₀ = initial angular speed of the flywheel = 69.1 rad/s
w = final angular speed of the flywheel = 471 rad/s
W = work done by the engine on the flywheel
Work done by the engine is given as
W = (0.5) I (w² - w₀²)
Inserting the values
W = (0.5) (0.525) (471² - 69.1²)
W = 5.7 x 10⁴ J
A long wire is known to have a radius greater than 10.0 mm and less 20 mm, carry a current uniformly distributed over its cross section. If the magnitude of the magnetic field is 3 mT at a point 6.0 mm from the axis of the wire and 1.50 mT at a point 20 mm from the axis, what is the radius of the wire?
Answer:
Radius = 7.75 mm
Explanation:
magnetic field inside a point in the wire is given as
[tex]B = \frac{\mu_0 i r}{2\pi R^2}[/tex]
now we have
[tex]3mT = \frac{\mu_0 i (6mm)}{2\pi R^2}[/tex]
now outside wire magnetic field is given as
[tex]B = \frac{\mu_0 i}{2\pi r}[/tex]
now we have
[tex]1.50 mT = \frac{\mu_0 i}{2\pi(20 mm)}[/tex]
now divide above two equations
[tex]2 = \frac{6 mm\times 20 mm}{R^2}[/tex]
[tex]R = 7.75 mm[/tex]
Heated lithium atoms emit photons of light with an energy of 2.961 × 10−19 J. Calculate the frequency and wavelength of one of these photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?
Answer:
4.5 x 10¹⁴ Hz
666.7 nm
1.8 x 10⁵ J
The color of the emitted light is red
Explanation:
E = energy of photons of light = 2.961 x 10⁻¹⁹ J
f = frequency of the photon
Energy of photons is given as
E = h f
2.961 x 10⁻¹⁹ = (6.63 x 10⁻³⁴) f
f = 4.5 x 10¹⁴ Hz
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of photon
Using the equation
c = f λ
3 x 10⁸ = (4.5 x 10¹⁴) λ
λ = 0.6667 x 10⁻⁶ m
λ = 666.7 x 10⁻⁹ m
λ = 666.7 nm
n = number of photons in 1 mole = 6.023 x 10²³
U = energy of 1 mole of photons
Energy of 1 mole of photons is given as
U = n E
U = (6.023 x 10²³) (2.961 x 10⁻¹⁹)
U = 1.8 x 10⁵ J
The color of the emitted light is red
The frequency and wavelength of a photon with energy of 2.961 × 10-19 J are approximately 4.468 x 10¹4 s¯¹ and 656.3 nm, respectively. The energy of one mole of these photons is approximately 1.823 x 10⁵ J mol-¹. The color of the light emitted by these photons is red.
Explanation:The energy of a photon of light can be calculated into both a frequency and wavelength using the Planck-Einstein relation E = hf, and the speed of light relation c = λf, where h is Planck's constant, c is the speed of light, and λ is wavelength.
Let's first find the frequency. Given that the energy E is 2.961 × 10-19 J and the value of Planck's constant h is 6.63 × 10-³4 J·s (approximately), you can solve for the frequency f = E/h, which results in approximately 4.468 × 10¹4 s¯¹.
To find the wavelength, you can use the light speed equation c = fλ. Given the speed of light c is approximately 3.00 × 10⁸ m/s and using the frequency calculated above, solve for λ = c/f, which results in approximately 656.3 nm.
To find the energy of one mole of these photons, use Avogadro's number (6.022 x 10²³ photons/mole) and the provided energy of a single photon. The total energy is then calculated as E = (2.961 × 10-19 J/photon) x (6.022 x 10²³ /mole) = 1.823 × 10⁵ J mol-¹.
Lastly, the color of the photon can be inferred from its wavelength. As the wavelength is 656.3 nm, this falls within the range of the visible light spectrum for the color red.
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