Answer:
Accuracy
Explanation:
Percent error is the ratio of the difference of the measured and actual value to the actual value multiplied by 100.
It gives the percent deviation of the value obtained from the actual value.
Accuracy is the measure of how close the readings are to the actual value or set standard and can be improved by increase the no. of readings in an experiment.
Precision is the measure of the closeness of the obtained values to one another.
Thus accuracy of the reading can be sensed by the percent error.
A charge Q= 2 C is distributed uniformly through out a bar of length L=2.5 m. The bar is placed horizontally in free space. A second charge q = 10-°C is placed along the line of the bar a distance d= 2m measured from the right end of the bar. What force is exerted on charge q by the charged bar?
Answer:
The force exerted by the charge q on the rod is [tex]5\times 10^{10}\ \rm N[/tex]
Explanation:
Given:
Charge on the rod=Q=2 CLength of the rod=L=2.5 mmagnitude of the point charge q=-10 CThe distance of the point charge from the right end of the rod d=2 mWe have to find the force exerted by the charge on the rod. this will be equal to the force exerted by the rod on the charge according to coulombs law.
The Electric field due the the rod at the location of the charge is given by
[tex]E=\dfrac{kQL}{d(d+L}\\E=\dfrac{9\times10^9\times2\times2.5}{2\times4.5}\\\\E=5\times 10^9\ \rm N/C\\[/tex]
Force between them is given by F
[tex]F=qE\\\\=10\times5\times10^9\\=5\times10^{10}\ \rm N[/tex]
A 243 mL cup of whole milk contains about 45 mg of cholesterol. Express the cholesterol concentration of the milk in kilograms per cubic meter (kg/m^3).
Answer: [tex]0.1851 \frac{kg}{m^{3}}[/tex]
Explanation:
We have this concentration in units of [tex]mg/ml[/tex]:
[tex]\frac{45 mg}{243 ml}[/tex]
And we need to express it in [tex]kg/m^{3}[/tex], knowing:
[tex]1 ml= 1 cm^{3}[/tex]
[tex]1 m^{3}= (100 cm) ^{3}[/tex]
[tex]1g = 1000 mg[/tex]
[tex]1 kg = 1000 g[/tex]
Hence:
[tex]\frac{45 mg}{243 cm^{3}} \frac{1 g}{1000 mg} \frac{1 kg}{1000 g} \frac{(100 cm) ^{3}}{1 m^{3}}=0.1851 \frac{kg}{m^{3}}[/tex]
The cholesterol concentration of the milk in kilograms per cubic meter is 0.1851 kg/m³
The calculation is done as follows;
1 mg= 10⁻⁶ kg
1ml= 10⁻⁶ m³
= 45 ÷243
= 0.1851 kg/m³
Hence the cholesterol concentration of the milk is 0.1851 kg/m³
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An emission spectrum shows : a. Brightly colored lines
b. A rainbow of colors merging into each other
c. Only red and orange bands
d. Only green and blue bands
Answer: a)
Explanation: The phenomenon of emission is related to electronic transitions with the atom so brightly emission lines can represent the most important electronics transitions. They cover the whole spectrum from UV to IR.
Material A has a specific heat of 100 J/kg-K and Material B has a specific heat of 200 J/kg-K. If the two materials start at the same temperature and have heat energy added at the same constant rate, which would you expect to reach a higher temperature first? Material A
Material B
Both the same
Cannot determine from the information given None of the above
Answer:
material A
Explanation:
Specific heat of material A = 100 J/kg - K
Specific heat of material B = 200 J/kg - K
The specific heat of a material is defined as the amount of heat required to raise the temperature of substance of mass 1 kg by 1 °C.
So, material A requires 100 J of heat to raise the temperature of mass 1 kg by 1°C.
So, material B requires 200 J of heat to raise the temperature of mass 1 kg by 1°C.
So, the temperature of material A is higher is more than the material B when same amount of heat is added.
The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position as x = kamtn, where k is a dimensionless constant. Show by dimensional analysis that this expression is satisfied if m = 1 and n = 2. (Submit a file with a maximum size of 1 MB.)
Answer:
Explanation:
[tex]x = k\times a^m\times t^n[/tex]
k is constant , it is dimension is zero. Using dimensional unit , we cal write the relation as follows
L = [tex](LT^{-2})^m(T)^n[/tex]
= [tex]L^mT^{-2m+n}[/tex]
Equating power of like items
m=1
-2m+n = 0
n = 2
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the center of the electrode?
Express your answer to two significant figures and include the appropriate units.
Answer:
The electric field strength is [tex]4.5\times 10^{4} N/C[/tex]
Solution:
As per the question:
Area of the electrode, [tex]A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}[/tex]
Charge, q = 50 nC = [tex]50\times 10^{- 9} C[/etx]
Distance, x = 2 mm = [tex]2\times 10^{- 3} m[/tex]
Now,
To calculate the electric field strength, we first calculate the surface charge density which is given by:
[tex]\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}[/tex]
Now, the electric field strength of the electrode is:
[tex]\vec{E} = \frac{\sigma}{2\epsilon_{o}}[/tex]
where
[tex]\epsilon_{o} = 8.85\times 10^{- 12} F/m[/tex]
[tex]\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}[/tex]
[tex]\vec{E} = 4.5\times 10^{4} N/C[/tex]
A flying hawk makes a noise of frequency 50 Hz that spreads evenly in every direction. The intensity at the ground, 60 m below, is measured at 70 dB. What is the power radiated by the hawk?
Answer:
The power radiated by the hawk is 0.452 Watt.
Explanation:
Given that,
Frequency = 50 Hz
Distance r=60 m
Level = 70 dB
We need to calculate the intensity
Using formula of intensity
[tex]dB=10 log(\dfrac{I}{I_{0}})[/tex]
Put the value into the formula
[tex]70=10 log(\dfrac{I}{10^{-12}})[/tex]
[tex]I=10^{7}\times10^{-12}[/tex]
[tex]I = 1\times10^{-5}\ W/m^2[/tex]
We need to calculate the power radiated by the hawk
Using formula of power
[tex]P = I\times 4\pi r^2[/tex]
Put the value into the formula
[tex]P=1\times10^{-5}\times4\pi\times(60)^2[/tex]
[tex]P=0.452\ W[/tex]
Hence, The power radiated by the hawk is 0.452 Watt.
Low voltage power system: You want to "supercharge"1 your brand new Tesla Model 3 at a rate of 100 kW at home. For safety reasons, you chose a 5 V direct current (DC) voltage source. a) Calculate the required current. b) Assuming a cylindrical conductor with a current density of 10 A/mm2 , what diameter conductor would you need? c) Assuming copper with a resistivity of rhoCu = 1.67×10−8 Ωm and a cable length
Answer:
(a) 20 kA (B) [tex]diameter=5.0035\times 10^{-2}m[/tex] (c) length = 29.94 m
Explanation:
We have given power P = 100 KW
Voltage V = 5 V
Current density [tex]10A/m^2=10\times 10^{6}A/m^2[/tex]
(A) We have to calculate current
We know that power P = voltage × current
So [tex]100\times 1000=5\times i[/tex]
[tex]i=20kA[/tex]
(b) We know that current density is given by
Current density [tex]=\frac{current}{area}[/tex]
So [tex]10\times 10^{6}A/m^2=\frac{20\times 10^3}{area}[/tex]
[tex]area=2\times 10^{-3}m^2[/tex]
We know that [tex]a=\pi r^2[/tex]
[tex]2\times 10^{-3}=3.14\times r^2[/tex]
[tex]r^2=6.34\times 10^{-4}[/tex]
[tex]r=2.517\times 10^{-2}m[/tex]
[tex]d=2 r=2\times 2.517\times 10^{-2}=5.035\times 10^{-2}m[/tex]
(c) We have to find the length of the cable
We know that resistance [tex]R=\frac{V}{I}=\frac{5}{20\times 10^3}=0.25mohm[/tex]
Resistance is given by [tex]R=\rho \frac{l}{A}[/tex]
So [tex]0.25\times 10^{-3}=1.67\times 10^{-8}\frac{l}{2\times 10^{-3}}[/tex]
l = 29.94 m
An electrostatic paint sprayer has a 0.8 m diameter metal sphere at a potential of 39 kV that repels paint droplets onto a grounded object. What charge is on the sphere? Give answer in terms of 10^-7 C.
Answer:
The charge on the sphere is [tex]1.735\times 10^{- 7} C[/tex]
Solution:
The electric potential on the surface of a sphere of radius 'b' and charge 'Q' is given by:
[tex]V_{sphere} = \frac{Q}{4\pi\epsilon_{o}b}[/tex]
According to the question:
Diameter, b = 0.8 m
Potential of sphere, [tex]V_{sphere} = 39 kV = 39000 V[/tex]
[tex]{4\pi\epsilon_{o}\frac{0.8}{2}\times 39000 = Q[/tex]
Q = [tex]1.735\times 10^{- 7} C[/tex]
The nucleus of a 125 Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm in diameter. It has 54 protons and charge q=+54e (1 fm = 1 femtometer = 1× 10 −15 m .) Hint: Treat the spherical nucleus as a point charge. Part A What is the electric force on a proton 3.0 fm from the surface of the nucleus? Express your answer in newtons. F nucleusonproton F n u c l e u s o n p r o t o n = nothing N SubmitRequest Answer Part B What is the proton's acceleration? Express your answer in meters per second squared. a proton a p r o t o n = nothing m/ s 2 SubmitRequest Answer Provide Feedback Next
Answer:
(a): 345.6 N.
(b): [tex]\rm 2.069\times 10^{29}\ m/s^2.[/tex]
Explanation:
Given:
Charge on the 125 Xe nucleus, [tex]\rm q = +54e.[/tex]Mass of the 125 Xe nucleus, [tex]\rm m = 125\ u.[/tex]Diameter of the 125 Xe nucleus, [tex]\rm d=6.0\ fm = 6.0\times 10^{-15}\ m.[/tex]Distance of the proton from the surface of the nucleus, [tex]\rm a=3.0\ fm = 3.0\times 10^{-15}\ m.[/tex]Part A:According to Coulomb's law, the electric field due to a charged sphere of charge Q at a point r distance away from its center is given by
[tex]\rm E=\dfrac{kQ}{r^2}.[/tex]
where, k is the Coulomb's constant, having value = [tex]9\times 10^9\ \rm Nm^2/C^2.[/tex]
Therefore, the electric field due to the 125 Xe nucleus at the proton is given by
[tex]\rm E=\dfrac{kq}{r^2}=\dfrac{(9\times 10^9)\times (+54 e)}{r^2}[/tex]
Here,
e is the elementary charge, having value = [tex]\rm 1.6\times 10^{-19}\ C.[/tex]
r is the distance of the proton from the center of the nucleus = [tex]\rm a + \dfrac d2 = 3.0+\dfrac{6.0}2=6.0\ fm = 6.0\times 10^{-15}\ m.[/tex]
Using these values,
[tex]\rm E=\dfrac{(9\times 10^9)\times (+54 \times 1.6\times 10^{-19})}{(6.0\times 10^{-15})^2}=2.16\times 10^{21}\ N/C.[/tex]
Now, the electric force on a charge q due to an electric field is given as
[tex]\rm F=qE[/tex]
For the proton, [tex]\rm q = e =1.6\times 10^{-19}\ C.[/tex]
Thus, the electric force on the proton is given by
[tex]\rm F = 1.6\times 10^{-19}\times 2.16\times 10^{21}=345.6\ N.[/tex]
Part B:According to Newton's second law,
[tex]\rm F=ma[/tex]
where, a is the acceleration.
The mass of the proton is [tex]\rm m_p=1.67\times 10^{-27}\ kg.[/tex]
Therefore, the proton's acceleration is given by
[tex]\rm a = \dfrac{F}{m_p}=\dfrac{345.6}{1.67\times 10^{-27}}=2.069\times 10^{29}\ m/s^2.[/tex]
(a) The electric force of the proton is 2.56 x 10⁻²⁹ N.
(b) The acceleration of the proton is 0.0153 m/s².
Electric force of the protonThe electric force of the proton is calculated using Coulomb's law of electrostatic force.
F = kq²/r²
F = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹)/(3)²
F = 2.56 x 10⁻²⁹ N
Acceleration of the protonThe acceleration of the proton is determined by applying Newton's second law of motion;
F = ma
a = F/m
a = (2.56 x 10⁻²⁹ )/(1.67 x 10⁻²⁷)
a = 0.0153 m/s²
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An object is undergoing uniform circular motion. Which of the following changes would increase the period of the motion? I. Increase the radius of the circular motion. II. Decrease the radius of the circular motion. III. Increase the speed of the object. IV. Decrease the speed of the object.
Answer:
See it in the pic
Explanation:
See it in the pic
Final answer:
Increasing the radius of the circular motion or decreasing the speed of the object would both increase the period of uniform circular motion. This is due to the direct proportionality of period to radius and inverse proportionality to speed.
Explanation:
To determine which changes would increase the period of an object's uniform circular motion, let's recall the relationship between period (T), speed (v), and radius (r) of the circle. According to the formula for calculating the centripetal acceleration of an object in uniform circular motion (a = v²/r or a = 4π²r/T²), we can deduce that the period T is directly proportional to the radius and inversely proportional to the speed. Therefore, increasing the radius (Option I) would increase the period since T is proportional to the square root of the radius when speed is constant. Decreasing the speed of the object (Option IV) would also increase the period because the period is inversely proportional to speed.
The correct answer to the question of which changes would increase the period of motion in uniform circular motion are: I. Increase the radius of the circular motion and IV. Decrease the speed of the object. Options II and III would decrease the period, not increase it.
A sparrow is flying around in a circle at a constant speed and height. there is air resistance. In what direction is the net force of the sparrow on the air as the sparrow flies. 1. outward and downward 2.Inward and Downward 3.Downward and Backward 4. outward, downward and backward
Final answer:
The sparrow exerts a net force on the air in the downward and backward directions as per Newton's Third Law, as it pushes against the air to get lift and move forward.
Explanation:
The sparrow flying around in a circle at a constant speed and height with air resistance is subject to several forces. However, when we consider the force exerted by the sparrow on the air, we need to apply Newton's Third Law of motion which states that for every action, there is an equal and opposite reaction.
Therefore, as the sparrow's wings push the air downward and backward to get lift and forward movement, the air will exert an equal and opposite force on the sparrow. The sparrow's net force on the air would be downward and backward because the sparrow pushes the air in these directions to maintain flight.
You and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The first 60 miles of the trip are driven at 55 mi/h but then you get stuck in a standstill traffic jam for 20 minutes. In order to make up time, you then proceed at 75 mi/h for the rest of the trip. What is the magnitude of your average velocity for the whole trip?
The magnitude of your average velocity for the whole trip is 102.19 miles per hour.
First, let's calculate the time taken for each segment of the trip:
1. For the first 60 miles at 55 mi/h:
Time = Distance / Speed
= 60 miles / 55 mi/h
= 1.0909 hours
2. During the traffic jam, you're not moving, so the time is 20 minutes, which needs to be converted to hours:
Time = 20 minutes / 60 minutes/hour
= 1/3 hours
3. For the last 40 miles at 75 mi/h:
Time = 40 miles / 75 mi/h
= 0.5333 hours
Now, calculate the total time for the trip:
Total Time = 1.0909 hours + 1/3 hours + 0.5333 hours
= 1.9572 hours
Since you traveled 100 miles to your aunt's house and then returned,
the total displacement = 2 * 100 miles = 200 miles.
Now, calculate the average velocity:
Average Velocity = Total Displacement / Total Time
= 200 miles / 1.9572 hours
= 102.19 mi/h
So, the average velocity is 102.19 miles per hour.
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If you and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The magnitude of your average velocity for the whole trip is 51.04 miles per hour.
What is the magnitude?To determine the magnitude of your average velocity for the whole trip, we can use the formula for average velocity:
Average Velocity = Total Displacement / Total Time
Time for the first 60 miles at 55 mi/h:
Time = Distance / Speed
= 60 miles / 55 mi/h
= 1.0909 hours
Time spent in the traffic jam:
20 minutes = 20 / 60
= 1/3 hours
Time for the remaining 40 miles at 75 mi/h:
Time = Distance / Speed
= 40 miles / 75 mi/h
= 0.5333 hours
Total Time = 1.0909 + 1/3 + 0.5333
= 1.9572 hours
Now we can calculate the average velocity:
Average Velocity = Total Displacement / Total Time
Average Velocity = 100 miles / 1.9572 hours
≈ 51.04 mi/h
Therefore the magnitude of your average velocity for the whole trip is 51.04 miles per hour.
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A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped? (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?
Answer:
a) 447.21m
b) -62.99 m/s
c)94.17 m/s
Explanation:
This situation we can divide in 2 parts:
⇒ Vertical : y =-200 m
y =1/2 at²
-200 = 1/2 *(-9.81)*t²
t= 6.388766 s
⇒Horizontal: Vx = Δx/Δt
Δx = 70 * 6.388766 = 447.21 m
b) ⇒ Horizontal
Vx = Δx/Δt ⇒ 70 = 400 /Δt
Δt= 5.7142857 s
⇒ Vertical:
y = v0t + 1/2 at²
-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²
v0= -7 m/s ⇒ it's negative because it goes down.
v= v0 +at
v= -7 + (-9.81) * 5.7142857
v= -62.99 m/s
c) √(70² + 62.99²) = 94.17 m/s
The horizontal distance at which the the package must be dropped and the vertical velocity are; 316.23 m and -22.08 m/s
What is the vertical velocity?
A)To fall 200 m, the time required is given by the formula;
t = √(2H/g)
t= √(200/9.8)
t = 4.518 seconds.
The supplies will travel forward by a distance of;
Δx = 4.518 * 70
Δx = 316.23 m
B) Time in the horizontal direction is;
t = Δx/V
t = 316.23/70
t = 4.52 s
In the vertical direction;
y = ut + ¹/₂at²
-200 = 4.52u + ¹/₂(-9.81)(4.52)²
100.21 - 200 = 4.52u
u = -99.79/4.52
u = -22.08 m/s
C) Speed at which the package lands in the latter case is;
v = √(70² + (-22.08²))
v = 73.4 m/s
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While following a treasure map, you start at an old oak tree. You first walk 825 m directly south, then turn and walk 1.25 km at 30.0° west of north, and finally walk 1.00 km at 32.0° north of east, where you find the treasure: a biography of Isaac Newton! (a) To return to the old oak tree, in what direction should you head and how far will you walk? Use components to solve this problem. (b) To see whether your calculation in part (a) is reason- able, compare it with a graphical solution drawn roughly to scale.
To return to the tree after finding the treasure, you need to calculate the total displacement vector from the tree to the treasure by adding up the vectors of each leg of the journey. The direction and magnitude of the opposite of this vector will provide the direction and distance to get back to the tree. A graphical representation of the vectors will help confirm this.
Explanation:The problem is about determining the direction and distance to return to the original location after walking certain distances at specific angles. This involves the understanding of vectors and how they are applied in real life.
Given the three legs of the journey:
1st leg: 825 m directly south, which can be represented as a vector with a magnitude of 825 m and a direction of 180 degrees.2nd leg: 1.25 km at a 30-degree angle west of north, which can be represented as a vector with north and west components. Using trigonometry, we get a north-component of 1.25 km * cos(30), and a west-component of 1.25 km * sin(30).3rd leg: 1.00 km at a 32-degree angle north of east, which can be represented with east and north components. Calculating similarly, we get the east-component as 1.00 km * cos(32), and the north-component as 1.00 km * sin(32).
Summing up these components for all the legs, we get the total displacement vector from the tree to the treasure. To get back to the tree, you would need to walk in the opposite direction of this vector. The magnitude of this vector gives the total distance to be covered.
Comparing this calculated result with a graphical solution will help validate the projected direction and distance.
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A mass of 222g of Helium gas at an initial temperature of 54.43°C and at an initial absolute pressure of 4.45 atm undergoes an isothermal expansion until its volume increases by a factor of 2.25. (a) What is the final pressure? (Pa)
(b) How much work is done on the gas?
(c) How much heat does the gas absorb?
(d) What is the change in the total internal energy of the gas?
Explanation:
Given that,
Mass of gas = 222 g
Temperature = 54.43°
Pressure = 4.45 atm
Final volume = 2.25 initial volume
For isothermal expansion
(a). We need to calculate the pressure
Using relation of pressure and volume
[tex]P_{i}V_{i}=P_{f}V_{f}[/tex]
[tex]P_{f}=\dfrac{P_{1}V_{i}}{V_{f}}[/tex]
Put the value into the formula
[tex]P_{f}=\dfrac{4.45\timesV_{i}}{2.25V_{i}}[/tex]
[tex]P_{f}=\dfrac{4.45}{2.25}[/tex]
[tex]P_{f}=1.97\ atm[/tex]
[tex]P_{f}=199610.3\ pa[/tex]
[tex]P_{f}=1.99\times10^{5}\ Pa[/tex]
The pressure is [tex]1.99\times10^{5}\ Pa[/tex]
(b). We need to calculate the work done
1 mole of Hg is 200.59 gram
222 g of Hg is
[tex]n =\dfrac{200}{200.59}[/tex]
[tex]n =0.997[/tex]
Using formula of work done
[tex]W=nRTln(\dfrac{V_{f}}{V_{i}})[/tex]
Put the value into the formula
[tex]W=0.997\times8.314\times(54.43+273)ln2.25[/tex]
[tex]W=2200.9\ J[/tex]
[tex]W=2.2009\ kJ[/tex]
The work done is 2.20 kJ.
(c). We need to calculate the gas absorb
Heat absorbed by the gas is the work done
[tex]Q=W[/tex]
[tex]Q=2.20 kJ[/tex]
The absorb heat is 2.20 kJ.
(d). We need to calculate the change in the total internal energy of the gas
Change in internal energy in an isothermal process is zero.
So, [tex]U=0 [/tex]
Hence, This is the required solution.
A student applies a horizontal force of 100 N to a 25 kg box causing the block to accelerate at a rate of 1.5 m/s^2. Which of the following is the best approximation of the magnitude of the friction force on the box?
Answer:
By Newton's law Friction= -62.5N
Explanation:
Applying second law of Newton (which is the the sum of the forces is equal to the mass of the object who feels the forces times the acceleration)
Push+Friction= [tex]m_{box}a_{box}[/tex]
[tex]100+Friction=25*1.5=37.5\\Friction=37.5-100=-62.5 N[/tex]
Which indicates that the friction is opposed to the push and it is really appreciable, that is to say, there is a lot of friction between the box and the floor.
Final answer:
The best approximation of the magnitude of the friction force on the box, when a 25 kg box is accelerated by a 100 N force at 1.5 m/s^2, is 62.5 N.
Explanation:
The student's question involves understanding Newton's second law of motion and the concept of friction. Since a horizontal force of 100 N is applied to a 25 kg box, causing it to accelerate at a rate of 1.5 m/s2, we can calculate the net force using Newton's second law (F = ma). The net force is the difference between the applied force (100 N) and the friction force which we are trying to find.
First, calculate the net force:
Fnet = m × a = 25 kg × 1.5 m/s2 = 37.5 N
Next, calculate the friction force:
Ffriction = Fapplied - Fnet = 100 N - 37.5 N = 62.5 N
Therefore, the best approximation of the magnitude of the friction force on the box is 62.5 N.
Air is compressed in a cylinder such that the volume changes from 100.0 to 10.0 in^3. The initial pressure is 50.0 psia and the temperature is held at a constant 100.0 °F. How much work is required to change the volume? (Hint you will need to identify what psia means)
Answer:
5953.42 J
Explanation:
Given:
Initial volume, [tex]V_i[/tex]= 100 in³
Final Volume, [tex]V_f[/tex] = 10 in³
Initial pressure = 50 psia
Temperature = 100° F = 310.93 K
For isothermal reversible process, work done is given as:
Work done = [tex]-2.303RTlog_{10}\frac{V_f}{V_i}[/tex]
Where,
R is the ideal gas constant = 8.314 J/mol.K
or
Work done = [tex]-2.303\times8.314\times310.93log_{10}\frac{10}{100}[/tex]
or
Work done = 5953.42 J
A rabbit is traveling at a steady 6 m/s and (without knowing it) passes a wolf. The wolf takes off at the instant the rabbit passes it, accelerating at a steady rate of 2.0 m/s^2. How much time elapses before the wolf is moving as fast as the rabbit? How far is the wolf from the rabbit when it reaches this speed?
Answer:
It takes the wolf 3 s to move as fast as the rabbit.
The wolf is 9 m from the rabbit when it reaches a speed of 6 m/s
Explanation:
The equations for the position and velocity of objects moving in a straight line are as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the object at time t
x0 = initial position of the object
t = time
v0 = initial speed
a = acceleration
v = velocity of the object at time t
For the first question, let´s use the equation of velocity of the wolf to find at which time its velocity is the same as the velocity of the rabbit ( 6 m/s):
v = v0 + a · t (v0 = 0 because the wolf starts at rest)
6 m/s = 0 + 2.0 m/s² · t
t = 3 s
Now, with this calculated time, let´s obtain the position of the wolf:
x = x0 + v0 · t + 1/2 · a · t² (Placing the center of the frame of reference at the point when the wolf starts running makes x0 = 0)
x = 1/2 · a · t²
x = 1/2 · 2.0 m/s² · (3 s)²
x = 9 m
Now, let´s calculate the position of the rabbit. Notice that a = 0. Then:
x = x0 + v · t x0 = 0
x = v · t
x = 6 m/s · 3 s = 18 m
The wolf is (18 m - 9 m) 9 m from the rabbit when it reaches a speed of 6 m/s
What is necessary to designate a position? A. a reference point B. a direction C. fundamental units D. motion E. all of these
Answer:
E. all of these
Explanation:
The designation of a point in space all the points that necessary
- reference point
- a direction
- fundamental units
- a direction
- motion
all are necessary to designate a point in space. Hence option E is correct.
For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.
To designate a position, the necessary elements are A. A reference point, B. A direction, C. Fundamental units, D. Motion. Therefore option E is correct.
A. A reference point: A reference point is needed to establish a starting point or a fixed location from which the position is measured. It serves as a point of comparison and allows for consistent measurements.
B. A direction: The direction specifies the orientation or path followed from the reference point to the object or location being described. It provides information on the relative position or displacement of the object.
C. Fundamental units: Fundamental units, such as meters or feet, are used to quantify and measure the distance or displacement between the reference point and the object. These units provide a standardized way to express the position.
D. Motion: While motion itself is not necessary to designate a position, it can be relevant in certain cases when describing the position of a moving object. In such situations, the position is defined with respect to both the reference point and the object's movement.
Therefore, the correct answer is E. All of these elements - a reference point, a direction, fundamental units, and in some cases, motion - are necessary to designate a position accurately.
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If two particles have equal kinetic energies, are their momenta necessarily equal? explain.
No, the momenta of two particles with equal kinetic energies are not necessarily equal.
Explanation:No, the momenta of two particles with equal kinetic energies are not necessarily equal. Momentum is given by the equation:
p = mv
Where p is momentum, m is mass, and v is velocity. Kinetic energy is given by the equation:
K = mv²/2
In order for two particles to have equal kinetic energies, their masses and velocities must satisfy the equation:
mv₁²/2 = mv₂²/2
But this does not necessarily mean that their momenta are equal, as the masses and velocities can still differ.
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A woman on a motorcycle moving uniformly at a rate of 20 m/s passes a truck at rest. At the instant the motorcycle passes the truck, the truck begins to accelerate at the rate of 5 m/s . (A) How long does it take the truck to catch up to the motorcycle? (B) How far has the motorcycle traveled? (C) What is the truck's speed at this point?
Answer:
A) It takes the truck 8 s to catch the motorcycle.
B) The motorcycle has traveled 160 m in that time.
C) The velocity of the truck is 40 m/s at that time.
Explanation:
The equations of the position and velocity of an object moving in a straight line are as follows:
x = x0 +v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
(A) When the the truck catches the motorcycle, both have the same position. Notice that the motorcycle moves at constant speed so that a = 0:
x truck = x motorcycle
x0 +v0 · t + 1/2 · a · t² = x0 + v · t
Placing the origin of the frame of reference at the point where the truck starts, both have an initial position of 0. The initial velocity of the truck is 0. Then:
1/2 · a · t² = v · t
solving for t:
t = 2 v/a
t = 2 · 20 m/s/ 5 m/s²
t = 8 s
It takes the truck 8 s to catch the motorcycle.
(B) Using the equation of the position of the motorcycle, we can calculate the traveled distance in 8 s.
x = v · t
x = 20 m/s · 8 s
x = 160 m
(C) Now, we use the velocity equation at time 8 s.
v = v0 + a · t
v = 0 m/s + 5 m/s² · 8 s
v = 40 m/s
To solve the problem, we equate the displacements of the truck and motorcycle to find when they catch up to each other. The motorcycle travels a total of 160 meters in 8 seconds, while the truck accelerates to a speed of 40 m/s in that time.
Explanation:The subject at hand involves concepts of uniform motion and constant acceleration, relevant to the study of Physics. We have a motorcycle moving at a constant speed of 20 m/s and a truck starting from rest and accelerating at a rate of 5 m/s².
To answer part (A) we use the equation for the displacement of each vehicle. Our goal is to find the time 't' where their displacements are equal. For the motorcycle, since it's moving at a constant speed, the equation for displacement is given by: x = Ut, where U = initial velocity is 20 m/s and x = displacement For the truck, since it is accelerating, the equation for displacement is 1⁄2at², where a = acceleration is 5 m/s².
Setting these displacements equal to each other gives: Ut = 1⁄2at² To solve for time 't', we obtain: t = 2U/a = 2(20 m/s)/(5 m/s²) = 8 seconds.
For part (B), we calculate how far the motorcycle has traveled using its equation for displacement: x = Ut = 20 m/s * 8 seconds = 160 meters.
Finally, for part (C), we find the truck's final speed using the equation v = U + at. Since the truck starts from rest, U = 0, thus: v = at = 5 m/s² * 8 seconds = 40 m/s.
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An aircraft cruises at an altitude of 10,700 meters above sea level. Estimate the atmospheric pressure in bar at cruise altitude. Assume the acceleration of gravity is constant at g = 9.8 m/s". The average specific volume of the air is 1.334 m®/kg.
Answer:
We can calculate the atmospheric pressure using the given formula:
[tex]P = P_{o} - \rho\times g\times h[/tex]
where;
[tex]P_{o} = 1.013 \times 10^{5} pa[/tex]
[tex]\rho = \frac{1}{V}[/tex]
[tex]\rho = \frac{1}{1.334} = 0.75 kg/m^{3}[/tex]
h = 10,700 m
equating the following variables in above equation, we get;
[tex]P = 1.013\times 10^{5} - 0.75 \times 9.8 \times 10700[/tex]
P = 0.227 bar
In a thundercloud there may be electric charges of +44.0 C near the top of the cloud and -44.0 C near the bottom of the cloud. These charges are separated by 1.60 km. What is the electric force on the top charge?
Answer:6.806 MN
Explanation:
Given
Charge on clouds is +44 C and -44 C
They are separated by 1.6 km
and electrostatic Force is given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
[tex]F=\frac{9\times 10^9\times 44\times 44}{1600^2}[/tex]
[tex]F=\frac{17,424\times 10^9}{256\times 10^4}[/tex]
F=6.806 MN
Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.3) 10 units 110 units 50 units 30 units
Answer:
the correct answer is option C which is 50 units.
Explanation:
given,
two vector of magnitude = 30 units and of 70 units
to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}
cos θ value varies from -1 to 1
so, resultant vector
=[tex]\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}[/tex]
a = 30 units and b = 70 units
= [tex]\sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}[/tex]
= 40 units to 100 units
hence, the correct answer is option C which is 50 units.
Describe the role of impurities in dielectrics. Why are sapphire and diamond transparent and how can impurities change this?
Answer:
Sapphire and diamond impurities change with their transparency and role of impurities in dielectrics is discussed below.
Explanation:
Temperature dependence of the dielectrics with different degrees of purity measured in the range of 233-313 K. This was observed that the impurities presence is strongly influences the dielectric constant, dipole moment and melting point.
SAPPHIRE
Sapphire is a very important gem stone. It's color is blue, but the naturally occurring sapphires have purple, yellow,orange and green. This variety in color is due to trace amount of presence of impurities like Iron,titanium, chromium, copper and Magnesium. It is the third hardest materials in the Mohs scale. The transparency of Sapphire is due to the impurities present in it.
DIAMOND
Diamond is the very precious stone. The transparency in the diamond is due to the band structure, that is the band gap is high in the diamonds, so they are transparent in nature. Diamond has many unequaled qualities and is very unique among the other minerals. It has highest refractive index of any natural minerals and is transparent over greatest number of wavelengths due to conduction present in it.
A 7.0- g bullet is fired into a 1.5- kg ballistic pendulum.The
bullet emerges from the block with a speed of 200 m/s, and theblock
rises to a maximum height of 12 cm. Find the initial speed ofthe
bullet.
Final answer:
The problem involves a ballistic pendulum scenario to determine the initial speed of a bullet, requiring an analysis that combines conservation of momentum and energy. However, the twist in this problem is that the bullet exits the block, requiring indirect methods to deduce the initial conditions.
Explanation:
The question involves finding the initial speed of a bullet by analyzing a ballistic pendulum scenario. This is a physics problem that integrates concepts of momentum and energy conservation. The solution requires two steps: first, finding the velocity of the bullet-block system right after the bullet exits, and second, using energy conservation to link this velocity to the height reached by the block.
Use the conservation of momentum to find the system's velocity right after the bullet exits the block. The exiting speed of the bullet and the mass details are essential for this step. However, an important observation is that the information given does not allow for the direct application of momentum conservation in the traditional sense, since the bullet exits the block, unlike the typical scenario where it remains lodged in.
Next, use the conservation of energy principle to relate the kinetic energy of the block right after the bullet exits to the potential energy at the maximum height. The height given allows calculation of the velocity of the block right after the bullet exits, which indirectly informs the system's dynamics at impact.
This problem has a twist as the bullet does not remain in the block, which complicates the direct application of the conservation of momentum to find the initial speed of the bullet. Instead, it involves a more nuanced approach using the given exit velocity of the bullet and the rise of the block to infer the initial conditions.
The initial speed of the bullet is approximately 23.68 m/s.
To solve this problem, we can use the principle of conservation of momentum and conservation of energy.
1. Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
[tex]\[m_b v_b = (m_b + m_p)v'\][/tex]
Where:
[tex]\(m_b = 7.0 \, \text{g} = 0.007 \, \text{kg}\)[/tex] (mass of the bullet)
\(v_b\) = initial speed of the bullet (which we need to find)
[tex]\(m_p = 1.5 \, \text{kg}\)[/tex] (mass of the ballistic pendulum)
\(v'\) = final velocity of the bullet and pendulum system
2. Conservation of energy:
The initial kinetic energy of the bullet is equal to the sum of the final kinetic energy of the bullet and the kinetic energy of the pendulum, plus the gravitational potential energy gained by the pendulum.
[tex]\[ \frac{1}{2} m_b v_b^2 = \frac{1}{2} (m_b + m_p) v'^2 + m_p g h\][/tex]
Where:
[tex]\(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)[/tex]
[tex]\(h = 12 \, \text{cm} = 0.12 \, \text{m}\) (maximum height reached by the pendulum)[/tex]
Let's solve these equations step by step:
Step 1: Conservation of momentum:
[tex]\[0.007 \, \text{kg} \times v_b = (0.007 \, \text{kg} + 1.5 \, \text{kg}) \times v'\][/tex]
[tex]\[0.007 \, \text{kg} \times v_b = 1.507 \, \text{kg} \times v'\][/tex]
[tex]\[v_b = \frac{1.507}{0.007} \times v' \][/tex]
[tex]\[v_b = 215.28 \times v'\][/tex] (Equation 1)
Step 2: Conservation of energy:
[tex]\[ \frac{1}{2} \times 0.007 \, \text{kg} \times v_b^2 = \frac{1}{2} \times (0.007 \, \text{kg} + 1.5 \, \text{kg}) \times v'^2 + 1.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.12 \, \text{m}\][/tex]
[tex]\[0.5 \times 0.007 \, \text{kg} \times v_b^2 = 0.5 \times 1.507 \, \text{kg} \times v'^2 + 1.47 \, \text{J}\][/tex]
Now, substitute [tex]\(v_b = 215.28 \times v'\)[/tex] from Equation 1 into the above equation:
[tex]\[0.5 \times 0.007 \, \text{kg} \times (215.28 \times v')^2 = 0.5 \times 1.507 \, \text{kg} \times v'^2 + 1.47 \, \text{J}\][/tex]
Simplify and solve for \(v'\):
[tex]\[0.5 \times 0.007 \times (215.28)^2 \times v'^2 = 0.5 \times 1.507 \times v'^2 + 1.47\][/tex]
[tex]\[ 160.49 \times v'^2 = 0.7535 \times v'^2 + 1.47\][/tex]
[tex]\[159.7365 \times v'^2 = 1.47\][/tex]
[tex]\[v'^2 = \frac{1.47}{159.7365}\][/tex]
[tex]\[v' = \sqrt{\frac{1.47}{159.7365}}\][/tex]
[tex]\[v' \approx 0.11 \, \text{m/s}\][/tex]
Finally, we can use this value of \(v'\) to find \(v_b\) using Equation 1:
[tex]\[v_b = 215.28 \times 0.11\][/tex]
[tex]\[v_b \approx 23.68 \, \text{m/s}\][/tex]
So, the initial speed of the bullet is approximately [tex]\(23.68 \, \text{m/s}\).[/tex]
Walk 42 miles due north, deviate 78 degrees to east, and walk 65miles. What is the displacement? ( magnitude and direction with respect to North). A) Show work through calculations for predictions
Answer:84.405m,[tex]\theta =48.876^{\circ}[/tex]
Explanation:
Given
Person walk 42 miles due to north so its position vector is
[tex]r_1=42\hat{j}[/tex]
Now he deviates [tex]78^{\circ}[/tex] to east and walk 65 miles
so its new position vector
[tex]r_2=42\hat{j}+65cos78\hat{j}+65sin78\hat{i}[/tex]
[tex]r_2=65sin78\hat{i}+\left ( 42+65cos78\right )\hat{j}[/tex]
So magnitude of acceleration is
[tex]|r_2|=\sqrt{\left ( 65sin78\right )^2+\left ( 42+65cos78\right )^2}[/tex]
[tex]|r_2|=\sqrt{63.58^2+55.514^2}[/tex]
[tex]|r_2|=84.405 m[/tex]
for direction
[tex]tan\theta =\frac{42+65cos78}{65sin78}[/tex]
[tex]tan\theta =0.8731 [/tex]
[tex]\theta =41.124^{\circ}with\ respect\ to\ east[/tex]
[tex]\theta =48.876^{\circ}with\ respect\ to\ North[/tex]
A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x = 0.60 m , y = 0. A. Calculate the magnitude of the net electric field at the origin due to these two point charges.
B. Calculate the direction of the net electric field at the origin due to these two point charges.
Part A: The magnitude of the electric field at the origin is approximately [tex]57.89 \text{ N/C}[/tex].
Part B: The direction of the net electric field at the origin is approximately [tex]249.44°[/tex] relative to the negative x-axis.
To solve this problem, we need to calculate the net electric field at the origin (0, 0) due to the two-point charges q₁ and q₂. Here’s a step-by-step breakdown of the process:
Given Information:
Charge q₁ = -4.00 nC located at (0.600 m, 0.800 m)
Charge q₂ = +6.00 nC located at (0.600 m, 0.0 m)
Part A: Calculate the Magnitude of the Electric Field at the Origin
Calculate the Distance from Each Charge to the Origin:
For q₁:
[tex]r_1 = \sqrt{(0 - 0.600)^2 + (0 - 0.800)^2} \\= \sqrt{0.36 + 0.64} \\= \sqrt{1.00} \\= 1.00 \text{ m}[/tex]
For q₂:
[tex]r_2 = \sqrt{(0 - 0.600)^2 + (0 - 0.0)^2} \\= \sqrt{0.36} \\= 0.600 \text{ m}[/tex]
Calculate the Electric Field due to Each Charge:
The formula for the electric field due to a point charge is:
[tex]E = \frac{k |q|}{r^2}[/tex]
Where [tex]k \approx 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2[/tex]
For q₁:
[tex]E_1 = \frac{8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times 4.00 \times 10^{-9} \text{ C}}{(1.00)^2} \\= 35.96 \text{ N/C}[/tex]
Direction of E₁ is towards q₁ (since it is negative) and therefore directed along the vector from (0.600, 0.800) to the origin. The components can be calculated as:
[tex]E_{1x} = -E_1 \cdot \frac{0.600}{1.00} = -35.96 \cdot 0.600 \\= -21.576 \text{ N/C}[/tex]
[tex]E_{1y} = -E_1 \cdot \frac{0.800}{1.00} \\= -35.96 \cdot 0.800 \\= -28.768 \text{ N/C}[/tex]
For q₂:
[tex]E_2 = \frac{8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times 6.00 \times 10^{-9} \text{ C}}{(0.600)^2} \\= 24.99 \text{ N/C}[/tex]
Direction of E₂ is away from the charge because it is positive, directed along the negative y-axis:
[tex]E_{2x} = 0[/tex]
[tex]E_{2y} = -E_2 = -24.99 \text{ N/C}[/tex]
Calculate the Net Electric Field at the Origin:
Combine the x and y components of the electric fields:
[tex]E_{net,x} = E_{1x} + E_{2x} \\= -21.576 + 0 \\= -21.576 \text{ N/C}[/tex]
[tex]E_{net,y} = E_{1y} + E_{2y} \\= -28.768 - 24.99 \\= -53.758 \text{ N/C}[/tex]
The total electric field magnitude is:
[tex]E_{net} = \sqrt{E_{net,x}^2 + E_{net,y}^2}[/tex]
[tex]E_{net} = \sqrt{(-21.576)^2 + (-53.758)^2} \\= \sqrt{465.33 + 2884.46} \\= \sqrt{3349.79} \\= 57.89 \text{ N/C}[/tex]
Final Answer for Part A: The magnitude of the electric field at the origin is approximately [tex]57.89 \text{ N/C}[/tex].
Part B: Calculate the Direction of the Net Electric Field
Calculate the Angle of the Electric Field:
Final Answer for Part B: The direction of the net electric field at the origin is approximately [tex]249.44°[/tex] relative to the negative x-axis.
Complete Question:
A point charge q1 = −4.00nC is at the point x = 0.600 meters, y = 0.800 meters, and a second point charge q2 = +6.00nC is at the point x = 0.600 meters, y = 0.
Part A
Calculate the magnitude E of the net electric field at the origin due to these two-point charges. (Express your answer in newtons per coulomb to three significant figures.)
Part B
What is the direction, relative to the negative x-axis, of the net electric field at the origin due to these two-point charges. (Express your answer in degrees to three significant figures).
Two point charges are brought closer together, increasing the force between them by a factor of 22. By what factor did their separation decrease? Hint: If the force increases, distance between charges must decrease. Force is indirectly proportional to the distance squared.
Answer:0.21
Explanation:
Given
Two point charges are brought closer together, increasing the force by a factor of 22
Let the original force be
[tex]F=\frac{kq_1q_2}{r^2}---1[/tex]
where [tex]q_1,q_2[/tex] are charges and r is the distance between them
new force [tex]F'=\frac{kq_1q_2}{r'^2}----2[/tex]
divide 1 & 2
[tex]\frac{F'}{F}=\frac{\frac{kq_1q_2}{r'^2}}{\frac{kq_1q_2}{r^2}}[/tex]
[tex]22=\frac{r^2}{r'^2}[/tex]
[tex]r'=\frac{r}{\sqrt{22}}\approx 0.213 r[/tex]
Distance between them is decrease by a factor of 0.21
The separation between two point charges decreased by a factor of 5.
Explanation:To find the factor by which the separation between two point charges decreased when the force between them increased by a factor of 25, we need to understand the relationship between force and distance. According to Coulomb's law, the force between two point charges is inversely proportional to the square of the distance between them. This means that when the distance decreases, the force increases, and vice versa.
Since the force increased by a factor of 25, we can find the factor by which the distance decreased by taking the square root of 25, which is 5.
Therefore, the separation between the two point charges decreased by a factor of 5.
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