Answer:
The kinetic energy of the merry-goround after 3.62 s is 544J
Explanation:
Given :
Weight w = 745 N
Radius r = 1.45 m
Force = 56.3 N
To Find:
The kinetic energy of the merry-go round after 3.62 = ?
Solution:
Step 1: Finding the Mass of merry-go-round
[tex]m = \frac{ weight}{g}[/tex]
[tex]m = \frac{745}{9.81 }[/tex]
m = 76.02 kg
Step 2: Finding the Moment of Inertia of solid cylinder
Moment of Inertia of solid cylinder I =[tex]0.5 \times m \times r^2[/tex]
Substituting the values
Moment of Inertia of solid cylinder I
=>[tex]0.5 \times 76.02 \times (1.45)^2[/tex]
=> [tex]0.5 \times 76.02\times 2.1025[/tex]
=> [tex]79.91 kg.m^2[/tex]
Step 3: Finding the Torque applied T
Torque applied T = [tex]F \times r[/tex]
Substituting the values
T = [tex]56.3 \times 1.45[/tex]
T = 81.635 N.m
Step 4: Finding the Angular acceleration
Angular acceleration ,[tex]\alpha = \frac{Torque}{Inertia}[/tex]
Substituting the values,
[tex]\alpha = \frac{81.635}{79.91}[/tex]
[tex]\alpha = 1.021 rad/s^2[/tex]
Step 4: Finding the Final angular velocity
Final angular velocity ,[tex]\omega = \alpha \times t[/tex]
Substituting the values,
[tex]\omega = 1.021 \times 3.62[/tex]
[tex]\omega = 3.69 rad/s[/tex]
Now KE (100% rotational) after 3.62s is:
KE = [tex]0.5 \times I \times \omega^2[/tex]
KE =[tex]0.5 \times 79.91 \times 3.69^2[/tex]
KE = 544J
determine the required values of the tension in ac and ad so that the resultant of the three forces applied at A is verticalFigure:A wire is coupled from three sides
Answer:
Fac = 21 KN
Fad = 64.29 KN
Explanation:
Step 1: Find the unit vectors in BA , CA, and DA directions
BA = -16 i + 48 j - 12 k
CA = -16 i + 48 j + 24 k
DA = 14 i + 48 j
The following magnitudes
mag (BA) = sqrt ( 16 ^2 + 48^2 + 12^2 ) = 52
mag (CA) = sqrt ( 16 ^2 + 48^2 + 24^2 ) = 56
mag (DA) = sqrt ( 14 ^2 + 48^2 ) = 50
Now compute unit vectors
unit (BA) = (-4 / 13) i + (12 / 13) j - (3 / 13) k
unit (CA) = (-2 / 7) i + (6 / 7) j - (3 / 7) k
unit (DA) = (7 / 25) i + (24 / 25) j
Step 2: Compute Force Vectors F(ml) dot unit( ML )
Fab = (-4*Fab / 13) i + (12*Fab / 13) j - (3*Fab / 13) k
Fac = (-2*Fac / 7) i + (6*Fac / 7) j - (3*Fac / 7) k
Fad = (7*Fad / 25) i + (24*Fad / 25) j
Step 3: Equilibrium force equations in x and z direction
Fr,x = 0 = 39*( -4 / 13 ) - Fac*( 2 / 7 ) + Fad*( 7 / 25 )
Fr,z = 0 = 39*( -3 / 13 ) + Fad*( 3 / 7 )
Step 4: Solve the above equations for Fca and Fda
Fca = 21 KN
Fda = 64.29 KN
When the balls are brought close together, without touching, the following observations are made: Balls B, C, and D are attracted to ball A. Balls B and D have no effect on each other. Balls B is attracted to ball C. What are the charge states (positive, negative, or neutral) of balls A, B, C, and D
Answer:
B and D is neutral. A and C have opposite charges.
Explanation:
Opposite charges attract each other and similar charges repel each other. Furthermore, a neutral object is attracted to a charged object whether negative or positive.
- If the balls B, C, and D is attracted to A, then their charge is either opposite to that of A or neutral.
- If the balls B and D have no effect on each other, then both balls are neutral.
- If B is neutral and attracted to C, then ball C have opposite charges than A.
Finally, B and D is neutral. A and C have opposite charges. Whether they are positive or negative cannot be determined by the information given in the question.
A rocket has a mass 280(103) slugs on earth. Specify its mass in SI units, and its weight in SI units.
To solve this problem we will apply the concepts related to the conversion of units for which we will have that 1 slug is equal to 14.59kg. At the same time we will use Newton's second law for which weight is defined as the product between mass and acceleration (Due to gravity). This is then
A: Using the conversion ratio of slug to kilogram we have to,
[tex]1 slug = 14.59kg[/tex]
Then
[tex]m = 280*10^3 slugs (\frac{14.59kg}{1slug})[/tex]
[tex]m = 4.09*10^6kg[/tex]
B: Using Newton's second law we have to,
[tex]W = mg[/tex]
[tex]W = (4.09*10^6)(9.8)[/tex]
[tex]W = 40034960N\approx 4*10^7N[/tex]
If the x-component of a force vector is 2.16 newtons and its y-component is 4.11 newtons, then what is its magnitude? F = newtons
Answer:
Magnitude of resultant vector will be 4.643 N
Explanation:
We have given x-component of the force vector is given [tex]F_{X}=2.16N[/tex]
And y component of the force vector is given [tex]F_{y}=4.11N[/tex]
We have to find the magnitude of resultant force F
Resultant force will be equal to vector sum of magnitude of x competent and y component of force vector
So F = [tex]\sqrt{F_X^2+F_Y^2}=\sqrt{2.16^2+4.11^2}=4.643N[/tex]
So magnitude of resultant vector will be 4.643 N
The 9-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point Clocated on the vertical wall. If the tension in the cable is 3 kN, determine the moment about A of the force exertedby the cable at B.
Answer:
27000 Nm
Explanation:
The boom end at A is fixed and end at B is subjected to a 3kN force. Boom AB has length of 9m. The moment about A is the product of force 3kN (3000 N) at B and the moment arm of 9m
M = FL = 3000 * 9 = 27000 Nm
So the moment about A is 27000 Nm
A baseball has a mass of about 155 g. What is the magnitude of the momentum of a baseball thrown at a speed of 87 miles per hour? (Note that you need to convert mass to kilograms and speed to meters/second. A mile is 1.6 kilometers or 1600 meters.)
Answer:
5.994 kgm/s
Explanation:
Momentum: This can be defined as the product of the mass of a body to the velocity of that body. The S.I unit of momentum is kgm/s.
Mathematically, momentum can be expressed as,
M = mv.................................... Equation 1
Where M = momentum of the baseball, m = mass of the baseball, v = velocity of the baseball.
Given: m = 155 g = (155/1000) kg = 0.155 kg. v = 87 miles per hour = 87(1600/3600) m/s = 38.67 m/s.
Substituting into equation 1
M = 0.155(38.67)
M = 5.994 kgm/s.
Thus the momentum of the baseball = 5.994 kgm/s
The magnitude of the momentum of a baseball with a mass of 155 g thrown at 87 mph is approximately 6.03 kg·m/s after converting units and applying the momentum equation.
Explanation:To calculate the magnitude of the momentum of a baseball thrown at a speed of 87 miles per hour, we use the formula p = mv, where p is the momentum, m is the mass in kilograms, and v is the velocity in meters per second. First, we need to convert the mass of the baseball from grams to kilograms (155 g to 0.155 kg) and the speed from miles per hour to meters per second (87 mph to 38.88 m/s, using the conversion factor 1 mile = 1600 meters and 1 hour = 3600 seconds).
Then, we substitute the values into the momentum equation to find the momentum: p = (0.155 kg)(38.88 m/s) = 6.0264 kg·m/s. Therefore, the magnitude of the momentum of the baseball is approximately 6.03 kg·m/s when rounded to two decimal places.
For tax and accounting purposes, corporations depreciate the value of equipment each year. One method used is called "linear depreciation," where the value decreases over time in a linear manner. Suppose that two years after purchase, an industrial milling machine is worth $820,000, and five years after purchase, the machine is worth $430,000. Find a formula for the machine value V.
Answer:
[tex]V=-130000x+1080000[/tex]
Explanation:
Linear Dependence
Some variables are known or assumed to have linear dependence which means the graph of the ordered pairs (x,V) is a straight line.
If we know two points of the line, we can come up with the exact equation and therefore make predictions for other values of x
The linear depreciation gives us these points (2,820000) and (5,430000)
The general equation of the line is
[tex]V=mx+b[/tex]
Where V is the machine value and x is the number of years after purchase. We need to find the values of m and b.
Replacing the first point
[tex]820000=m(2)+b[/tex]
[tex]2m+b=820000[/tex]
Replacing the second point
[tex]5m+b=430000[/tex]
Subtracting them
[tex]-3m=390000[/tex]
[tex]m=-130000[/tex]
Replacing in any of the equations, say, the first one
[tex]2(-130000)+b=820000[/tex]
Solving for b
[tex]b=820000+260000[/tex]
[tex]b=1080000[/tex]
The formula for the machine value V is
[tex]\boxed{V=-130000x+1080000}[/tex]
Formula for the machine value V in x year is v = -130000(x) + 1,080,000
Liner depreciation based problem:What information do we have?
Cost of machine after 2 year = $820,000
Cost of machine after 5 year = $430,000
Liner depreciation equation
v = mx + b
After 2 year
820,000 = m(2) + b
820,000 = 2m + b ......... Eq1
After 5 year
430,000 = m(5) + b
430,000 = 5m + b ......... Eq2
Eq2 - Eq1
3m = -390,000
m = -130,000
From Eq
430,000 = 5m + b
430,000 = 5(-130,000) + b
b = 1,080,000
Liner equation of tha cost.
Amount of machine = mx + b
Amount of machine = -130000(x) + 1,080,000
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Another droplet of the same mass falls 8.4 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by the droplet.
Answer:
The charge carried by the droplet is [tex]1.330\times10^{-19}\ C[/tex]
Explanation:
Given that,
Distance =8.4 cm
Time = 0.250 s
Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude [tex]5.92\times10^4\ N/C[/tex] points straight down and if the mass of the droplet is [tex]2.93\times10^{-15} kg[/tex]
We need to calculate the acceleration
Using equation of motion
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2[/tex]
[tex]a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}[/tex]
[tex]a=2.688\ m/s^2[/tex]
We need to calculate the charge carried by the droplet
Using formula of electric filed
[tex]E=\dfrac{F}{q}[/tex]
[tex]q=\dfrac{ma}{E}[/tex]
Put the value into the formula
[tex]q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}[/tex]
[tex]q=1.330\times10^{-19}\ C[/tex]
Hence, The charge carried by the droplet is [tex]1.330\times10^{-19}\ C[/tex]
given that a car's distance from a stop sign t seconds since it started moving is represented by d=t2-2t determine the car's average speed over each interval of elapsed time
Answer:
Average speed =t-2t
Explanation:
Equation given for Average speed = Total distance/Elapsed time
Given: Total distance=d=t^2-2t
Elapsed time= t
Substituting into the equation
Average speed = t^2-2t/t
Dividing both numerator an denomination with t
Average speed=t-2t
To seven significant figures, the mass of a proton is 1.672623 x10^{-27} kg. Which of the following choices demonstrates correct rounding?
Answer:
[tex]1.67\times 10^{-27}kg[/tex]
Explanation:
We are given that mass of proton
[tex]1.672623\times 10^{-27}kg[/tex]
There are seven significant figures.
We have to round off.
If we round off to three significant figures
The thousandth place of given mass of proton is less than five therefore, digits on left side of thousandth place remains same and digits on right side of thousandth place and thousandth replace by zero
Therefore, the mass of proton can be written as
[tex]1.67\times 10^{-27}kg[/tex]
Hence, the mass of proton is [tex]1.67\times 10^{-27}kg[/tex]
To round the mass of a proton to seven significant figures, the correct way is to round up the last significant figure if it is 5 or greater, and if it is less than 5, simply drop the remaining digits.
Explanation:To round the mass of a proton, we look at the digit right after the desired number of significant figures, which in this case is the seventh significant figure. If this digit is 5 or greater, we round up the last significant figure. If it is less than 5, we simply drop the remaining digits.
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Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is needed to make your meaning clear. For example, 1⋅105 is not the same as 1105. When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol ×. For example, for the expression ma, typing m ⋅ a would be correct, but mxa would be incorrect.Enter the expression ma .
Final answer:
Type m · a to correctly enter the multiplication of m and a, ensuring clarity in expressions, especially when dealing with scientific notation which simplifies multiplication and division by manipulating coefficients and powers of ten separately.
Explanation:
When entering the expression m times a, you should use proper notation to differentiate between multiplication and adjacent variables or numbers. Instead of using 'x' as the multiplication symbol, as you might on paper, use the multiplication dot, represented by typing m · a with your keyboard. This avoids confusion, especially in contexts like scientific notation, where clarity is crucial. For example, 1·105 is not the same as 1 followed by 105. So, it is important to insert the multiplication dot to clarify that you are multiplying 1 by 10 raised to the 5th power.
In scientific notation, multiplication and division are simplified by separately manipulating the coefficients and the powers of ten. For instance, to multiply two numbers in scientific notation, such as (3 × 105) × (2 × 10°), you would multiply 3 by 2 to get 6, and then add the exponents (5 + 0) to get 5, resulting in an answer of 6 · 105. This simplifies calculations incredibly, especially with large numbers.
Water is pumped from a lake to a storage tank 15 m above at a rate of 90 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, answer the questions that follow.
a) the overall efficiency of the pump motor unit.
b) the pressure difference between the inlet and the exit of the pump.
Answer:
B
Explanation:
Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at the top of the ride? Express your answer with the appropriate units.
Answer:
422.36 N
Explanation:
given,
time of rotation = 4.30 s
T = 4.30 s
Assuming the diameter of the ring equal to 16 m
radius, R = 8 m
[tex]v = \dfrac{2\pi R}{T}[/tex]
[tex]v = \dfrac{2\pi\times 8}{4.30}[/tex]
v = 11.69 m/s
now, Force does the ring push on her at the top
[tex]- N - m g = \dfrac{-mv^2}{R}[/tex]
[tex] N + m g = \dfrac{mv^2}{R}[/tex]
[tex] N = \dfrac{mv^2}{R}- m g[/tex]
[tex] N = m(\dfrac{v^2}{R}- g)[/tex]
[tex] N = 58\times (\dfrac{11.69^2}{8}- 9.8)[/tex]
N = 422.36 N
The force exerted by the ring to push her is equal to 422.36 N.
The force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]
What will be the force does the ring push on her at the top of the ride?It is Given that
Time rotation T= 4.30 s
Mass m= 58 kg
Now the Velocity will be calculated as
[tex]V=\dfrac{2\pi r}{T} =\dfrac{2\pi 8}{4.30} =11.69 \frac{m}{s}[/tex]
Now by balancing the forces
[tex]N=\dfrac{mv^2}{R} -mg[/tex]
[tex]N=m(\dfrac{v^2}{R} -g)[/tex]
[tex]N=58\times (\dfrac{11.69^2}{8} -9.8)[/tex]
[tex]N=422.36 \ Newton[/tex]
Thus the force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]
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A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 9.0 m/s
A:What is the speed in the second segment?m/s
B:What is the speed in the third segment?m/s
C:What is the volume flow rate through the pipe?L/s
The speed in the second segment is 36 m/s, in the third segment is 144 m/s, and the volume flow rate through the pipe is 3.38 L/s.
A: Using the principle that the volume flow rate remains constant in an incompressible fluid, we can calculate the speed in the second segment to be 36 m/s and the speed in the third segment to be 144 m/s.
B: The volume flow rate through the pipe can be calculated by applying the equation Q = Av, where Q represents the flow rate, A is the cross-sectional area, and v is the velocity of the fluid. Given the diameter changes, the volume flow rate is 3.38 L/s.
C: The primary concept involved in this problem is the relationship between cross-sectional area and fluid velocity as the diameter changes along the pipe, affecting the flow rate.
How do the properties of an electromagnetic wave change as a result of increasing the period of the wave?
As the period of an electromagnetic wave increases, the wavelength of the wave also increases. However, the frequency of the wave remains the same. The energy carried by the wave is inversely proportional to its wavelength, so as the wavelength increases, the energy of the wave decreases.
Explanation:Electromagnetic waves have properties such as wavelength and frequency. As the period of an electromagnetic wave increases, the wavelength of the wave also increases. This means that the distance between successive crests or troughs of the wave becomes larger. However, the frequency of the wave remains the same. The energy carried by the wave is inversely proportional to its wavelength, so as the wavelength increases, the energy of the wave decreases.
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On a summer day in Narragansett, Rhode Island, the air temperature is 74 F and the barometric pressure is 14.5 lbf/in . Estimate the air density in kg/m.
There are some mistakes in the question as units pressure and air density are not written correctly.The correct question is here
On a summer day in Narragansett, Rhode Island, the air temperature is 74°F and the barometric pressure is 14.5 lbf/in². Estimate the air density in kg/m³.
Answer:
p=1.175 kg/m³
Explanation:
Given data
Temperature =74 F
Barometric pressure=14.5 lbf/in²
To find
Air density
Solution
From ideal gas law
[tex]pV=mRT\\p=(m/V)RT\\[/tex]
As mass/volume is pressure So
[tex]p=\frac{P}{RT}[/tex]
First we need to convert barometric pressure lbf/in² to N/m²
So
[tex]Pressure=(14.5lbf/in^{2} )*(\frac{6894.76N/m^{2} }{lbf/in^{2} } )\\Pressure=99974.02N/m^{2}[/tex]
Now Substitute the given value and pressure to find air density
[tex]p=\frac{P}{RT}\\ p=\frac{99974.02N/m^{2} }{(287J/kg.K)(\frac{74^{o}F-32}{1.8}+273.15 )}\\ p=\frac{99974.02N/m^{2} }{(287J/kg.K)*296.48K}\\ p=1.175kg/m^{3}[/tex]
Organ pipe A, with both ends open, has a fundamental frequency of 270 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. How long are___________.
(a) pipe A
(b) pipe B
Answer
Organ Pipe A has both end open.
Organ Pipe B has one end open.
speed of sound, v = 343 m/s
The fundamental frequency = 270 Hz
wavelength of the pipe when both end open
λ = 2 L
we know,
[tex]\lambda = \dfrac{v}{f}[/tex]
now,
[tex]L = \dfrac{v}{2f}[/tex]
inserting all the values
[tex]L = \dfrac{343}{2\times 270}[/tex]
L_A = 0.635 m
length of pipe A is equal to 0.635 m
b) since third harmonic of pipe B is equal to second harmonic of pipe A
[tex]f_B = f_A[/tex]
[tex]\dfrac{n_BV}{4L_B} = \dfrac{n_AV}{2L_A} [/tex]
[tex]L_B= \dfrac{2n_BL_A}{4n_A}[/tex]
[tex]L_B= \dfrac{2\times 3 \times 0.635}{4\times 2}[/tex]
L_B = 0.476 m
length of pipe B is equal to 0.476 m.
A kicker punts a football from the very center of the field to the sideline 39 yards downfield.
What is the magnitude of the net displacement of the ball? (a,football field is 53 yards wide)
What is the angle between the direction of the net displacement of the ball and the central line of the feild.
Answer:
L = 47.15 yards
∅ = 55.80°
Explanation:
Given
y = 39 yards
x = 53/2 yards
L = ?
We apply the formula
L = √(x²+y²)
L = √((53/2 yards)²+(39 yards)²)
L = 47.15 yards
The angle between the direction of the net displacement of the ball and the central line of the feild is obtained as follows
∅ = tg⁻¹(y/x)
∅ = tg⁻¹(39/(53/2)) = 55.80°
A circular-shaped object of mass 10 kg has an inner radius of 13 cm and an outer radius of 26 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 12 N, 26 N, and 16 N act on the object, as shown. The force of magnitude 26 N acts 30❦ below the horizontal. 12 N 16 N 26 N 30❦ ϝ Find the magnitude of the net torque on the wheel about the axle through the center of the object. Answer in units of N · m
The net torque on the circular object, considering the three forces acting on it, is calculated to be 1.46 N × m in an anti-clockwise direction.
To find the net torque on the circular object about its axle, we will first consider each of the three forces acting on it separately and determine the torque produced by each force. Torque (Τ) is defined by the equation Τ = r × F × sin(θ), where r is the radius at which the force is applied, F is the magnitude of the force, and θ is the angle between the force and the direction of the radius.
The 12 N force acts on the outer radius, so r = 0.26 m (26 cm converted to meters). Since the force is perpendicular to the radius, θ = 90 degrees, and the sin(90) = 1. Therefore, the torque from this force is Τ = 0.26 m × 12 N × 1 = 3.12 N × m.
The 16 N force also acts on the outer radius at 90 degrees to the radius, so its torque is Τ = 0.26 m × 16 N × 1 = 4.16 N × m.
The 26 N force acts at an angle 30 degrees below the horizontal, so it makes an angle of 60 degrees with the radius. Hence, the torque from this force is Τ = 0.26 m × 26 N × sin(60) = 5.82 N × m (rounded to two decimal places).
To find the net torque, we need to consider the direction of each torque. Both the 12 N and 16 N forces produce torque in the same direction (let's say clockwise), whereas the 30 degree component of the 26 N force produces torque in the opposite direction (counter-clockwise).
Net torque = Torque from 12 N + Torque from 16 N - Torque from 26 N = 3.12 N × m + 4.16 N × m - 5.82 N × m = 1.46 N × m (anti-clockwise).
The specific volume of 5 kg of water vapor at 1.5 MPa, 440°C is 0.2160 m3 /kg. Determine (a) the volume, in m3 , occupied by the water vapor, (b) the amount of water vapor present, in gram moles, and (c) the number of molecules.
(a) The volume occupied by the water vapor is V = 1.08 m³.
(b) The amount of water vapor present is 277.78 gram-moles.
(c) The number of molecules is [tex]\rm\(1.67305 \times 10^{26}\)[/tex] molecules.
Given:
Specific volume (v) = 0.2160 m³/kg
Mass of water vapor (m) = 5 kg
One gram-mole of water vapor = 18 g
Avogadro's number [tex]\rm (\(N_A\))[/tex] = [tex]\rm \(6.023 \times 10^{23}\)[/tex]
(a) The volume occupied by the water vapor (V) can be calculated using the formula: [tex]\rm \[ V = m \times v \][/tex]
Substitute the values:
[tex]\rm \[ V = 5 \, \text{kg} \times 0.2160 \, \text{m³/kg} \]\\\rm V = 1.08 \, \text m\³[/tex]
(b) The amount of water vapour present in gram moles can be calculated using the formula:
[tex]\rm \[ \text{Amount of water vapor} = \frac{m}{\text{One gram-mole of water vapor}} \][/tex]
Substitute the value of one gram-mole of water vapor [tex](\(18 \, \text{g}\))[/tex]:
[tex]\[ \text{Amount of water vapor} = \frac{5 \, \text{kg} \times 1000}{18 \, \text{g}} \]\\\\\ \text{Amount of water vapor} = 277.78 \, \text{gram-moles} \][/tex]
(c) The number of molecules can be calculated using Avogadro's number:
[tex]\rm \[ \text{Number of molecules in 1 gram-mole} = N_A \\= 6.023 \times 10^{23} \, \text{molecules} \][/tex]
So, the number of molecules in 1 gram of water vapor is [tex]\rm \(\frac{N_A}{18}\)[/tex] molecules.
The number of molecules in [tex]\rm \(5 \, \text{kg}\)[/tex] of water vapor is:
[tex]\rm \[ \text{Number of molecules} = 5 \times 1000 \times \frac{N_A}{18} \, \text{molecules} \][/tex]
Now, calculate the number of molecules:
[tex]\rm \[ \text{Number of molecules} = 1.67305 \times 10^{26} \, \text{molecules} \][/tex]
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Two point charges are released at rest in otherwise empty space. As time goes on, the magnitude of the force exerted on each of the particles decreases. Which of the following must be true?a. Both particles are negatively chargedb. Not enough infoc. The particles may be either both positively charged or both negatively chargedd. One particle is positively charged and the other is negatively chargede. Both particles are positively charged
Answer:
c.)
Explanation:
Assuming no other forces acting on the charges, if as time goes on, the magnitude of the force exerted on each of the particles decreases, as the electric force is inversely proportional to the square of the distance between charges, the only explanation is that this distance is increasing, so the force on both charges is repulsive.
When the force is repulsive, all we can say, is that both charges are of the same type, so they can be both positively charged or both negatively charged, as stated by the choice c).
You are pulling a box so it moves at increasing speed. Compare the work (W1) you need to do to accelerate it from 0 m/s to speed v to the work (W2) needed to accelerate it from speed v to the speed of 2v in the presence of friction. Compare the work () you need to do to accelerate it from 0 to speed to the work () needed to accelerate it from speed to the speed of 2 in the presence of friction. W2 > W1 W1 > W2 W1
Answer:
Explanation:
In absence of friction
Work done by force is given by change in kinetic energy of box
According work Energy theorem change in kinetic Energy of object is equal to work done by all the forces
[tex]W_1=\frac{1}{2}mv^2-0[/tex]
[tex]W_1=\frac{1}{2}mv^2[/tex]
where m=mass of object
[tex]W_2=\frac{1}{2}m(2v)^2-\frac{1}{2}mv^2[/tex]
[tex]W_2=\frac{3}{2}mv^2[/tex]
[tex]\frac{W_2}{W_1}=\frac{\frac{3}{2}mv^2}{\frac{1}{2}mv^2}[/tex]
[tex]W_2>W_1[/tex]
so work done in second case is three times of first case
(b)In Presence of friction
[tex]W_f+W_1=\frac{1}{2}mv^2[/tex]
suppose [tex]f_r[/tex] is the friction force and d is the displacement
so [tex]W_f=f_r\cdot d\cos 180[/tex]
[tex]W_f=-f_r\cdot d[/tex]
[tex]W_1=\frac{1}{2}mv^2+f_r\cdot d[/tex]
for second case when speed increases v to 2v
[tex]W_2=\frac{3}{2}mv^2+f_r\cdot d[/tex]
thus [tex]W_2>W_1[/tex] when friction is present
The work done to accelerate the box from zero is lesser than the work done to accelerate the box from velocity [tex]v[/tex].
From Work- energy theorem,Work can be defined as the change in the kinetic energy of the object.
So,
In the first Scenario,
[tex]W_1 = \dfrac 12 mv^2- 0\\\\W_1 = \dfrac 12 mv^2[/tex]
In the second scenario,
[tex]W_2 = \dfrac 12 m(2v^2) -\dfrac 12 mv^2\\\\W_2 = \dfrac 32 mv^2[/tex]
Now compare both scenarios,
[tex]\dfrac {W_1}{W_2}=\dfrac{ \dfrac 32 mv^2}{\dfrac12 mv^2}[/tex]
So, [tex]W_1 <W_2[/tex]
Therefore, the work done to accelerate the box from zero is lesser than the work done to accelerate the box from velocity [tex]v[/tex].
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A block with mass m=1.50 kg is initially at rest on a horizontal frictionless surface at x =0 , where x is the horizontal coordinate. A horizontally directed force is then applied to the block. The force is not constant: instead, its magnitude as a function of position is described by the relationship F(x)=(α-βx 2 )i , where x is given in units of meters, α = 2.50 N , β = 1.00 N/m2 , and i is the unit vector in the x direction.
a) What is the kinetic energy of the block as it passes through x=2.00 m?
b) What is the maximum speed of the block in the interval during which it moves from its initial position to x=2.00 m?
Answer:
(a) 2.33 J.
(b) 1.87 m/s.
Explanation:
(a)
[tex]F(x) = (2.5 - x^2)\^i[/tex]
The force as a function of position is given above. Since the force is a function of position, we can assume that we will use work-energy theorem.
[tex]W = \Delta K = K_2 - K_1\\\int\limits^2_0 {F(x)} \, dx = \frac{1}{2}mv_2 - 0\\\int\limits^2_0 {(2.5-x^2)} \, dx = (2.5x - \frac{x^3}{3})\left \{ {{x=2} \atop {x=0}} \right. = (5 - 8/3) = 7/3[/tex]
Therefore, the kinetic energy of the block is 2.33 J.
(b) In order to find the maximum speed in this interval, we need to investigate the acceleration of the block. Since acceleration is the derivative of velocity, velocity is at its maximum when acceleration is zero.
From Newton's Second Law:
[tex]F = ma\\a(x) = F(x)/m = 1.66 - 0.66x^2[/tex]
In order this to be zero:
[tex]1.66 - 0.66x^2 = 0\\x = 1.58~m[/tex]
The velocity of the block at x = 1.58 m can be found by work-energy theorem.
[tex]\int\limits^{1.58}_0 {(2.5-x^2)} \, dx = \frac{1}{2}(1.5)v^2\\ (2.5x-\frac{x^3}{3})\left \{ {{x=1.58} \atop {x=0}} \right. = 2.63 J\\2.63 = \frac{1}{2}(1.5)v^2\\v = 1.87~m/s[/tex]
What is the general relationship between earthquake epicenters and volcano positions along the Pacific coast of Mexico and Central America? What sort of plate boundary does this relationship indicate?
Geographically throughout this area of Mexico, Central America Caribbean is located the Cocos plate. This area is scientifically known as the Central American subduction zone.
In order for a volcano to form, there is usually a clash between the technical plates that generates the elevation of the ground and the connection with ducts that release the magma from the earth. If this entire area is a subduction area, it will also be a land stress release area where volcano lines will be formed, that is, it is a convergent plate boundary area
The relationship between the epicenters of the earthquake and the position of the volcano should be explained below.
What is Central American subduction zone?When the Mexico area should be geographically located, so the Caribbean of central America should be located on the Cocos plate. So this area should be called as the subduction zone. At the time when the volcano should be created so there is normally clash that lies between the technical plats where it generated the ground elevation and the linked with the ducts due to this, it releases the magma from the earth.
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A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 5 m/s. The bag is released with the same upward velocity of 5 m/s when t = 0 and hits the ground when t = 8 s.
(a) Determine the speed of the bag as it hits the ground.
(b) Determine the altitude of the balloon when the bag hits the ground
Answer:
a) 73.48 m/s
b) 313.92 m
Explanation:
Data provided in the question:
ascending velocity = - 5 m/s [ negative sign depicts upward movement]
Time taken by bag to hit the ground, t = 8 s
a) from the Newton's equation of motion
we have
[tex]s=ut+\frac{1}{2}at^2[/tex]
where,
u is the initial speed
a is the acceleration = 9.81 m/s² (since it is a case of free fall )
s is the distance
thus,
[tex]s=(-5)(8)+\frac{1}{2}(9.81)(8)^2[/tex]
s = - 40 + 313.92
s = 273.92 m
from
v = u + at
v is the final speed
v = -5 + (9.81)(8)
or
v = 73.48 m/s
b) Distance traveled by balloon = Speed × Time
= 5 × 8
= 40 m
Therefore,
Altitude of the balloon
= Distance traveled by bag + Distance traveled by balloon
= 273.92 m + 40 m
= 313.92 m
The speed of the bag as it hits the ground is 5 m/s. The altitude of the balloon when the bag hits the ground is 40 meters.
Explanation:To determine the speed of the bag as it hits the ground, we can use the equation for velocity:
v = u + at
Since the bag is dropped with an initial velocity of 5 m/s and there is no acceleration in the vertical direction, the final velocity of the bag as it hits the ground is also 5 m/s.
To determine the altitude of the balloon when the bag hits the ground, we can calculate the distance traveled by the bag using the equation for distance:
s = ut + (1/2)at^2
Plugging in the values, we get:
s = 5(8) + (1/2)(0)(8^2) = 40 meters
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Total lung capacity of a typical adult is approximately 5.0L. Approximately 20% of the air is oxygen, as air is 20% oxygen. At sea level and at anaverage body temperature of 37°C. How many moles ofoxygen do the lungs contain at the end of an inflation?
Answer:
n=0.03928 moles
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles
Explanation:
The amount of oxygen which lung can have is 20% of 5 L which is the capacity of lungs
Volume of oxygen in lungs =V=5*20%= 1 L=[tex]1*10^{-3} m^3[/tex]
Temperature=T=[tex]37^oC=273+37=310K[/tex]
Pressure at sea level = P= 1 atm=[tex]1.0125*10^5 Pa[/tex]
R is universal Gas Constant =8.314 J/mol.K
Formula:
[tex]n=\frac{PV}{RT}\\n=\frac{(1.0125*10^5) *(1*10^{-3})}{(8.314)*310} \\n=0.03928 mol[/tex]
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles
A +6.00 μC point charge is moving at a constant 9.00 ×10^6m/s in the + y-direction, relative to a reference frame.
At the instant when the point charge is at the origin of this reference frame, what is the magnetic field B it produces at the following points?
A) x=0.500m,y=0, z=0
B) x=0, y=-0.500m, z=0
C) x=0, y=0, z=+0.500m
D) x=0, y=-0.500m, z=+0.500m
Explanation:
We know magnetic field due to a charge q moving with a velocity v [tex]=\dfrac{\mu_oq}{4\pi r^3}(v \times r)[/tex].
Case A:
r=0.5i
Putting value of r, q and v .
We get, [tex]B=-2.34\times 10^{-5} \ k.[/tex] [tex](\ j\times i=-k)[/tex]
Case B:
r=0.5 j
B=0 [tex]( j\times j=0)[/tex]
Case C:
We get, [tex]B=2.34\times 10^{-5} \ i\ (Since\ , j \times k=i)[/tex]
Case D:
We get , [tex]B=8.28\times 10^{-6} \ T \ i[/tex].
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Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having the dimension of time. Determine the dimen- sions of the constants A and B. (b) Determine the dimen- sions of the derivative dx/dt 5 3At2 1 B.
Answer:
(a) A = m/s^3, B = m/s.
(b) dx/dt = m/s.
Explanation:
(a)
[tex]x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s[/tex]
Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.
(b) [tex]\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s[/tex]
This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.
If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?
Answer:
Star A is brighter than Star B by a factor of 2754.22
Explanation:
Lets assume,
the magnitude of star A = m₁ = 1
the magnitude of star B = m₂ = 9.6
the apparent brightness of star A and star B are b₁ and b₂ respectively
Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: [tex](m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})[/tex]
The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.
We need to find the factor by which star A is brighter than star B. Using the equation given above,
[tex](9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})[/tex]
[tex]\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})[/tex]
[tex]\log_{10}(b_{1}/b_{2}) = 3.44[/tex]
Thus,
[tex](b_{1}/b_{2}) = 2754.22[/tex]
It means star A is 2754.22 time brighter than Star B.
Star A is brighter than Star B by approximately 2512 times.
Explanation:The brightness of stars is measured using the magnitude scale. The smaller the magnitude, the brighter the star. In this case, Star A has a magnitude of 1.0 and Star B has a magnitude of 9.6. Since Star A has a smaller magnitude, it is brighter than Star B. The difference in magnitude between the two stars can be calculated using the formula:
difference = 2.512^(m₁ - m₂)
where m₁ is the magnitude of Star A and m₂ is the magnitude of Star B. Plugging in the values, we get:
difference = 2.512^(1.0 - 9.6) ≈ 2512
Therefore, Star A is approximately 2512 times brighter than Star B.
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Why is a frequency distribution useful? It allows researchers to see the "shape" of the data. It tells researchers how often the mean occurs in a set of data. It can visually identify the mean. It ignores outliers.
Answer: Frequency distribution helps to understand data. some complex data needs to be converted into intervals to see its frequency of occurrence. it helps to find the mean median and mode of the data. It helps in the analysis of data and its practical implementation.
Explanation: