Answer:
197.263157895 m/s
169.491525424 m/s
Explanation:
x Denotes position
t Denotes time
Average velocity is given by
[tex]v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-63}{4.75}\\\Rightarrow v_a=197.263157895\ m/s[/tex]
The average velocity is 197.263157895 m/s
[tex]v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-0}{5.9}\\\Rightarrow v_a=169.491525424\ m/s[/tex]
The average velocity is 169.491525424 m/s
The magnitude of the average velocity of the rocket during the 4.75-second part of the flight is 197.3 m/s, while for the first 5.90 seconds of the flight, the average velocity is 169.5 m/s.
Explanation:To find the magnitude of the average velocity of the rocket, we use the formula average velocity = displacement / time.
(a) The displacement during the 4.75-second part of the flight is 1.00 km - 63 m = 937 m (we converted km to m to keep units consistent). Hence, the average velocity in this part of the flight is 937 m / 4.75 s = 197.3 m/s.
(b) For the first 5.90 seconds of flight, the displacement is 1.00 km = 1000 m (the height above the ground) while time is 5.90 s. Therefore, for this duration, the average velocity is 1000 m / 5.90 s = 169.5 m/s.
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Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with + 90° corresponding to an initial velocity straight up and – 90° straight down.
The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this? (a) 250 frames/s; (b) 2500 frames/s; (c) 25,000 frames/s; (d) 250,000 frames/s.
Answer:
trajectory
Explanation:
according to issac newton plants are important
What are the units of the following properties? Enter your answer as a sequence of five letters separated by commas, e.g., A,F,G,E,D. Note that some properties listed may have the same units. 1) mass 2) heat 3) density 4) energy 5) molarity (A) g (B) J (C) mol (D) K (E) g/mol (F) mol/L (G) mol/K (H) g/mL (I) J/K (J) J/K*mol (K) J/K*g (L) kJ/L
Answer:
The sequence is A,B,H,B,F
Explanation:
The Standard International unit is Kilogram (kg) and the mass of a body can also be expressed in gram (g).Heat is a form of energy and the unit for energy is joule (J), thus the unit of heat is also joule (J).Density is mass per unit volume where the unit of mass is gram (g) and the unit of volume can be taken as milli-liter (mL). Thus g/mL is the unit of density.The unit of energy is joule (J).Molarity is number of solute in mol dissolved in 1 liter of solution. Thus mol/L is the the unit of molarity.The maximum lift-to-drag ratio of the World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine fails, how far can it glide in terms of distance measured along the ground?
The related concepts to solve this problem is the Glide Ratio. This can be defined as the product between the height of fall and the lift-to-drag ratio. Mathematically, this expression can be written as,
[tex]R = h (\frac{L}{D})_{max}[/tex]
Replacing,
[tex]R = 5000ft (7.7)[/tex]
[tex]R = 38500ft[/tex]
Converting this units to miles.
[tex]R = 38500ft (\frac{1mile}{5280ft})[/tex]
[tex]R = 7.2916miles[/tex]
Therefore the glide in terms of distance measured along the ground is 7.2916miles
The far that can it glide in terms of distance measured along the ground is 38,500 ft.
Given that,
World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine fails.Based on the above information, the calculation is as follows:
[tex]\frac{lift}{drag} = \frac{Distance}{height}\\\\7.7 = Distance \div 5500\\\\[/tex]
So, the distance is 38,500 ft.
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A railroad car with mass m is moving with an initial velocity v when it collides and connects with a second railroad car with a mass of 3m, which is initially at rest. How do the speed and momentum of the connected car system compare with those of the car with mass m before the collision
Answer:
Same momentum but speed is reduced 4 times.
Explanation:
According to the law of momentum conservation, the total momentum of the system before and after the collision is the same
Before the collision, the bigger car is at rest, only the 1st car of mass m is moving at speed v
p = mv
After the collision, both cars are connected and moving at speed V
P = (m + 3m)V = 4mV
These 2 momentum are equal
p = P
mv = 4mV
V = v/4
So after the collision, they have the same momentum but the speed decreased 4 times
Final answer:
After the collision of a moving car with mass m and a stationary car with mass 3m, the final velocity of the connected system is v/4, a quarter of the original car's velocity. The momentum of the system remains unchanged due to the conservation of momentum, but the total mechanical energy is typically not conserved as some is lost to other forms of energy.
Explanation:
Conservation of Momentum and Collision of Railway Cars
When a railroad car of mass m moving with an initial velocity v collides with and connects to another car at rest with a mass of 3m, the conservation of momentum principle applies. Given that neither external forces nor friction are significant, the total system momentum before and after the collision remains constant. The initial momentum of the system, which is mv (since the second car is at rest), must equal the final momentum of the now combined mass (4m) moving at the new velocity v'. To find the final velocity, we use the conservation of momentum equation:
mv = (m + 3m)v'
Which simplifies to:
v' = (mv)/(4m) = v/4
This equation shows that the speed of the combined railroad cars after the collision is a quarter of the initial speed of the moving car. While the speed of the connected car system is reduced compared to the initial speed of the car with mass m, the total momentum remains the same because the mass has increased fourfold.
Energy Conservation After the Collision
As for energy conservation, the total mechanical energy may not be conserved in an inelastic collision (like when cars connect after collision). Although the momentum is conserved, some kinetic energy is usually lost as other forms of energy such as heat, sound, or deformation of the vehicles.
A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density of water is 62.4 lb/ft3).
Final answer:
The force of water on a rectangular dam is given by the formula F = pgh²L/2, where p is the density of water, h is the depth at the dam, and L is the length of the dam. Using the given values, we can calculate the force of the water on the dam to be approximately 1,130,946 lb.
Explanation:
The force of water on a rectangular dam is given by the formula F = pgh²L/2, where p is the density of water, h is the depth at the dam, and L is the length of the dam. We are given the values p = 62.4 lb/ft³, h = 35 ft, and L = 101 ft.
Substituting these values into the formula, we have F = (62.4 lb/ft³) * (35 ft)² * (101 ft) / 2.
Simplifying the expression, we get F = 1,130,946 lb. Therefore, the force of the water on the dam is approximately 1,130,946 lb.
You spot a plane that is 1.37 km north, 2.71 km east, and at an altitude 4.65 km above your position. (a) How far from you is the plane? (b) At what angle from due north (in the horizontal plane) are you looking? °E of N (c) Determine the plane's position vector (from your location) in terms of the unit vectors, letting î be toward the east direction, ĵ be toward the north direction, and k be in vertically upward. ( km)î + ( km)ĵ + ( km) k (d) At what elevation angle (above the horizontal plane of Earth) is the airplane?
Answer:
Explanation:
a ) Position of the plane with respect to observer ( origin ) is
R = 2.71 i + 1.37 j + 4.65 k
magnitude of R = √ (2.71² + 1.37² + 4.65²)
√(7.344 + 1.8769 + 21.6225)
=√30.8434
= 5.55 km
b ) angle with north
cos Ф = 1.37 / 5.55
= .2468
Ф = 75°
c )
R = 2.71 i + 1.37 j + 4.65 k
=
If a 0.10−mL volume of oil can spread over a surface of water of about 47.0 m2 in area, calculate the thickness of the layer in centimeters. Enter your answer in scientific notation.
Answer:
[tex]2.1276595745\times 10^{-7}\ cm[/tex]
Explanation:
Volume of oil = 0.1 mL = 0.1 cm³
Area of oil = [tex]47\ m^2[/tex]
Converting to [tex]cm^2[/tex]
[tex]1\ m=10^2\ cm[/tex]
[tex]1\ m^2=10^4\ cm^2[/tex]
[tex]47\ m^2=47\times 10^4\ cm^2[/tex]
Volume is given by
[tex]V=At\\\Rightarrow t=\dfrac{V}{A}\\\Rightarrow t=\dfrac{0.1}{47\times 10^4}\\\Rightarrow t=2.1276595745\times 10^{-7}\ cm[/tex]
The thickness of the layer is [tex]2.1276595745\times 10^{-7}\ cm[/tex]
A movie stuntwoman drops from a helicopter that is 30.0 m above the ground and moving with a constant velocity whose components are 10.0 m/s upward and 15.0 m/s horizontal and toward the south. Ignore air resistance. (a) Where on the ground (relative to the position of the helicopter when she drops) should the stuntwoman have placed foam mats to break her fall? (b) Draw x-t, y-t, vx-t, and vy-t graphs of her motion.
The stuntwoman should place the foam mats approximately 37 meters south of the helicopter's position. Calculations are based on her dropping from a height of 30.0 m with an initial horizontal velocity of 15 m/s, taking into account gravitational acceleration and ignoring air resistance.
Explanation:To solve this problem, we need to calculate two things: how long the stuntwoman is in the air and how far she will travel horizontally during this time. Given the constant velocity components are 10.0 m/s upward and 15.0 m/s horizontal towards the south and ignoring air resistance, we'll tackle part (a) of the question first.
Part (a) - Determining the landing spot of the stuntwoman
Firstly, we acknowledge that the upward component of the helicopter's velocity will momentarily counteract gravity for the stuntwoman, but since air resistance is ignored, this effect is instantaneously null once she begins her descent. The primary considerations then are the height of 30.0 m and the horizontal component of 15.0 m/s.
To calculate the time (t) it takes for the stuntwoman to hit the ground, we use the equation for vertical motion under gravity:
h = 1/2gt^2
Where h is the height (30.0 m) and g is the acceleration due to gravity (~9.8 m/s^2). Solving for t, we get:
t = sqrt((2*h)/g) = sqrt((2*30)/9.8) ≈ 2.47 s
With the time in air known, we calculate the horizontal displacement (d) using:
d = v*t
Where v is the horizontal velocity (15.0 m/s). Thus, d ≈ 15.0 m/s * 2.47 s ≈ 37.05 m.
This means the stuntwoman should place the foam mats around 37 meters south of the helicopter's position at the moment she drops.
Part (b) - Drawing the graphs
For simplicity, the explanation of graph drawing is summarized: The x-t graph will show a linear increase in displacement over time, illustrating constant velocity in the horizontal direction. The y-t graph will depict a parabola, indicating acceleration (deceleration up then acceleration down) due to gravity in the vertical component. The vx-t graph will be a horizontal line showing constant horizontal velocity, and the vy-t graph will start at a positive value, decrease to zero at the peak of her motion, and then increase negatively as she accelerates downwards.
A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the instant the ball is thrown. If the boy throws the ball horizontally at 8.50 m/s, (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball?
Answer:
Explanation:
Height covered = 12m
time to fall by 12 m
s = 1/2 gt²
12 = 1/2 g t²
t = 1.565 s
Horizontal distance of throw
= 8.5 x 1.565
= 13.3 m
This distance is to be covered by dog during the time ball falls ie 1.565 s
Speed of dog required = 13.3 / 1.565
= 8.5 m /s
b ) dog will catch the ball at a distance of 13.3 m .
At a certain time a particle had a speed of 87 m/s in the positive x direction, and 6.0 s later its speed was 74 m/s in the opposite direction. What was the average acceleration of the particle during this 6.0 s interval?
Answer:
The average acceleration during the 6.0 s interval was -27 m/s².
Explanation:
Hi there!
The average acceleration is defined as the change in velocity over time:
a = Δv/t
Where:
a = acceleration.
Δv = change in velocity = final velocity - initial velocity
t = elapsed time
The change in velocity will be:
Δv = final velocity - initial velocity
Δv = -74 m/s - 87 m/s = -161 m/s
(notice the negative sign of the velocity that is in opposite direction to the direction considered positive)
Then the average acceleration will be:
a = Δv/t
a = -161 m/s / 6.0 s
a = -27 m/s²
The average acceleration during the 6.0 s interval was -27 m/s².
A package falls out of an airplane that is flying in a straight line at a constant altitude and speed. If you ignore air resistance, what would be the path of the package as observed by the pilot? As observed by a person on the ground?
Answer:
Explanation:
If you ignore air resistant, then nothing affects the package horizontal motion. In Newton's first law it would keep the package at a constant speed, the speed of the airplane.
So to the eyes of the pilot who is also moving at the same horizontal speed, the lateral position of the package does not change. He can only perceive that the package is getting further away from him as it's dropping vertically.
To a person on the ground then the package is travelling in a parabolic path, where its horizontal speed is constant but vertical speed is increasing toward the ground at the rate of g.
The package would follow a straight line for the pilot and a parabolic curve for an observer on the ground.
Explanation:
The path of the package as observed by the pilot would be a straight line parallel to the airplane's motion. This is because the package falls with the same horizontal velocity as the airplane due to the absence of air resistance. As observed by a person on the ground, the path of the package would be a parabolic curve. This is because the package has an initial horizontal velocity but is acted upon by the force of gravity, causing it to follow a curved trajectory.
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A 500-Hz whistle is moved toward a listener at a speed of 10.0 m/s. At the same time, the listener moves at a speed of 20.0 m/s in a direction away from the whistle. What is the apparent frequency heard by the listener? (The speed of sound is 340 m/s.)
Answer:
f' = 485 Hz
Explanation:
given,
Frequency of whistle,f = 500 Hz
speed of source, v_s = 10 m/s
Speed of observer, v_o - 20 m/s
speed of sound,v = 340 m/s
Apparent frequency heard = ?
Using Doppler's effect formula to find apparent frequency
[tex]f' = (\dfrac{v-v_0}{v-v_s})f[/tex]
[tex]f' = (\dfrac{340-20}{340-10})\times 500[/tex]
[tex]f' = 0.9696\times 500[/tex]
f' = 485 Hz
Hence, the apparent frequency is equal to 485 Hz.
To determine the apparent frequency heard by the listener is equal to 545.45 Hz.
Given the following data:
Observer velocity = 20.0 m/sFrequency of sound = 500 HzSource velocity = 10.0 m/sSpeed of sound = 340 m/sTo determine the apparent frequency heard by the listener, we would apply Doppler's effect of sound waves:
Mathematically, Doppler's effect of sound waves is given by the formula:
[tex]F_o = \frac{V \;+ \;V_o}{V\; - \;V_s} F[/tex]
Where:
V is the speed of a sound wave.F is the actual frequency of sound.[tex]V_o[/tex] is the observer velocity.[tex]V_s[/tex] is the source velocity.[tex]F_o[/tex] is the apparent frequency.Substituting the given parameters into the formula, we have;
[tex]F_o = \frac{340 \;+ \;20}{340\; - \;10} \times 500\\\\F_o =\frac{360}{330} \times 500\\\\F_o =1.0909 \times 500[/tex]
Apparent frequency = 545.45 Hz.
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In uniform circular motion, the acceleration is perpendicular to the velocity at every instant. Is this true when the motion is not uniform—that is, when the speed is not constant?
Answer:
No.
Explanation:
Through uniform We can presume you mean constant in both that is in magnitude as well as in direction ⠀
When you apply a steady, non-zero acceleration to a resting body, the velocity can be parallel to the acceleration and not always perpendicular at any time later.
We can do the same for an object with an initial velocity in a direction which is different than that of acceleration, and the direction of its velocity will asymptotically approach to that of acceleration over time.
In both cases the motion of a object undergoing a non-zero persistent acceleration can not be uniform That is the concept of acceleration: the velocity change over time.
You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testtube. The speed of sound in air is 344 m/s and the test tube actsas a stopped pipe.
(a) If the length of the air column in the test tubeis 11.0 cm, what is thefrequency of this standing wave?
kHz
(b) What is the frequency of the fundamental standing wave in theair column if the test tube is half filled with water?
kHz
Answer:
(a). The frequency of this standing wave is 0.782 kHz.
(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.
Explanation:
Given that,
Length of tube = 11.0 cm
(a). We need to calculate the frequency of this standing wave
Using formula of fundamental frequency
[tex]f_{1}=\dfrac{v}{4l}[/tex]
Put the value into the formula
[tex]f_{1}=\dfrac{344}{4\times0.11}[/tex]
[tex]f_{1}=781.81\ Hz[/tex]
[tex]f_{1}=0.782\ kHz[/tex]
(b). If the test tube is half filled with water
When the tube is half filled the effective length of the tube is halved
We need to calculate the frequency
Using formula of fundamental frequency of the fundamental standing wave in the air
[tex]f_{1}=\dfrac{v}{4(\dfrac{L}{2})}[/tex]
Put the value into the formula
[tex]f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}[/tex]
[tex]f_{1}=1563.63\ Hz[/tex]
[tex]f_{1}=1.563\ kHz[/tex]
Hence, (a). The frequency of this standing wave is 0.782 kHz.
(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.
Your bedroom has a rectangular shape, and you want to measure its area. You use a tape that is precise to 0.001 and find that the shortest wall in the room is 3.547 long. The tape, however, is too short to measure the length of the second wall, so you use a second tape, which is longer but only precise to 0.01 . You measure the second wall to be 4.79 long. Which of the following numbers is the most precise estimate that you can obtain from your measurements for the area of your bedroom?
If your bedroom has a circular shape, and its diameter measured 6.32 , which of the following numbers would be the most precise value for its area?
a)30 m^2
b) 31.4 m^2
c)31.37 m^2
d)31.371 m^2
Answer:
The question is incomplete.the complete question is giving below "College Physics 10+5 pts
Your bedroom has a rectangular shape, and you want to measure its area. You use a tape that is precise to 0.001 and find that the shortest wall in the room is 3.547 long. The tape, however, is too short to measure the length of the second wall, so you use a second tape, which is longer but only precise to 0.01 . You measure the second wall to be 4.79 long. Which of the following numbers is the most precise estimate that you can obtain from your measurements for the area of your bedroom?
A. 17.0m^2
B. 16.990m^2
C.16.99m^2
D.16.9m^2
E. .16.8m^2
b. If your bedroom has a circular shape, and its diameter measured 6.32 , which of the following numbers would be the most precise value for its area?
a)30 m^2
b) 31.4 m^2
c)31.37 m^2
d)31.371 m^2
Answers:
A. 17.0m²
B. 31.4m²
Explanation:
First, we determine the area of the room using the measured parameters I.e
Length=4.79 long
Breadth= 3.547 long
Hence area is calculated as
A=length *Breadth
A=4.79*3.547
A=16.99013m²
From the question the different measurements have different precision, the answer should match the number of significant figures of the least precisely known number in the calculation which is 3 significant digits.
Hence the correct answer will be 17.0m²
B. The are of a circle is express as
A=πd²/4
A=π(6.32)²/4
A=31.37069m²
The π and 4 are exact number they have no effect on the accuracy on the accuracy of the area. Hence we express the answer to the same significant figure as giving in the question..
The correct answer will be 31.4m²
The most precise estimate for the area of the rectangular bedroom is 17.004613 units. The most precise value for the area of the circular bedroom is 31.373968 units.
Explanation:To find the most precise estimate for the area of your rectangular bedroom, you need to consider the precision of the measurements. The first wall measurement is 3.547, which has a precision of 0.001. The second wall measurement is 4.79, which has a precision of 0.01. Since the second measurement has a lower precision, it will determine the precision of the final estimate. Therefore, the most precise estimate for the area of your bedroom is obtained by multiplying the two measurements: 3.547 x 4.79 = 17.004613.
To find the most precise value for the area of your circular bedroom, you need to consider the precision of the diameter measurement. The diameter is measured as 6.32, which has a precision of 0.01. The formula to calculate the area of a circle is A = πr^2, where r is the radius (half of the diameter). So, the radius is 6.32/2 = 3.16. Therefore, the most precise value for the area of your circular bedroom is obtained by calculating the area using the radius: A = π(3.16)^2 = 31.373968.
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The volume of a fluid in a tank is 0.25 m3 . of the specific gravity of the fluid is 2.0. Determine the mass of the fluid. Given the density of water is 1000 kg/m3. Express the answer in Kg.
Answer:
Explanation:
Given
volume of Tank [tex]V=0.25\ m^3[/tex]
Specific gravity [tex]=2[/tex]
specific gravity is the defined as the ratio of density of fluid to the density of water
Density of water [tex]\rho _w=1000\ kg/m^3[/tex]
Density of Fluid [tex]\rho =2\times 1000=2000\ kg/m^3[/tex]
We know mass of a fluid is given by the product of density and volume
[tex]m=\rho \times V[/tex]
[tex]m=2000\times 0.25[/tex]
[tex]m=500\ kg[/tex]
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep constant pressure on the mixture of 1 atm. The cylinder is also submerged in a large insulated water bath. The temperature of the water bath is monitored, and it is determined from this data that 133.0 kJ of heat flows into the system during the reaction. The position of the piston is also monitored, and it is determined from the data that the piston does 241.0 kJ of work on the system during the reaction.
(1) Does the temperature of the water bath go up or down?
(2) Does the piston move in or out?
(3) Does heat flow into or out of the gaseous mixture?
(4) How much heat flows?
Answer:
1) Temperature of the water bath rises
2) the piston moves out
3) 133 kJ of heat flows
Explanation:
Given:(Isobaric process)
Heat absorbed by the system,[tex]Q_{in}=133000\ J[/tex]work done on the system, [tex]W=241000\ J[/tex]pressure on the system, [tex]P=1\ atm=101325\ Pa[/tex]1)
Since the whole setup is isolated and the heat is absorbed by the system therefore the the temperature of the water will go down.
2)
When the system absorbs heat its pressure has to be constant so the piston needs to move up outwards giving the inside gases more volume.
As ideal gas law equation:
[tex]P.V=m.R.T[/tex]
[tex]P=[/tex]absolute pressure of the gas
[tex]V=[/tex] volume of the gas
[tex]m=[/tex] mass of the gas
[tex]R=[/tex] universal gas constant
[tex]T=[/tex] absolute temperature of the gas
Since pressure, mass and gas constants are the constant value we observe that: [tex]T\propto V[/tex]
3)
According to the given data the heat that flows is 133 kilo-joule in quantity.
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have3C+4D=52C+5D=2None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?a. C=53−43Db. C=1−52Dc. D=25−25Cd. D=54−34C
Answer:
According to the instructions given, only options a and b are correct.
That is,
C = (5/3) - (4D/3)
C = 1 – (5D/2)
D= -4/7
C= 17/7
Explanation:
3C + 4D = 5 and 2C + 5D = 2
So, following the instructions from the question,
1) we'll pick the variable with the smallest coefficient and isolate it.
In eqn 1, C has the smallest coefficient,
3C = 5 - 4D (isolated!)
In eqn 2, C still has the smallest coefficient,
2C = 2 - 5D
2) Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient.
For eqn 1,
3C = 5 - 4D, divide through by the coefficient of C,
C = (5/3) - (4D/3)
This matches option a perfectly.
For eqn 2,
2C = 2 - 5D, divide through by the coefficient of C,
C = (2/2) - (5D/2) = 1 - (5D/2)
This matches option b perfectly!
Further solving the equations now,
Since C = C
(5/3) - (4D/3) = 1 - (5D/2)
(5D/2) - (4D/3) = 1 - (5/3)
(15D - 8D)/6 = -2/3
7D/6 = -2/3
D = -4/7
Substituting this into one of the eqns for C
C = 1 - (5D/2)
C = 1 - (5/2)(-4/7) = 1 - (-10/7) = 1 + (10/7) = 17/7.
QED!
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal toward the cliff. (a) What must the minimum muzzle velocity be for the shell to clear the top of the cliff? (b) The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?
Answer:
a) v₀ = 32.64 m / s , b) x = 59.68 m
Explanation:
a) This is a projectile launching exercise, we the distance and height of the cliff
x = v₀ₓ t
y = [tex]v_{oy}[/tex] t - ½ g t²
We look for the components of speed with trigonometry
sin 43 = v_{oy} / v₀
cos 43 = v₀ₓ / v₀
v_{oy} = v₀ sin 43
v₀ₓ = v₀ cos 43
Let's look for time in the first equation and substitute in the second
t = x / v₀ cos 43
y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²
y = x tan 43 - ½ g x² / v₀² cos² 43
1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²
v₀² = g x² / [(x tan 43 –y) 2 cos² 43]
Let's calculate
v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]
v₀ = √ (35280 / 33.11)
v₀ = 32.64 m / s
.b) we use the vertical distance equation with the speed found
y = [tex]v_{oy}[/tex] t - ½ g t²
.y = v₀ sin43 t - ½ g t²
25 = 32.64 sin 43 t - ½ 9.8 t²
4.9 t² - 22.26 t + 25 = 0
t² - 4.54 t + 5.10 = 0
We solve the second degree equation
t = (4.54 ±√(4.54 2 - 4 5.1)) / 2
t = (4.54 ± 0.46) / 2
t₁ = 2.50 s
t₂ = 2.04 s
The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled
x = v₀ₓ t
x = v₀ cos 43 t
x = 32.64 cos 43 2.50
x = 59.68 m
The minimum muzzle velocity of a cannon to clear a 25.0 m tall cliff from 60.0 m away involves using projectile motion equations to find the initial velocity components, while ensuring the shell reaches the required height and distance.
Explanation:To solve for the minimum muzzle velocity for a cannon to shoot a shell past a 25.0 m cliff from 60.0 m away, we can use projectile motion equations. First, we separate the initial velocity into its horizontal (vx) and vertical (vy) components:
vx = v0 × cos(θ)vy = v0 × sin(θ)The shell must reach a height of at least 25.0 m to clear the cliff. We use the equation of motion in the vertical direction, considering the initial vertical velocity (vy) and the displacement (s = 25 m), to find the time (t) it takes to reach the top of the cliff. Then, using the horizontal velocity (vx), we can calculate how far the shell would travel horizontally in that time, ensuring that it covers at least 60.0 m. With the horizontal distance (d) and time (t) determined, we can calculate the shell's trajectory past the cliff edge.
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Speed of a rocket At t sec after liftof, the height of a rocket is 3t 2 ft. How fast is the rocket climbing 10 sec after liftof?
Final answer:
The speed of the rocket 10 seconds after liftoff is 60 ft/s.
Explanation:
To find the speed of the rocket 10 seconds after liftoff, we can differentiate the equation for the height of the rocket with respect to time. In this case, the equation for the height of the rocket is given as h = 3t^2. Taking the derivative of this equation with respect to time will give us the rate of change of height with respect to time, which is the speed of the rocket. The derivative of 3t^2 with respect to t is 6t. Plugging in t = 10 into the derivative, we get 6(10) = 60.
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation 6.0 m above the ground. The ball lands 25 m away.
What is his pitching speed?
To solve this problem we will apply the concepts related to the kinematic equations of linear motion. Punctually we will verify the vertical displacement and the horizontal displacement from their respective components. We will start by calculating the time it took to reach the objective and later with that time, we will find the horizontal velocity launch component. The position can be written as,
[tex]h= v_{0y}t+\frac{1}{2}a_yt^2[/tex]
Here,
h = Height
[tex]v_{0y}[/tex]= Initial velocity in vertical direction
[tex]a_{y}[/tex] = Vertical acceleration (At this case, due to gravity)
[tex]t[/tex] = Time
There is not vertical velocity because the ball was thrown horizontally), then we have that
[tex]6= (0)t+\frac{1}{2}(9.8)t^2[/tex]
[tex]t = 1.1065s[/tex]
Now using the equation of horizontal motion we have with this time that the initial velocity was,
[tex]x = v_{ox}t+\frac{1}{2}a_xt^2[/tex]
Here,
[tex]v_{0x}[/tex]= Horizontal initial velocity
[tex]t[/tex] = Time
[tex]a_x[/tex] = Acceleration in horizontal plane
There is not acceleration in horizontal plane, only in vertical plane, then we have
[tex]25= v_{0x}(1.1065)+\frac{1}{2}(0)(1.1065)^2[/tex]
[tex]v_{0x} = 22.5938m/s[/tex]
Therefore the pitching speed is 22.5938m/s
to what circuit element is an ideal inductor equivalent for circuits with constant currents and voltages?
Answer:
Short circuit
Explanation:
In an ideal inductor circuit with constant current and voltage, it implies that the voltage drop in the circuit is zero (0).
Also, In circuit analysis, a short circuit is defined as a connection between two nodes that forces them to be at the same voltage.
In an ideal short circuit, this means there is no resistance and thus no voltage drop across the connection. That is voltage drop is zero (0).
Therefore, the circuit element is short circuit.
For circuits with constant currents and voltages, an ideal inductor is equivalent to a short circuit or a piece of wire with no resistance, as it does not affect the circuit voltage or resistance.
An ideal inductor in a circuit with constant currents and voltages acts as if it is effectively a wire with no resistance, since an ideal inductor does not dissipate energy when the current is constant. However, if we consider the energy transfer in a circuit with an inductor and another circuit element, often referred to as a 'black box', which could be a resistor, we notice energy is exchanged only when there is a change in current. Using the lumped circuit approximation, the inductor's magnetic fields are assumed to be completely internal, meaning it only interacts with other components via the current flowing through the wires, not by overlapping magnetic fields in space.
Given the Kirchhoff's voltage law, which states that the sum of the voltage drops in a closed loop must be zero, an inductor with a constant current will have a voltage drop that is zero, making it equivalent to a short circuit in terms of its impact on the voltage in the circuit. Therefore, for circuits with constant currents and voltages, an ideal inductor is equivalent to a short circuit or a piece of wire with no resistance, as it does not contribute any voltage drop or generate heat like a resistor would.
(2 points) A ring weighing 9.45 g is placed in a graduated cylinder containing 25.3 mL of water. After the ring is added to the cylinder the water rises to 27.4 mL. What metal is the ring made out of? Assume the ring is a single metal.
Answer:
Titanium.
Explanation:
Density of a metal is defined as the mass of the metal and its volume.
Volume of the metal = total volume- volume of initial water
= 27.4 - 25.3
= 2.1 ml
Mathematically,
Density = mass/volume
= 0.00945 kg/0.0000021 m3
= 4500 kg/m3.
The metal is Titanium.
Waves travel along a 100-m length of string which has a mass of 55 g and is held taut with a tension of 75 N. What is the speed of the waves?
Answer:
[tex]v=369.27\frac{m}{s}[/tex]
Explanation:
The speed of the waves in a string is related with the tension and mass per unit length of the string, as follows:
[tex]v=\sqrt\frac{T}{\mu}[/tex]
First, we calculate the mass per unit length:
[tex]\mu=\frac{m}{L}\\\mu=\frac{55*10^{-3}kg}{100m}\\\mu=5.5*10^{-4}\frac{kg}{m}[/tex]
Now, we calculate the speed of the waves:
[tex]v=\sqrt\frac{75N}{5.5*10^{-4}\frac{kg}{m}}\\v=369.27\frac{m}{s}[/tex]
Given values,
Mass, [tex]m = 55 \ g[/tex]or, [tex]= 55\times 10^{-3} \ kg[/tex]
Length, [tex]L = 100 \ m[/tex]As we know,
→ [tex]v = \sqrt{\frac{T}{\mu} }[/tex]
or,
→ [tex]\mu = \frac{m}{L}[/tex]
By putting the values,
[tex]= \frac{55\times 10^{-3}}{100}[/tex]
[tex]= 5.5\times 10^{-4} \ kg/m[/tex]
hence,
The speed of wave:
→ [tex]v = \sqrt{\frac{75}{5.5\times 10^{-4}} }[/tex]
[tex]= 369.27 \ m/s[/tex]
Thus the above response is correct.
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When measuring weight on a scale that is accurate to the nearest 0.1 pound, what are the real limits for the weight of 145 pounds?
When measuring weight on a scale that is accurate to the nearest 0.1
pound, the real limits for the weight of 145 pounds is 144.9- 145.1
ScaleThis is an instrument which is used to measure the weight of objects. There
may be differences in the measurement as a result of air interference and
other factors.
We were told that the accuracy is to the nearest 0.1 pound which means
= 145± 0.1
= (145-0.1) - (145+0.1)
= 144.9 - 145.1
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The real limits for the weight of 145 pounds on a scale accurate to the nearest 0.1 pound are 144.95 to 145.05 pounds.
Explanation:When measuring weight on a scale that is accurate to the nearest 0.1 pound, the real limits for the weight of 145 pounds would be 144.95 to 145.05 pounds. This is because the scale is accurate to the tenths place, meaning it can measure weights up to one decimal place. In this case, the weight of 145 pounds would be rounded to 145.0 pounds. Therefore, the real limits would be 144.95 pounds (145.0 - 0.05) to 145.05 pounds (145.0 + 0.05).
A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm wide-that is, the two first-order diffraction minima are separated by 1.40cm
What is the distance between the two second-order minima?
Answer:
2.8 cm
Explanation:
[tex]y_1[/tex] = Separation between two first order diffraction minima = 1.4 cm
D = Distance of screen = 1.2 m
m = Order
Fringe width is given by
[tex]\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm[/tex]
Fringe width is also given by
[tex]\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}[/tex]
For second order
[tex]\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1[/tex]
Distance between two second order minima is given by
[tex]y_2=2\beta_2[/tex]
[tex]\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm[/tex]
The distance between the two second order minima is 2.8 cm
The distance between the two second-order minima is 2.80 cm.
The central maximum's width given is 1.40 cm (this is the distance between the first-order minima, m = ±1). So, the distance from the center to the first-order minimum (m = ±1) is 0.70 cm.
Using the formula for the first-order minimum (m = 1):
a sin(θ₁) = λ
We know that the distance to the first minima (y₁) with the screen distance (L) is given by:
tan(θ₁) ≈ sin(θ₁) = y₁ / L
so, for the first-order minima:
y₁ = Lλ / a
We have y₁ = 0.70 cm, L = 1.20 m. Solving for aλ:
a = Lλ / 0.007
Next, for the second-order minima (m = 2), we use:
y₂ = 2Lλ / a
Thus, the distance between the second-order minima will be:
2y₂ = 2 × 2Lλ / a = 2 × 1.40 cm = 2.80 cm
So, the distance between the two second-order minima is 2.80 cm.
The height, h , in meters of a dropped object after t seconds can be represented by h ( t ) = − 4.9 t 2 + 136 . What is the instantaneous velocity of the object one second after it is dropped?
Answer:
The velocity of the object 1 s after it is dropped is -9.8 m/s.
Explanation:
Hi there!
The instantaneous velocity is defined by the change in height over a very small time. Mathematically, it is expressed as the derivative of the function h(t):
instantaneous velocity = dh/dt
dh/dt = h´(t) = -2 · 4.9 · t
h´(t) = -9.8 · t
Now we have to eveluate the function h´(t) at t = 1 s:
h´(1) = -9.8 · (1) = -9.8
The velocity of the object 1 s after it is dropped is -9.8 m/s.
Final answer:
The instantaneous velocity of the object one second after it is dropped is 9.8 m/s, moving downwards.
Explanation:
The question asks for the instantaneous velocity of a dropped object after a specific time has elapsed, which is a physics concept related to mechanics and motion. Given the height equation h(t) = -4.9t2 + 136, the instantaneous velocity at time t can be found by taking the derivative of the height equation with respect to time, which gives us the velocity as a function of time v(t) = dh/dt. At t = 1 second, the derivative of the height equation is v(1) = -9.8(1) = -9.8 m/s. The negative sign indicates that the object is moving downwards. However, when we speak about the speed or magnitude of the velocity (instantaneous velocity), we typically refer to the positive value, which would be 9.8 m/s.
Water flows at speed of 5.9 m/s through a horizontal pipe of diameter 3.1 cm . The gauge pressure P1 of the water in the pipe is 1.5 atm . A short segment of the pipe is constricted to a smaller diameter of 2.1 cm. What is the gauge pressure of the water flowing through the constricted segment?
Final answer:
Using Bernoulli's equation and the continuity equation, we find that as the diameter of a pipe decreases, the velocity of the water increases, and to conserve the total mechanical energy, gauge pressure in the constricted segment decreases.
Explanation:
The situation described in the question can be analyzed using the principle of conservation of energy, specifically Bernoulli's equation for incompressible fluid flow. We are given that water flows at a speed of 5.9 m/s through a horizontal pipe of diameter 3.1 cm, and the gauge pressure is 1.5 atm. With the constriction of the pipe's diameter to 2.1 cm, we want to find the new gauge pressure of the water.
According to Bernoulli's equation, the total mechanical energy per unit volume is the same at all points along the streamline, i.e.,
[tex]P + \(\frac{1}{2}\)\(\rho\)v^2 + \(\rho\)gh = constant[/tex]
where P is the pressure, \(\rho\) is the density of the fluid, v is the flow velocity, g is acceleration due to gravity, and h is the elevation. For the situation described, h remains constant (horizontal pipe), and there is no change in the gravitational potential energy. Therefore, we can simplify the equation to:
[tex]P1 + \(\frac{1}{2}\)\(\rho\)v1^2 = P2 + \(\frac{1}{2}\)\(\rho\)v2^2[/tex]
Since the fluid is incompressible and the flow rate must be conserved, the velocity v2 in the constricted segment is determined by the continuity equation:
[tex]A_1v_1 = A_2v_2[/tex]
To solve this problem, we'll use the principle of continuity, which states that the product of cross-sectional area and velocity is constant for an incompressible fluid flowing through a tube. Mathematically, it can be expressed as:
Step 1: Find [tex]\( v_2 \)[/tex]
Using the continuity equation:
[tex]\[ A_1 V_1 = A_2 V_2 \][/tex]
We can find [tex]\( v_2 \)[/tex] as:
[tex]\[ V_2 = \frac{\frac{\pi D_1^2}{4}}{\frac{\pi D_2^2}{4}} \times V_1 \]\[ V_2 = \frac{D_1^2}{D_2^2} \times V_1 \][/tex]
Since [tex]\( A = \frac{\pi D^2}{4} \)[/tex], we can rewrite the equation as:
[tex]\[ V_2 = \frac{\frac{\pi D_1^2}{4}}{\frac{\pi D_2^2}{4}} \times V_1 \]\[ V_2 = \frac{D_1^2}{D_2^2} \times V_1 \][/tex]
Now, let's plug in the values:
[tex]\[ V_2 = \frac{(0.031 \, \text{m})^2}{(0.021 \, \text{m})^2} \times 5.9 \, \text{m/s} \]\[ V_2 = \frac{0.000961 \, \text{m}^2}{0.000441 \, \text{m}^2} \times 5.9 \, \text{m/s} \]\[ V_2 = \frac{0.000961}{0.000441} \times 5.9 \, \text{m/s} \]\[ V_2 \approx 12.84 \, \text{m/s} \][/tex]
Step 2: Calculate [tex]\( P_2 \)[/tex]
We'll use Bernoulli's equation, which states that the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant along any streamline of flow. In this case, we'll ignore changes in height (assuming horizontal flow), and the equation simplifies to:
[tex]\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \][/tex]
Where:
- [tex]\( P_1 \)[/tex] is the initial pressure
- [tex]\( P_2 \)[/tex] is the pressure in the constricted segment
- [tex]\( \rho \)[/tex]is the density of the fluid (we'll assume water,[tex]\( \rho = 1000 \, \text{kg/m}^3 \))[/tex]
-[tex]\( v_1 \) and \( v_2 \)[/tex] are initial and final velocities, respectively.
We rearrange this equation to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2 \][/tex]
Now, let's plug in the values:
[tex]\[ P_2 = P_1 + \frac{1}{2} \times 1000 \times (5.9)^2 - \frac{1}{2} \times 1000 \times (12.84)^2 \]\[ P_2 = P_1 + 17421.5 - 82626.24 \]\[ P_2 = P_1 - 65204.74 \][/tex]
Step 3: Convert [tex]\( P_2 \)[/tex] to atmospheres (gauge pressure)
To express the pressure in atmospheres and considering atmospheric pressure as the baseline, we need to subtract the atmospheric pressure from [tex]\( P_2 \)[/tex]. Assuming atmospheric pressure is [tex]\( 101.3 \, \text{kPa} \) (or \( 101300 \, \text{Pa} \))[/tex]:
[tex]\[ P_{\text{gauge}} = \frac{P_2 - \text{atmospheric pressure}}{\text{atmospheric pressure}} \]\[ P_{\text{gauge}} = \frac{-65204.74 - 101300}{101300} \]\[ P_{\text{gauge}} \approx -1.64 \][/tex]
So, the gauge pressure in the constricted segment is approximately [tex]\( -1.64 \)[/tex] atmospheres.
int[] numList = {2,3,4,5,6,7,9,11,12,13,14,15,16}; int count=0; for(int spot=0; spot
Hi, there is not much information about what do you need to do, but base on the C++ code you need to complete it to count the number of items in the array, using the instructions already written.
Answer:
#include <iostream>
using namespace std;
int main()
{
int numList [] = {2,3,4,5,6,7,9,11,12,13,14,15,16};
int count=0;
for(int spot=0; spot < (sizeof(numList)/sizeof( numList[ 0 ])); ++spot)
{
cout << numList[spot];
cout << "\n";
++count;
}
cout << "The number of items in the array is: ";
cout << count;
return 0;
}
Explanation:
To complete the program we need to finish the for statement, we want to know the number of items, we can get it by using this expression: (sizeof(numList)/sizeof( numList[ 0 ])), sizeof() function returns the number of bytes occupied by an array, therefore, the division between the number of bytes occupied for all the array (sizeof(numList)) by the number of bytes occupied for one item of the array (sizeof( numList[ 0])) equal the length of the array. While iterating for the array we are increasing the variable count that at the end contains the result that we print using the expression cout << "The number of items in the array is: "
Can the sum of the magnitudes of two vectors ever be equal to the magnitude of the sum of the same two vectors? If no, why not? If yes, when?
Answer:
They cannot be equal
Explanation:
Let 2 vector equal in magnitude and opposite direction. One of them is 1 and the other is -1
The sum of their magnitudes is 1 + 1 = 2
The magnitude of their sum of 2 vector is 0 since the 2 are in opposite directions, they cancel out each other and their final magnitude is thus 0.
So the sum of the magnitudes of two vectors cannot be equal to the magnitude of the sum of the same two vectors.