A traffic light hangs from a pole as shown in (Figure 1). The uniform aluminum pole AB is 7.20 m long and has a mass of 11.0 kg . The mass of the traffic light is 22.0 kg .
-Determine the tension in the horizontal massless cable CD.
-Determine the vertical component of the force exerted by the pivot A on the aluminum pole.
-Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.
Answers
There is a 409.30 N force in the horizontal massless cable CD.
The vertical component of the pivot A's force on the aluminum pole, which is 323.73N.
The force that the pivot A applies to the aluminum pole has a horizontal component of 409.30 N.
Force is any influence to the motion of the body .
Basically the product of mass and acceleration.Length of the pole = 7.20 m
Mass of the pole = m = 11.0 kg
Mass of the traffic light = M = 22.0 kg
Let the length of the rod AD be L
Also [tex]Lsin\alpha = 3.8[/tex]
[tex]\alpha = 37[/tex]°
[tex]Sin\alpha = 0.602[/tex]
L = [tex]\frac{3.8}{0.602}[/tex]
L = 6.312 m
In order to determine the tension (T) in the cable, the free body diagram will give the detailed information.
On balancing all the forces in both the 'x' and 'y' direction, i.e. summation of all the forces in 'x' and 'y' direction must equal to zero.∑[tex]F_{x}[/tex] = 0
[tex]R_{x}[/tex] - T = 0
[tex]R_{x}[/tex] = T
∑[tex]F_{y}[/tex] = 0
[tex]R_y- W_p - W_l = 0[/tex]
[tex]W_p[/tex] = 9.81 * 11 = 107.91 kg
[tex]W_l\\[/tex] = 9.81 * 22 = 215.82 kg
[tex]R_y\\[/tex] = [tex]W_p + W_l[/tex]
[tex]R_y\\[/tex] = 107.91 + 215.82
[tex]R_y\\[/tex] = 323.73 N
The vertical component of the force exerted by the pivot A on the aluminum pole is 323.73 N
Taking moment along x-axis
M = 0
[tex]Th-mgCos\alpha* \frac{1}{2} - mglCos\alpha[/tex] = 0
T = [tex]\frac{glCos\alpha }{h} *\frac{m}{2}*M[/tex]
T = 409.30 N
The tension in the horizontal massless cable CD is 409.30 N
[tex]R_{x}[/tex] = T = 409.30 N
The horizontal component of the force exerted by the pivot A on the aluminum pole is 409.30 N.
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Related Questions
21. An object m is tied to one end of a string, moves in a circle with a constant speed v
on a horizontal frictionless table. The second end of the string is connected to a big
mass M and goes through a small hole in the table. What is the value of M if it stays
in equilibrium?
I
(B) v²/rmg
(A) mv²/rg
(C) rg/mv²
(D) mv²r/g
Answers
The value of M that goes through a small hole in the table if it stays in equilibrium is mv²/rg ( A )
The force acting on object m is the centripetal force.
[tex]F_{c}[/tex] = m v² / r
Mass = m
Velocity = v
Radius = r
The force acting on object M is the gravitational force,
[tex]F_{g}[/tex] = M g
g = Acceleration due to gravity
Since the system is at equilibrium,
[tex]F_{c}[/tex] - [tex]F_{g}[/tex] = 0
m v² / r = M g
M = m v² / r g
Therefore, the value of M if it stays in equilibrium is mv²/rg ( A )
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carts, bricks, and bands
2. Which of the following conclusions are specifically supported by the data in Table 1?
a. A constant mass causes the acceleration value to increase.
b. An increase in the number of bricks causes the acceleration to decrease.
c. An increase in the length of the rubber band causes the acceleration to increase.
d. An increase in the number of rubber bands causes an increase in the acceleration.
Answers
The conclusions that are specifically supported by the data in Table 1 is that An increase in the number of rubber bands causes an increase in the acceleration. That is option D.
What is acceleration?Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).
From the table given,
Trial 1 ----> 1 band = 0.24m/s²
Trial 2 ----> 2 bands = 0.51 m/s²
Trial 3 ----> 3 bands = 0.73 m/s²
Trial 4 -----> 4 bands = 1.00 m/s²
This clearly shows that increase in the number of bands increases the acceleration of one brick that was placed on the cart.
This is because increasing the number of rubber bands has the effect of doubling the force leading to an effective increase in velocity of the moving cart.
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part 2 of 2 ASSUME BOTH snowballs are thrown with the same initial speed 39.9 m/s. the first snowball is thrown at an angle of 51 degrees above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? how many seconds after the first snowball should you throw the second so that they arrive on target at the same time?
Answers
Explanation
Step 1
Let
a) for ball 1
[tex]\begin{gathered} \text{ Initial sp}eed=v_0=33.9\text{ }\frac{m}{s} \\ \text{ Angle=51 \degree} \end{gathered}[/tex]the formula for the distance is given by:
[tex]x=\frac{v^2_0\sin(2\theta)}{g}[/tex][tex]\begin{gathered} \text{hence, let v}_0=39.9,\text{ angle= 51 \degree , g=9.8 } \\ \text{replace to solve for x } \\ x=\frac{(39.9)^2\sin(2\cdot51)}{9.8} \\ x=158.9\text{ m} \\ \end{gathered}[/tex]hence, the horizontal distance reached by the ball 1 is 158.9 meters
Step 2
as the ball started from the same point at the same initial speed, the only way to make the second ball hits the same point as the first is thworing the second ball at the same angle, it is 51 °
sin ([3pi over 2] + x) + sin ([3pi over 2] + x) = -2must show work also
Answers
ANSWER:
[tex]x=0\text{\degree}+360\text{\degree{}n}[/tex]STEP-BY-STEP EXPLANATION:
We have the following equation:
[tex]sin\: \mleft(\mleft[\frac{3\pi}{2}\mright]+x\mright)\: +\: sin\: \mleft(\mleft[\frac{3\pi\:}{2}\mright]+x\mright)\: =\: -2[/tex]Solving for x:
[tex]\begin{gathered} 2\cdot sin\: (\lbrack\frac{3\pi}{2}\rbrack\: +\: x)\: \: =\: -2 \\ sin\: (\lbrack\frac{3\pi}{2}\rbrack\: +\: x)=-\frac{2}{2} \\ sin\: (\lbrack\frac{3\pi}{2}\rbrack\: +\: x)=-1 \\ \frac{3\pi}{2}+\: x=\arcsin (-1) \\ \frac{3\pi}{2}+\: x=\frac{3\pi}{2}+\: 2\pi n \\ x=2\pi n \\ x=0\text{\degree}+360\text{\degree{}n} \\ \text{for n = 0} \\ \end{gathered}[/tex]A car is being tested for safety by colliding it with a brick wall.The 1300 kg car is initially driving towards the wall at a speedof 15 m/s, and after colliding with the wall the car movesaway from the wall at 2 m/s. If the car is in contact with thewall for 0.5 s, calculate the average force exerted on the carby the wall.
Answers
Answer:
44200 N
Explanation:
To calculate the average force exerted on the car, we will use the following equation
[tex]\begin{gathered} Ft=\Delta p \\ Ft=m(v_f-v_i) \end{gathered}[/tex]Where F is the average force, t is the time, m is the mass, vf is the final velocity and vi is the initial velocity of the car.
Replacing t = 0.5s, m = 1300 kg, vf = -2 m/s, and vi = 15 m/s and solving for F, we get
[tex]\begin{gathered} F(0.5s)=(1300\text{ kg\rparen\lparen-2 m/s - 15 m/s\rparen} \\ F(0.5s)=(1300\text{ kg\rparen\lparen-17 m/s\rparen} \\ F(0.5s)=-22100\text{ kg m/s} \\ F=\frac{-22100\text{ kg m/s}}{0.5\text{ s}} \\ F=-44200\text{ N} \end{gathered}[/tex]Therefore, the average force exerted on the car by the wall was 44200 N
What are the answers for a, b and c in MJ?
Answers
Given:
The orbital height of the satellite, h=94 km=94000 m
The mass of the satellite, m=1045 kg
The new altitude of the satellite, d=207 km=207000 m
To find:
a) The energy needed.
b) The change in the kinetic energy.
c) The change in the potential energy.
Explanation:
The radius of the earth, R=6.37×10⁶ m
The mass of the earth, M=6×10²⁴ kg
a) The orbital velocity is given by,
[tex]v=\sqrt{\frac{GM}{r}}[/tex]Where G is the gravitational constant and r is the radius of the satellite from the center of the earth.
Thus the initial orbital velocity of the earth,
[tex]\begin{gathered} v_1=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37×10^6+94000)}} \\ =7868.43\text{ m/s} \end{gathered}[/tex]The orbital velocity after changing the altitude is,
[tex]\begin{gathered} v_2=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37\times10^6+207000)}} \\ =7800.5\text{ m/s} \end{gathered}[/tex]Thus the total energy needed is given by,
[tex]E=(\frac{1}{2}mv_2^2-\frac{GMm}{(R+d)})-(\frac{1}{2}mv_1^2-\frac{GMm}{(R+h)})[/tex]On substituting the known values,
[tex]\begin{gathered} E=1045[(\frac{1}{2}\times7868.43^2-\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37\times10^6+207000)})-(\frac{1}{2}\times7800.5^2-\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37×10^6+94000)})] \\ =623\text{ MJ} \end{gathered}[/tex]b)
The change in the kinetic energy is given by,
[tex]\begin{gathered} KE=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 \\ =\frac{1}{2}m(v_2^2-v_1^2) \end{gathered}[/tex]On substituting the known values,
Two points ___________ create a line.A. sometimesB. neverC. alwaysD. not enough information
Answers
According to the Euclidean Postulates, a straight line segment can be drawn joining any two points.
Therefore, the answer is:
[tex]C)\text{ Always}[/tex]A 77 kg student traveling in a car with a constant velocity has a kinetic energy of 1.7 104 J. What is the speedometer reading of the car in km/h? answer with:_____km/h
Answers
ANSWER
75.65 km/h
EXPLANATION
Given:
• The student's mass, m = 77 kg
,• The kinetic energy of the student in the car, KE = 1.7 x 10⁴ J
Find:
• The speed read in the speedometer of the car, which is the speed of the student, v (in km/h)
The kinetic energy of an object with mass m, traveling at a speed v, is,
[tex]KE=\frac{1}{2}mv^2[/tex]Solving for v,
[tex]v=\sqrt{\frac{2KE}{m}}[/tex]Replace the known values and solve,
[tex]v=\sqrt{\frac{2\cdot1.7\cdot10^4J}{77kg}}\approx21.013m/s[/tex]Note that because the kinetic energy is given in Joules - which is equivalent to kg*m²/s², the speed we found is in m/s. Now, knowing that there are 3600 seconds in 1 hour and that 1 km is equivalent to 1000 m, we can convert this to km/s,
[tex]v=21.013\frac{m}{s}\cdot\frac{3600s}{1h}\cdot\frac{1km}{1000m}\approx75.65km/h[/tex]Hence, the speedometer reading of the car is 75.65 km/h, rounded to the nearest hundredth.
25. A student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m. What is the average braking force?
Answers
If a student bikes at a pace of 5.0 m/s down a straight route. The student's bicycle weighs 120 kg in total. The pupil puts on the brakes and comes to a stop. The average braking force of the automobile would be 150 newtons if the stopping distance were 10 meters.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
Work done by the braking force of the cycle = change in kinetic energy of the student
Force × distance = 1/2 × mass × velocity²
F = 0.5 x 120 x 5² / 10
F = 150 Newtons
Thus, the average braking force of the cycle would be 150 Newtons.
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Two sounds of 610Hz and 620Hz are played together. The frequency you hear will be ___. The beat frequency is ___
Answers
Explanation
The beat frequency is equal to the complete value of the alteration in the frequency of the two waves,it can be calculated by using the expression
[tex]f_{beat}=\lvert{{f_1}-f_2}\rvert[/tex]so
Step 1
a)let
[tex]\begin{gathered} f_1=610\text{ Hz} \\ f_2=620\text{ Hz} \end{gathered}[/tex]b) replace and calculate
[tex]\begin{gathered} f_{beat}=\lvert{{f_1}-f_2}\rvert \\ f_{beat}=\lvert610\text{ Hz-620Hz}\rvert \\ f_{beat}=10\text{ Hz} \end{gathered}[/tex]so
the beat frequency is 10 HZ
Step 2
the frequency yuo will detect is the difference of the frequency, so , teh frequency beat, so
the frequency you hear will be 10 Hz
I hope this helps you
read through the different scenarios below and choose what kind of energy transfer is happening for each one. note: each answer will be used 1 time.
Answers
1. A ceiling fan plugged into the wall socket.
The input is the electrical energy and the output is the mechanical energy from the rotation of the fan blades.
So, the energy transfer must be electrical to mechanical.
2. A stove heating a pot of water for pasta
The input is the light energy from the stove and the output is the thermal energy to heat the water.
So, the energy transfer must be light to thermal.
3. A wind turbine generating power
The input is the mechanical energy from the rotation of the turbine blades and the output is the electrical energy that is being generated.
So, the energy transfer must be mechanical to electrical.
4. A person eats a snack when they are hungry and has the energy to take a walk.
The input is the chemical energy that the person gets from eating food and the output is the mechanical energy (doing physical work).
So, the energy transfer must be chemical to mechanical.
5. A solar panel is used to create power
The input is the light energy from the Sun and it is being converted into electrical energy to generate power.
So, the energy transfer must be light to electrical.
6. A microwave is used to heat up a burrito.
The input to the microwave is the electrical energy and the output is the thermal energy to heat up the burrito.
So, the energy transfer must be electrical to thermal.
7. A lamp is plugged into the wall and shines down on a room
The input to the lamp is electrical energy which is being converted into light energy.
So, the energy transfer must be electrical to light.
8. A plant grows by absorbing the sun's rays
The input to the plant is the nuclear energy from the sun and the output is the chemical energy.
So, the energy transfer must be nuclear to chemical.
9. A game controller uses batteries to power it.
The batteries provide chemical energy as input and the output is the electrical energy to power the controller.
So, the energy transfer must be chemical to electrical.
10. The sun heats up a person sunbathing at the beach.
The input is the nuclear energy from the sun and output is the thermal energy by absorbing the heat.
So, the energy transfer must be nuclear to thermal.
Timmy walks 5 m North, 3m West, and finally 1 m South. What is his displacement from his starting point?
Answers
Timmy walks 5 m North, 3m West, and finally 1 m South then his displacement from the starting point would be 5 meters in the northwest direction.
What is displacement?Displacement describes this shift in location and it is calculated with the help of the initial and the final position of the object.
As given in the problem If Timmy walks 5 m North, 3m West, and finally 1 m South ,
The resultant displacement of the Timmy = √(4² + 3²)
= 5 meters
Thus, the resultant displacement of the Timmy would be 5 meters
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Which of the following is an appropriate measure of electric power on a toaster label?220 W55 Ω110 V2.0 A
Answers
A Watt is the unit of electrical power
. Chloe performed an experiment on the amount of a certain gas that can be dissolved in water at different temperatures. Her data from the experiment is shown on the graph below. What type of function would best fit this set of data?
Answers
The values on the curve corresponding to the y axis are the y values
The values on the curve corresponding to the x axis are the x values
We would pick corresponding x and y values on the graph and compare them
On the graph, each small square represents 1 unit on both axes.
If x = 1, y = 1
If x = 4, y = 2
if x = 9, y = 3
We know that the square root of 1 is 1, the square root of 4 is 2 and the square root 9 is 3. This means that
y is the square root of x
The function is
[tex]y\text{ = }\sqrt[]{x}[/tex]Thus, it is a square root function
The first option is correct
Mechanical energy is the form of energy associated with the ,or of an object.
Answers
Mechanical energy is the form of energy associated with the motion or position of an object.
Explanation:Mechanical energy can be defined as the energy possessed by an object as a result of its motion or position.
Mechanical energy is divided into two.
Kinetic energy and potential energy
Kinetic energy is the energy possessed by an object as a result of its motion
Potential energy is the energy possessed by an object as a result of its position.
Therefore, we can conclude that mechanical energy is the form of energy associated with the motion or position of an object.
Hello please could you help me with the physics ?
Answers
In order to calculate the amount of energy needed, we can use the formula below:
[tex]Q=m\cdot c\cdot\Delta T[/tex]Where Q is the amount of energy in Joules, m is the mass in kg, c is the specific heat coefficient and DeltaT is the change in temperature.
So, for m = 0.8, c = 4.2*10^3 and DeltaT = 85 (from 15°C to 100°C), we have:
[tex]\begin{gathered} Q=0.8\cdot4.2\cdot10^3\cdot85 \\ Q=285600\text{ J} \end{gathered}[/tex]Then, for the energy needed to evaporate the water, we have a similar formula:
[tex]Q=m\cdot L[/tex]Where L is the latent heat. So, for L = 2.3 * 10^6, we have:
[tex]\begin{gathered} Q=0.8\cdot2.3\cdot10^6 \\ Q=1840000\text{ J} \end{gathered}[/tex]A hydroelectric plant takes energy from water and turns it into electrical energy.What are the transformations of energy in the water molecules that are used in theprocess of generating electricity this way?The water particles initially have kinetic energy due to their motion. This kineticenergy is transformed into potential energy due to the position of the water andthen this energy is used to produce electricity.The water particles initially have potential due to their position. This potentialenergy is transformed into thermal energy and then this energy is used toproduce electricity.The water particles initially have chemical energy due to the bonds in water. Thischemical energy is transformed into thermal and then the thermal energy is usedto produce electricity.The water particles initially have potential due to their position. This potentialenergy is transformed into kinetic energy due to the motion of energy and thenthe kinetic energy is used to produce electricity.
Answers
Answer:
The last option.
Explanation:
The principle of hydroelectric dams is the following:
Basically, we utilizie the energy of the falling wat
Instructions: 1) write out the question 2) work out the solution 3) Explain in words how you would know to do that1. What is the momentum of a 5.0 g bullet with a velocity of 500 m/s?
Answers
Given data:
The mass of bullet is m=5.0 g.
The velocity fo bullet is v=500 m/s.
The formula for the momentum is given by,
[tex]p=mv[/tex]Substitute the given values in above equation,
[tex]\begin{gathered} p=(5g\times\frac{1kg}{1000\text{ g}})(\frac{500m}{s}) \\ p=\frac{2.5kgm}{s} \end{gathered}[/tex]Thus, the momentum of the bullet is 2.5 kgm/s.
According to the definition of momentum, "It is the product of mass and velocity". Therefore, the momentum of moving body can be calculated by multiplying the mass of the body and velocity at which the body is moving.
So, that is how we know how to calculate the momentum of the body.
What total energy (in J) is stored in the capacitors in the figure below (C1 = 0.900 µF, C2 = 16.0 µF) if 1.80 10-4 J is stored in the 2.50 µF capacitor?
Answers
The total energy stored in the capacitors is determined as 2.41 x 10⁻⁴ J.
What is the potential difference of the circuit?The potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
where;
C is capacitance of the capacitorV is the potential differenceFor a parallel circuit the voltage in the circuit is always the same.
The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
2U = CV²
V = √2U/C
V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)
V = 12 V
The equivalent capacitance of C1 and C2 is calculated as follows;
1/C = 1/C₁ + 1/C₂
1/C = (1)/(0.9 x 10⁻⁶) + (1)/(16 x 10⁻⁶)
1/C = 1,173,611.11
C = 1/1,173,611.11
C = 8.52 x 10⁻⁷ C
The total capacitance of the circuit is calculated as follows;
Ct = 8.52 x 10⁻⁷ C + 2.5 x 10⁻⁶ C
Ct = 3.35 x 10⁻⁶ C
The total energy of the circuit is calculated as follows;
U = ¹/₂CtV²
U = ¹/₂(3.35 x 10⁻⁶ )(12)²
U = 2.41 x 10⁻⁴ J
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Car A is traveling with a constant velocity of 18 [m/s]. Car B speeds up from 0 [m/s] to 10 [m/s] in 4 seconds. Which car has a greater acceleration?Car ACar BCar A and Car B have the same accelerationNeither car is accelerating
Answers
Given:
Car A is traveling with a constant velocity of,
[tex]18\text{ m/s}[/tex]The initial speed of car B is,
[tex]v_i=0\text{ m/s}[/tex]After t=4 s, carB's speed is,
[tex]v_f=10\text{ m/s}[/tex]To find:
Which car has a greater acceleration
Explanation:
The acceleration of Car A is Zero as there is no change in velocity with time.
The acceleration of car B is,
[tex]\begin{gathered} a_B=\frac{v_f-v_i}{t} \\ =\frac{10-0}{4} \\ =2.5\text{ m/s}^2 \end{gathered}[/tex]So, Car B has greater acceleration.
Hence, Car B has greater acceleration.
A light, inextensible cord passes over alight, frictionless pulley with a radius of15 cm. It has a(n) 18 kg mass on the left and a(n) 2.6 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 1.5 m apart.The acceleration of gravity is 9.8 m/s².
At what rate are the two masses accelerating when they pass each other answer in units of m/s^2
Answers
Answer:
quizlet
Explanation:
they help
2. During which Epoch did humans first appear?Holocene3. How many million years ago did humans first appear?2.8 million years ago4. What are the three periods of the Mesozoic Era?Cretaceous, Jurassic, Triassic5. When did dinosaurs first appear?Triassic period6. In what period did birds first appear?Jurassic7. During what period did the dome-like uplift of the Adirondack region begin?
Answers
(1)
The three eras within the Phanerozoic eon are
1) The Paleozoic era (541 million to 252 million years ago)
2) The Mesozoic era (252 million to 66 million years ago)
3) The Cenozoic era (66 million years ago to the present)
A 4-kg ball traveling westward at 25 m/s hits a 15-kg ball at rest. The 4-kg ball bounces east at 8.0 m/s. What is the speed and direction of the 15-kg ball? What is the impulse of the second ball?
Answers
Given:
The mass of the first ball is,
[tex]m_1=4\text{ kg}[/tex]The initial velocity of the first ball towards West is,
[tex]u_1=25\text{ m/s}[/tex]The mass of thr second ball is,
[tex]m_2=15\text{ kg}[/tex]the second object is initially at rest.
The final velocity of the first ball is,
[tex]v_1=-8.0\text{ m/s}[/tex]we are taking West as positive.
Applying momentum conservation principle we can write,
[tex]m_1u_1+m_2\times0=m_1v_1+m_2v_2[/tex]Substituting the values we get,
[tex]\begin{gathered} 4\times25+0=4\times(-8.0)+15\times v_2 \\ v_2=\frac{100+32}{15} \\ v_2=8.8\text{ m/s} \end{gathered}[/tex]THe final velocity of the second ball is towards East and the magnitude is 8.8 m/s.
The impulse of the Second ball is,
[tex]\begin{gathered} I=m_2v_2-m_2\times0 \\ =15\times8.8 \\ =132\text{ kg.m/s} \end{gathered}[/tex]what is the maximum efficiency that a heat engine could have when operating between the normal boiling and freezing temperatures of water
Answers
The maximum efficiency that a heat engine could have when operating between the normal boiling and freezing temperatures of water is 26.8 %
η = ( [tex]T_{H}[/tex] - [tex]T_{C}[/tex] ) / [tex]T_{H}[/tex] * 100
η = Efficiency
[tex]T_{H}[/tex] = Hottest temperature
[tex]T_{C}[/tex] = Coldest temperature
Hottest temperature = Boiling point
Coldest temperature = Freezing point
[tex]T_{H}[/tex] = 100 °C = 373 K
[tex]T_{C}[/tex] = 0 °C = 272 K
η = ( 373 - 273 ) / 373 * 100
η = 100 / 373 * 100
η = 26.8 %
In a heat engine, the heat energy is converted into mechanical energy which will be used to do mechanical work like pushing a piston out from the cylinder.
Therefore, the maximum efficiency that a heat engine could have when operating between the normal boiling and freezing temperatures of water is 26.8 %
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Look at Figure 13-17 to answer the question.Will the acceleration of the piano be greater in A or in B? Use Newton's second law ofmotion to explain your answer.
Answers
Answer:
Explanation:
Newton's second law states that the applied force is directly proportional to the rate of change of momentum
momentum = mass x velocity
Rate of change in momentum = (final momentum - initial momentum)/time
Rate of change in momentum = m(v - u)/t
where
v is final velocity
u is initial velocity
t is times
Recall,
Acceleration = (v - u)/t
Thus, the equation becomes
Force = ma
This means that as the force increases, the acceleration increases.
Considering the given scenarios,
First piano is pushed by force from the man
For second piano, there is additional force from the woman and the mass of the piano remains the same. Since there is more force, there is more acceleration. Thus,
Acceleration is greater in B
A student on skateboard pushes off from the top of small hill with a apees of 2.0m/s, and then geos down the hill with a constant acceleration of 0.5 m/s2
After traveling a distancie 12.0m, how fast is the student going?
Answers
The final velocity of the student after travelling 12 m is 4 m/s.
What is the final velocity of the student?
The final velocity of the student is determined by applying the following Kinematic equation.
v² = u² + 2as
where;
u is the initial velocity of the studentv is the final velocity of the studenta is the acceleration of the students is the distance travelled by the studentv² = (2)² + 2(0.5)(12)
v² = 16
v = √16
v = 4 m/s
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find an equation of the line with y intercept (0,7) and the slope of 1/2
Answers
Consider that a general way of writing a line equation is:
y = mx + b
where m is the slope of the line and b is the y coordinate of the y-intercept of the line.
Then, by using the given information:
m = 1/2
b = 7
You have the following equation of line:
y = 1/2*x + 7
The density of copper is 8.94 g/cm^3. What is the mass of a rectangular sheet of copper: 10 cm wide, 45 cm long, and 0.2 cm thick? I have no idea how to solve for mass or where to start. :(
Answers
Mass = 804.6 g
Explanation:The dimension of the rectangular copper = 10 cm wide, 45 cm long, and 0.2 cm thick
The volume of the rectangular copper = 0.1m x 0.45m x 0.002m
The volume of the rectangular copper = 0.00009 m³
The density of copper = 8.94 g/cm³ = 8.94 x 1000 kg/m³
The density of copper = 8940 kg/m³
Mass = Density x Volume
Mass = 8940 x 0.00009
Mass = 0.8046 kg
Mass = 804.6 g
What is the difference in the path of a ball tossed straight up in the air by a passenger on a a bus from the view point of the passenger, and the person on the street?
Answers
The difference in the path of a ball tossed straight up in the air by a passenger on a a bus from the view point of the passenger, and the person on the street is that the ball would appear to the passenger to be making an up and down movement when the ball is thrown up, while the stationary observer will actually see the ball moving along a parabolic path.
What is a parabolic path?A parabolic path is described as a Kepler orbit with the eccentricity equal to 1 and is an unbound orbit that is exactly on the border between elliptical and hyperbolic.
The Parabolic path is also defined as the angle of trajectory of a projectile.
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Determine the speed of the Earth in its motion around the Sun using Newton's Law of Universal Gravitation and centripetal force. Look up the values of the Earth's mass, the Sun's mass, and the average distance of Earth from the Sun; other than G, nothing else is needed
Answers
In order to determine the speed of the Earth, proceed as follow:
Consider that the centripetal force must be equal to the gravitational force between the Earth and the Sun (because guarantees the stability of the system):
[tex]F_g=F_c[/tex]Fg is the gravitational force and Fc the centripetal force. The expressions for each of these forces are:
[tex]\begin{gathered} F_g=\text{G}\frac{\text{mM}}{r^2} \\ F_c=ma_c=m\frac{v^2}{r} \end{gathered}[/tex]where,
G: Cavendish's constant = 6.67*10^-11 Nm^2/kg^2
m: Earth's mass = 5.97*10^24 kg
M: Sun's mass = 1.99*10^30kg
v: speed of Earth around the Sun = ?
r: distance between the center of mass of Earth and Sun = 1.49*10^8km = 1.49*10^11 m
Equal the expressions for Fg and Fc, solve for v, replace the previous values of the parameters and simplify:
[tex]\begin{gathered} \text{G}\frac{\text{mM}}{r^2}=m\frac{v^2}{r} \\ v^{}=\sqrt[]{\frac{GM}{r}} \\ v=\sqrt[]{\frac{(6.67\cdot10^{-11}N\frac{m^2}{\operatorname{kg}^2})(1.99\cdot10^{30}kg)}{1.49\cdot10^{11}m}} \\ v\approx29846.7\frac{m}{s} \end{gathered}[/tex]Hence, the speed of the Earth around the Sun is approximately 29846.7m/s
