ANSWER:
a. 9.09cm
STEP-BY-STEP EXPLANATION:
Given:
Focal length = 100 cm
Distance image = -10 cm
We use the following formula to calculate the value of the distance of the object:
[tex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/tex]We substitute each value and calculate the distance of the object just like this:
[tex]\begin{gathered} \frac{1}{d_o}+\frac{1}{-10}=\frac{1}{100} \\ \\ \frac{1}{d_{o}}=\frac{1}{100}+\frac{1}{10} \\ \\ \frac{1}{d_o}=\frac{11}{100} \\ \\ d_o=\frac{100}{11} \\ \\ d_o=9.09\text{ cm} \end{gathered}[/tex]Therefore, the correct answer is option a. 9.09cm
find the pressure increase in a fluid when a force of 25 N is exerted on a closed syringe where the piston radius is 2 cm.
Given:
• Force, F = 25 N
,• Radius, r = 2 cm
Let's find the pressure increase.
To find the pressure increase, apply the formula:
[tex]\Delta P=\frac{Force}{Area}=\frac{F}{\pi r^2}[/tex]Where:
F = 25 N
r is the radius in meters = 2 cm
P is the pressure
Convert the radius from cm to meters.
Where:
100 cm = 1 m
2 cm = 2 /100 = 0.02 m
Hence, we have:
[tex]\begin{gathered} \Delta P=\frac{25}{\pi(0.02)^2} \\ \\ \Delta P=\frac{25}{\pi(0.0004)} \\ \\ \Delta P=\frac{25}{0.00125664} \\ \\ \Delta P=19894.4\text{ Pa}\approx1.99\times10^4\text{ Pa} \end{gathered}[/tex]Therefore, the pressure increase is 1.99 x 10⁴ Pa.
ANSWER:
1.99 x 10⁴ Pa
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
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.
An installation consists of a 30-kVA, 3-phase transformer, a 480-volt primary, and a 240-volt secondary. Calculate the largest standard size circuit breaker permitted for primary-only protection to be used without applying Note 1 of Table 450.3(B).
Answer: 45 A
Explanation:
Primary only protection 3-phase
I = 3 phase kVA / ( 1.723 * V)
I = 30000 / ( 1.732 * 480 ) = 36.085 A
Table 450.3(B)
Currents of 9A or more column
primary only protection = 125%
Max OCPD pri = 125% of I = 1.25 * 36.085 = 45.11 A
Table 450.3(B) Note 1 does not apply, use next smaller Table 240.6(A)
Next smaller = 45 A
Sound travels faster on a cold day than on a warm day. Is this true or false?
ANSWER
False
EXPLANATION
Sound waves travel by causing vibrations in particles in the direction of the wave and these vibrations can travel through any state of matter (solid, liquid, gas).
This means that if the vibrations occur faster, the sound waves can also be transmitted faster. On a warm day, the temperature is higher, which means that molecules of air vibrate faster, resulting in faster transmission of sound waves.
Therefore, it is false that sound travels faster on a cold day than on a warm day.
How much heat in kcal must be added to 0.68 kg of water at room temperature (20°C) to raise its temperature to 45°C?answer:____ kcal
Given:
Mass, m = 0.68 kg
Initial temperature, T1 = 20°C
Final temperature, T2 = 45°C
Let's find the amount of heat needed.
Apply the Specific Heat Capacity formula:
[tex]\begin{gathered} Q=mc\Delta T \\ \\ Q=mc(T_2-T_1) \end{gathered}[/tex]Where:
m is the mass = 0.68 kg
c is the specific heat of water = 4.187 J/kg °C−1
T1 = 20°C
T2 = 45°C
Plug in the values and solve for Q:
[tex]\begin{gathered} Q=0.68*4.187*(45-20) \\ \\ Q=2.84716(25) \\ \\ Q=71.179\text{ kJ} \end{gathered}[/tex]Where:
1 kJ = 0.239 kCal
Since the answer is to be in kCal, we have:
[tex]Q=71.179*0.239=17.01\text{ kCal}[/tex]Therefore, the amount if heat added is 17.01 kCal.
ANSWER:
17.01 kCal
What is the kinetic energy of the ocean liner ?
In order to calculate the kinetic energy, we can use the formula:
[tex]E_k=\frac{mv^2}{2}[/tex]Where m is the mass in kg and v is the speed in m/s.
So, for the bullet, we have:
[tex]E_k=\frac{0.0014\cdot408^2}{2}=116.5\text{ J}[/tex]And for the liner, we have:
[tex]E_k=\frac{59000000\cdot12^2}{2}=4248000000\text{ J}[/tex]So the ocean liner has greater kinetic energy.
Two parallel wires carry currents of 20 A and 10 A, respectively, in the opposite direction and have a length of 5 m each. Determine the force between the wires if the distance between the wires is 1.0 cm.0.010 N, attraction0.010 N, repulsion0.020 N, attraction0.020 N, repulsion
Given:
The currents, I₁=20 A
I₂=10 A
The length of the wires, L=5 m
The distance between two wires, d=1.0 cm=0.01 m
To find:
The magnetic force between the wires.
Explanation:
As the wires are conducting the current in opposite directions, the nature of the force between the wires is repulsive.
The magnitude of the force between two wires is given by,
[tex]F=\frac{\mu_0I_1I_2L}{2\pi d}[/tex]Where μ₀=4π×10⁻⁷ H/m is the permeability of free space.
On substituting the known values,
[tex]\begin{gathered} F=\frac{4\pi\times10^{-7}\times20\times10\times5}{2\pi\times1\times10^{-2}} \\ =0.02\text{ N} \end{gathered}[/tex]Final answer:
The magnetic force between wires is 0.020 N, repulsive.
Hello, I am having a hard time understanding this question, is it possible for you to help me?
Answer: h = 14 m
Explanation:
From the information given,
weight of rocket = 10N
Recall, weight = mg
Thus, mg = 10
where
m = mass of object
g = acceleration due to gravity and its values is 9.8 m/s^2
At the exact top of the trajectory, the total mechanical energy = 140 J
Recall,
total mechanical energy = kinetic energy + potential energy
At the exact top, kinetic energy = 0
Thus,
140 = potential energy
Recall, potential energy = mgh
where h is the height of the object(in this case, h is the height at the exact top and its value is 140J). Thus,
140 = mgh
Substituting mg = 10, we have
140 = 10h
h = 140/10
h = 14 m
Jason is pulling a box across the room. He is pulling with a force of 24 newtons and his arm is making a 44 angle with the horizontal, what is the horizontal component of the forcehe is pulling with
The picture below shows a person swinging a toy plane attached to a string in
uniform circular motion.
Which vector points in the direction of the centripetal acceleration of the
plane?
According to the image A vector points in the direction of the centripetal acceleration of the plane.
The correct option is C.
What is centripetal acceleration?Centripetal acceleration is a property of an object's motion along a circular path. Centripetal force refers to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.
Briefing:A person is seen in the image swinging a toy plane on a string in a smooth, circular motion. The velocity direction is tangent to the circular orbit and perpendicular to the direction of the position.
The direction of the velocity and the speed may both vary as an object travels in a circular orbit. The velocity's direction is continually shifting. While acceleration is constantly moving uniformly in a circular path toward the recent, tangent at each point indicates the direction of velocity at that place.
As a result, vector A is pointed in the plane's centripetal acceleration's direction.
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The complete question is -
The picture below shows a person swinging a toy plane attached to a string in uniform circular motion.
Which vector points in the direction of the centripetal acceleration of the
plane?
A-B
B-D
C-A
D-C
PLEASE HELP :((
A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
Energy effectiveness would be a term that refers to the proportion of input power over output. Power generation, as well as simply energy utilization, is the process of reducing the amount of energy used to produce goods and services.
Considering that,
A cookie contains 54.0 kcal of energy. An athlete utilizes the 54.0 kcal inside this cookie from input energy.
The following diagram illustrates the relationship among input as well as output energy:
Efficiency = output energy / input energy...(i)
Output energy = efficiency × input energy
By using equation (i)
⇒ output energy = 0.25 × 54 kcal = 13.5 kcal.
The lifting exercise has been performed n times for the output energy.
In terms of potential energy, such output energy could be written as follows:
Mass × gravity ×height.
So, energy per repetition = mgh = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J = 0.059 kcal.
So, Count of repetitions = sim of output energy / energy per repetition..(ii)
By using equation (ii)
Count of repetitions = 13.5 kcal / 0.059 kcal =229 repetitions.
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Neptune circles the Sun at a distance of 4.50 × 1012 m once every 164 years. Saturn circles the Sun at a distance of 1.43 × 1012 m. What is the orbital period of Saturn?
The orbital time period of the Saturn is 29.6 years
We are given that,
Distance from Sun to Saturn is = a = 1.43 × 10¹²
The mass of the Sun is = M =1.99 × 10³⁰kg
The Gravitational constant = G = 6.67 × 10⁻¹¹N-m²kg⁻²
To find the orbital period of Saturn we can use the equation ,
[tex]T^{2} = \frac{4\pi }{GM}a^{3}[/tex]
Where, T is the orbital time period of the of the Saturn , M is the mass of the sun , G is the gravitational constant.
Therefore, after putting the value in above equation we can get,
[tex]T^{2} = \frac{4(\(3.14)^{2} }{(6.67*10)^{-11} )N-m^{2} kg^{-2}}(1.43*10^{12}) ^{3}m[/tex]
[tex]T^{2} = \sqrt{8.688*10^{17} } s[/tex]
[tex]T = 932094415.818s[/tex]
So that , from above to convert the orbital time period of Saturn from second into year i.e. above seconds divided by seconds (1 sec = 3.154 ×10⁷ Earth years)
Thus, the orbital time period can be ,
[tex]T = 29.6 years[/tex]
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Please help me, it's asking me here how does static energy work?
Static energy is the energy due to motionless state of a particle. If a particle is at rest it possess static energy also known as potential energy. When some force is applied to the particle then the static energy gets converted into kinetic energy of the particle.
The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208 m is mounted on the same rotating shaft as the second pulley.
Answer: It should be the answer beginning like this
The linear spread is that
The radial acceleration of locations along the blade's outer edge is approximately 17580 [tex]m/s^2[/tex].
What is Radial acceleration?Radial acceleration describes the acceleration of an object travelling on a circular path towards the circle's center. It can be defined as the rate of change of tangential velocity with regard to time and is also known as centripetal acceleration.
Given:
A table saw's engine rotates at 3450 revolutions per minute.A V-belt connects a pulley that's attached to the motor shaft to a second pulley half the diameter.A 0.208 m circular saw blade is installed on the same rotating shaft as the second pulley.We know that the motor is rotating at 3450 rev/min. One revolution is equal to 2π radians, so we can convert the motor speed to radians per minute:
ω₁ = (3450 rev/min) x (2π rad/rev) = 21675π rad/min
The second pulley is half the diameter of the first pulley, so its angular speed, ω₂, is twice that of ω₁:
ω₂ = 2ω₁ = 43350π rad/min
The circular saw blade is mounted on the same shaft as the second pulley, so it also rotates at the same angular speed:
ω = ω₂ = 43350π rad/min
We can now calculate the linear speed of the small piece of wood moving at the same rate as the rim of the circular saw blade, indicated by v. The circumference of the circle is supplied by the rim of the circular saw blade:
C = πd = π(0.208 m) = 0.6548 m
The linear speed of the little piece of wood is equal to the tangential speed of the circular saw blade's rim:
v = ωr
where r is the circular saw blade's radius, given by half its diameter:
r = d/2 = 0.208/2 = 0.104 m
By substituting the values, we obtain:
v = r = (43350 rad/min) x (0.104 m) x (1/60) = approx. 23.0 m/s
As a result, the linear speed of the little piece of wood moving at the same rate as the rim of the circular saw blade is about 23.0 m/s.
Next, compute the radial acceleration of locations on the blade's outer edge, represented by. The radial acceleration is calculated as follows:
α = rω²
By substituting the values, we obtain:
r2 = (0.104 m) x (43350 rad/min)2 x (1/602) = 17580 m/s2 (approximate)
Therefore, the radial acceleration of points on the outer edge of the blade is approximately 17580 m/s². This high radial acceleration explains why sawdust doesn't stick to the teeth of the saw blade.
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Your question is incomplete, most probably the complete question is:
The motor of a table saw is rotating at 3450{\rm rev/min}. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208{\rm m}is mounted on the same rotating shaft as the second pulley.
The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed?
v =_______________________ m/s
Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.
\alpha=______________________m/s2
Which one of the following is false? Select one: A. Wires resist the flow of current a little, components such as bulbs resist more B. Adding bulbs in the circuit means greater resistance C. Conductors allow current to flow relatively easily D. Metal and plastic are poor conductors E. The greater the resistance the less current flows
Metals are good conductor of electricity while plastics are bad conductors of electricity.
Thus, the statement D Metals and plastic are poor conductors is false.
Hence the correct option is D.
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.
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.
Use Newton’s Law of Universal Gravitation and Newton’s Second Law to
Find g = acceleration due to gravity
Show g is independent of mass
By equating the two forces, acceleration due to gravity g is obtained which is independent of mass
What is Newton’s Second Law ?The law state that the rate of change of momentum is directly proportional to the force applied.
From Newton’s Second Law, Force F = mg. And from Newton’s Law of Universal Gravitation, Force F = GMm/r²
Where m is the mass of the satellite or the body revolving round the earth.
Equate the two forces.
mg = GMm/r²
The two m cancelled out leaving
g = GM/r²
Where
g = Acceleration due to gravityM = Mass of the earthG = Universal gravitational constantr = Distance between them.Therefore, since the mass of the satellite m has cancelled out, acceleration due to gravity g is independent of mass.
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Pls quick will mark brainliest.
Lance is working in a library using a trolley to carry books. As he stacks books from the trolley on shelves, it gets easier to push or pull the trolley. Which part of Newton's laws of motion explains the increased ease of moving the trolley?
A. Objects at rest tend to stay at rest.
B. Objects in motion tend to stay in motion.
C. Every action has an equal and opposite reaction.
D. Larger objects require greater amounts of force to move.
Answer: D
Explanation:
Larger objects require greater amounts of force to move.
As he stacks books from the trolley on shelves, it gets easier to push or pull the trolley because the trolley gets lighter when the books are shifted from the trolley to the shelves. So, larger objects require greater amounts of force to move. Hence, option D is correct.
What is inertia?Inertia is a property of bodies that prevents them from moving or, if they are already moving, causes them to change the speed or direction of their motion. The inertia of a body is a passive quality that prohibits it from acting in any way besides opposing active agents like forces and torques.
A moving body continues to move not because of its inertia but rather because no force exists to stop it, alter its direction, or accelerate it.
According to the question, when there are books on the trolley they require a greater amount of force to move but when the books are shifted on the shelves it's easier to move the trolley. So inertia is proportional to mass.
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What the difference between velocity and speed
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.
Calculate the depth in the ocean at which thepressure is three times atmospheric pressure.Atmospheric pressure is 1.013 x 10^5 Pa. Theacceleration of gravity is 9.81 m/s^2and them/sdensity of sea water is 1025 kg/m^3Answer in units of m.
In order to determine the depth in the ocean, use the following equation:
[tex]h=\frac{P-P_o}{\rho g}[/tex]h: depth
P: pressure = 3*Po
Po: atmospheric pressure = 1.013*10^5Pa
g: gravitational acceleration constant = 9.8m/s^2
p: density of water = 1025 kg/m^3
Replace the previoua values into the formula for h and simplify:
[tex]\begin{gathered} h=\frac{3P_o-P_o}{\rho g}=\frac{2P_o}{\rho g} \\ h=\frac{2(1.013\cdot10^5Pa)}{(1025\frac{kg}{m^3})(9.8\frac{m}{s^2})}\approx20.17m \end{gathered}[/tex]Hence, the depth in the ocean is approximately 20.17m
Which is increased when the string of a stringed instrument is tightened?1) timbre2) pitch3) wavelength4) loudness
The pitch of a note is the same as its frequency.
Remember that te timbre depends on the shape and other properties of the vibrating object. The loudness is related to the amplitude of the wave, and the wavelength of sound is inversely proportional to the frequency.
On the other hand, the frequency of the sound of a string is proportional to the square root of the tension in the string. Then, if the tension increases (if the string is tightened), the frequency will also increase (and if the frequency increases, the wavelength decreases).
Therefore, the parameter that increases when a string of a stringed instrument is tightened, is the pitch.
Therefore, the answer is:
[tex]2)\text{ Pitch}[/tex]3. A rescuer jumped from an airship in the ocean 1.20 x 102 m above the water's surface. Whatwas her kinetic energy at the moment she was 30.0 m from the water's surface? What was herspeed at that moment assuming her mass is 60.0 kg?
Given data,
The initial velocity of the body is zero.
The distance travelled by the rescuer upto the height of 30 m from the water surface is,
[tex]\begin{gathered} S=102-30 \\ S=72\text{ m} \end{gathered}[/tex]The final velocity of the rescuer at the height 30 m is,
[tex]v^2-u^2=2gS[/tex]where g is the acceleration due to gravity.
Substituting the known values,
[tex]\begin{gathered} v^2=2\times9.8\times72 \\ v^2=1411.2 \\ v=37.6ms^2 \end{gathered}[/tex]Thus, the kinetic energy of the rescuer is,
[tex]K=\frac{1}{2}mv^2[/tex]Substituting the known values,
[tex]\begin{gathered} K=\frac{1}{2}\times60\times1411.2 \\ K=42336 \\ K=42.3\text{ KJ} \end{gathered}[/tex]Thus, the kinetic energy of the rescuer is 42.3 KJ and speed of the rescuer is 37.6 meter per second square.
A student has a cannon that can fire a cannonball at speeds up to 97.0mph. The students wants to determine the maximum range of the cannon and if she could hit a target on the ground as shown. Neglect drag and the initial height of the cannonball.
Answer:
88 ft / sec = 60 mph
Thus 97/60 * 88 = 142 ft/sec maximum speed of cannonball
R = V^2 sin 2 θ / g = 142^2 / 32 = 630 ft
Using 3.28 ft / m
630 ft / 3.28 f/m = 192 m is the maximum range of the cannonball
Vy = 142 ft / sec * sin 45 = 100 ft/sec vertical speed at 45 deg
Tup = 100 ft/sec / 32 ft/sec^2 = 3.12 sec time to reach height
T = 2 * 3.12 = 6.24 total time in air when fired at 45 deg
The hydrogen-to-helium ratio is an important piece of evidence for the Big Bang Theory. What is the minimum hydrogen-helium ratio observed?A. 4:1B. 3:1C. 2:1
Correct answer is option B.
Answer:
3:1
Explanation:
A child playing by the side of a well is throwing a stone vertically into the well with an initial velocity of 5 m/s. if the stone falls into the water 3 s later, find the height, h, of the point which the stone is thrown from the surface of the water. (g=10 m/s2)
a) 15m
b)30m
c)45m
d)60m
The height of the well is 30m
We are given that ,
The initial velocity of the stone = u = 5m/s
The stone fall into well in time = t = 3s
The acceleration due to gravity = g = 10 m/s²
To get the height of the stone which is thrown from the surface of the water in the well by equation,
h = ut + 1/2 gt²
Where, u is initial velocity in m/s, h is height in m, t is time in secs, g is acceleration due to gravity which is vertically upward direction i.e. (-10m/s²)
h = (5m/s) (3s) + 1/2 (-10m/s²)(3s)²
h = (15m) - (5m/s²)(9s²)
h = -30 m
Therefore, the height of the stone which is thrown upward is 30 m . And here, we can neglect the negative sign due to height is never be in negative.
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A plane's average speed between 2 cities is 800 km/hr. If the trip takes 4.5 hours howfar does the plane fly?
Considering the average velocity can be written as:
[tex]v=\frac{\Delta s}{\Delta t}[/tex]We can isolate the distance, and we get:
[tex]\Delta s=v*\Delta t\rightarrow\Delta s=800*4.5=3600km[/tex]Then, the distance the plane flew is 3600km
what is the maximum efficiency that a heat engine could have when operating between the normal boiling and freezing temperatures of water
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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Why isn't geothermal energy used more often?Question 12 options:It's hard to find geothermal energy sources that are close to the surface.It produces too much pollution.It adds a large amount of water into the water cycle through evaporation.It is not a renewable energy source.
Geothermal energy can be used only if an energy source is present.
Mining is required to use geothermal energy.
The depth of mining increases earthquake risk.
Thus, it is difficult to find geothermal energy sources that are close to the surface.
Acceleration problems use the 4 Step method to solve each of the following. a wagon has a beginning speed of 10 km/hr and it reaches 75 km/hr in 7 sec
Vi = initial speed = 10 km/h
Vf= final speed = 75 km/h
t = time = 7 sec
a = acceleration
First, express km/h in m/s
Vi = 2.78 m/s
Vf= 20.83 m/s
Apply
a = (vf-vi) / t
replace:
a = (20.83 - 2.78 ) / 7 = 2.58 m/s^2
In the setup of a cart pulled by a hanging mass, without friction, we have the following information: the mass of the cart is 0.53 Kg., and the hanging mass is 0.077 Kg.
Determine the acceleration of the cart (in m/s2).
The acceleration of the cart
a=1.2431m/s^2
What is acceleration?Generally, The equation for Newton's second law of motion is
2nd law of motion,
Fnet=m a
on hanging mass,
m_1 g-T=m_1 a
m_1 g-m_2 a=m_1 a
Acceleration, [tex]$a=\frac{m_1 g}{m_1+m_2}$[/tex]
[tex]&a=\frac{0.077 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^2}{0.077 \mathrm{~kg}+0.53 \mathrm{~kg}} \\[/tex]
a=1.2431m/s^2
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