2. Two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one of the objects is located at a distance of r away from the center of a uniform spherical planet with mass M (see figure). M r m L m Assume that m is very small so that you can ignore the gravitational force between the two small (green) objects. Is the rod being compressed, stretched, or neither? If compressed/stretched, calculate the magnitude of this deformation force on the rod. If neither, explain clearly why.
Answers
The rod is stretched due to the force of gravitational attraction.
The magnitude of the deformation force on the rod is -GMm×((1/r² + 1/(L+r)²)
What is gravity?The force of attraction felt by a person which is directed at the center of a planet or Earth is called as the gravity.
The force of attraction is directly proportional to the product of masses of the object and inversely proportional to the square of distance between them.
F = GMm/R²
Given are two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one of the objects is located at a distance of r away from the center of a uniform spherical planet with mass M. Assume that m is very small so that you can ignore the gravitational force between the two small (green) objects.
Let the mass closer to the planet is A and the other is B
FA = Force on planet A = -GMm/r²
FB = Force on planet B = -GMm/(L+r)²
Net Force on the rod is given by the addition of the individual forces.
Fnet= FA + FB = -GMm/r² -GMm/(L+r)²
Fnet = -GMm×((1/r² + 1/(L+r)²)
Thus, the magnitude of the deformation force on the rod is derived above.
A experience more force than B, so A will stretch out from B. Hence the force is stretching the rod.
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Related Questions
. A vertical electric field is set up in space to compensate for the gravitational force on a point charge. What is the required magnitude and direction of the field when the point charge is: (a) an electron? (b) a proton? Comment on the obtained values.
Answers
(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
Magnitude of electric fieldThe magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
E is the electric fieldm is mass of the particleg is acceleration due to gravityq is charge of the particleFor an electronE = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
For protonE = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
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A grocery cart is pushed with a force of 21.4 N. If 1,974 J of work is done in pushing the grocery cart through the store, what is the total distance that the grocery cart has traveled?
Answers
Answer:
Given - Force = 21.4 N
Work done = 1974 J
To find - Distance
Solution -
Work done = Force * displacement
1974 j = 21.4 N * displacement
1974/21.4 = displacement
92.24
[tex]\\ \rm\Rrightarrow Work=Force\times Displacement [/tex]
[tex]\\ \rm\Rrightarrow 1974=21.4×Displacement[/tex]
[tex]\\ \rm\Rrightarrow Displacement=92.24m[/tex]
What is the magnetic force on a proton that is moving at 5.2 x 107 m/s to the
right through a magnetic field that is 1.4 T and pointing away from you? The
charge on a proton is 1.6 × 10-19 C. Use F = qvx B sin(e)
Answers
Hello!
We can use the following equation for magnetic force on a charged particle:
[tex]F_B = qv \times B[/tex]
[tex]F_B[/tex] = Magnetic force (N)
q = Charge of particle (1.6 × 10⁻¹⁹ C)
v = velocity of particle (5.2 × 10⁷ m/s)
B = Magnetic field strength (1.4 T)
This is a cross-product, so the equation can be rewritten to F = qvBsinφ where φ is the angle between the magnetic field and particle velocity vectors.
Since the proton's velocity vector and the magnetic field vector are perpendicular, sin(90) = 1. We can reduce the equation to:
[tex]F_B = qvB[/tex]
Plug in the known values.
[tex]F_B = (1.6*10^{-19})(5.2*10^7)(1.4) = \boxed{1.1648 *10^{-11} N}[/tex]
HELP!! ASAP!!
2. Describe the difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl. Which molecule contains a polar covalent bond and which molecule contains a nonpolar covalent bond? Explain your reasoning alongside describing the differences between the types of bonds.
3. How is the metallic bonding different than ionic or covalent bonding? What are some properties of metals that result from this type of bonding? Explain/connect how the nature of the bonding leads to the properties of metallic substances.
Answer:
Answers
The difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl are HCl is a polar covalent compound because the chloride ion is extra electronegative than the hydrogen ion.
Why HCl is polar and H2 is now no longer?HCl is a polar molecule. This is due to the fact the Chlorine (Cl) atom withinside the HCl molecule is extra electronegative and does now no longer proportion the bonding electrons similarly with Hydrogen (H). But H2 And Cl2 are nonpolar because of comparable electronegativity of each the atoms withinside the molecule H2 And Cl2 .
Hydrogen chloride is a diatomic molecule, such as a hydrogen atom H and a chlorine atom Cl related with the aid of using a polar covalent bond. The chlorine atom is an awful lot extra electronegative than the hydrogen atom, which makes this bond polar.
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Consider the baby being weighed in Figure 4.25.
Figure 4.25
(a) What is the mass of the child and basket if a scale reading of 104 N is observed?
kg
(b) What is the tension T in the cord attaching the child to the scale?
N
(c) What is the tension T' in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg?
N
(d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible. (Do this on paper. Your instructor may ask you to turn in this work.)
Answers
The mass and tension due to the system are as follows:
The mass of the child and scale = 10.6 kgThe tension T, in the cord attaching the child to the scale = 104N The tension T', in the cord attaching the scale to the ceiling T' = 108.9 NWhat is tension?Tension is a type of pulling force due transmitted by means of a string or cable.
Force = mass * acceleration due to gravitya) The mass of the child and scale = 104/9.81 = 10.6 kg
b) The tension T, in the cord attaching the child to the scale = scale reading = 104N
c) The tension T', in the cord attaching the scale to the ceiling = scale reading + weight of scale
T' = 104 + (0.5 * 9.81)
T' = 108.9 N
d) The sketch is attached in the picture
In conclusion, the tension is force exerted on the cord due to the weight of the scale and the baby.
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22. On a day the wind is blowing toward the south at 3m/s, a runner jogs west at 4m/s. What is the velocity of the air relative to the runner?
Answers
The velocity of the air relative to the runner is 5 m/s.
What is the relative velocity?We must recall that velocity is a vector quantity and the relative velocity must be obtained vectorially. Thus we know that;
Velocity of the runner = 4m/s. due west
Velocity of the wind = 3m/s due south
The relative velocity is;
Vr = √(4)^2 + (3)^2
Vr = 5 m/s
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If the distance between two objects is cut in half, what happens to the
gravitational force between them?
A. It decreases to 1/2 its original magnitude.
B. It decreases to 1/4 its original magnitude.
O
C. It increases to 4 times its original magnitude.
D. It increases to 2 times its original magnitude.
Answers
An object is released from an aeroplane which is diving at an angle of 30° from the horizontal with a speed of 50m/s. If the plane is at a height of
4000m from the ground when the object is released, find
(a) the velocity of the object when it hits the ground.
(b) the time taken for the object to reach the ground.
please I need the solution urgently.
Answers
a)The velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec
b) The time taken for the object to reach the ground will be 11.4 × 10³ sec.
What is projectile motion?The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.
The velocity in the x-direction;
[tex]\rm v_x = (vcos \theta)^2 +2gh \\\\ v_x = (50 cos 30)^2+ 2 \times 9.81 \times 4000 \\\\ v_x = 80355 m/sec[/tex]
Velocity in y-direction;
[tex]\rm v_ y = (vsin \theta)^2 + 2gh \\\\ v_ y = (25)^2+2 \times 9.81 \times 4000 \\\\ v_y =79105 \ m/sec[/tex]
The resultant velocity is found as 1.12 × 10⁵ m/sec.
The time taken to reach the ground is found as;
[tex]\rm v = u+gt \\\\ 1.12 \times 10^5 \ m/sec = 50 \ m/sec + 9.81 \times t \\\\ t = 11.4 \times 10^3 \ sec[/tex]
Hence, the velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec, and the time taken for the object to reach the ground will be 11.4 × 10³ sec.
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a 1kg ball is bening pushed by the rod to move in horizontal groved smooth slot if it startes from angle teta = zero degree . determaine the force the rod exertes on the ball at teta is =15 dgree if ai this instant the rod moves at angular speed of teta = 1 rad per sec end with angular acceleration theta = 2 rad persec and square the ball is only in contact with the outer side of the slot
Answers
The force the rod exerts on the ball at the given angle is determined as 3.94 N.
Force exerted on the rod by the ball
The force exerted is calculated as follows;
F = ma
F = mv²/r
F = mω²r
where;
m is mass of the ballω is angular speed of the ballr is radius of the pathr = 2cosθ
Angular speed when the ball moves 15 degreesωf² = ωi² + 2αθ
where;
θ is angular displacement in radians, 15⁰ = 15 x π/180 radωf² = (1)² + 2(2)(15 x π/180)
ωf² = 2.04
ωf = √2.04
ωf = 1.428 rad/s
F = mω²(2cosθ)
F = (1)(1.428)²(2 x cos15)
F = 3.94 N
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5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.
(a) How much charge is stored in the 5.00-μF capacitor?
(b) What is the potential difference across the 10.0-μF capacitor?
Answers
(a) The charge stored in the 5.00-μF capacitor is 37.2 μC.
(b) The potential difference across the 10.0-μF capacitor is 3.72 V.
What is capacitor?The capacitance of a capacitor is defined as the ratio of the charge stored and the potential difference between the capacitor.
The capacitance of a capacitor is denoted by C and expressed as
C = Q/V
Given, 5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.
(a) The equivalent capacitance is
1 / Ceq = 1 / C₁ +1 / C₂ + 1/ C₃
Substitute the values, we get
Ceq = 3.1 μF
The charge stored in 5.00-μF capacitor is
Q = Ceq x V
Q = 3.1 μF x 12 V
Q = 37.2 μC
Thus, the charge stored in the 5.00-μF capacitor is 37.2 μC
(b) The potential difference across the 10.0-μF capacitor is given by
V = Q/C₂
Put the values, we get
V = 37.2 / 10
V = 3.72 V
Thus, the potential difference across the 10.0-μF capacitor is 3.72 V.
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if you were to draw a 3rd harmonic of a tube open at both ends, what would you draw at the ends of the tube?
Answers
A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.
In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.
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a 10.00mf parallel-plate capacitor is connected to a 24.0v battery. after the capacitor is fully charged ,the
battery is disconnected without loss of any of charge on the plates.
a) a voltmeter is connected across the two plates without discharging them.what does it read
b)what would the voltmeter read if the plate separationwere doubled?
Answers
The answer to the first part of the question is 24 V and the answer to the second part of the question is 48 V.
The formula which relates Charge, Capacitance and Voltage is
Charge = Capacitance × Voltage
Q = CV
where Q denotes the charge, C denotes the capacitance and V denote the voltage.
Using the above formula the charge on the capacitance will be
Q = CV
C = 10 mF
V = 24 V
Q = 240 mC
Charge on capacitance is 240 mC.
Now after we disconnect the battery
C = 10 mF (will remain same)
Q = 240 mC
V = Q/C
V = 240/10 V
V = 24 V
So after we removed the battery, The voltage will remain same.
Now we know that parallel plate capacitor formula is
C = ε(A/d)
from here
C ∝ 1/d, and as there is no loss of charge we can say that V ∝ 1/d
Form C ∝ 1/d and V ∝ 1/C we can state that
V ∝ d {where d is the measurement of separation between the plates}
so if we double the distance between the plates then, the voltage will also get double.
Previously our voltage was 24 V, now if we double the distance between the plates the voltage will also get double, and it will become 48 V.
So, the voltmeter will take a reading of 24 V if voltmeter is connected across the two plates without discharging them, and if we double the plate separation then it will take 48 V as reading.
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A bumblebee
is flying towards a flower in a
straight line at 4.09 m/s when it begins to
accelerate at 1.01 m/s².
How long does it take the bee to reach the
flower if it is 23.4 m away?
Answers
Answer:
given -
initial velocity = 4.09 m/s
acceleration = 1.01 m/s²
distance = 23.4 m
time = ?
using second formula of motion,
s = ut + 1/2 at².
where, s = distance
u = initial velocity
t = time
a = acceleration
23.4 = 4.09(t) + 1/2(1.01)(t) ²
23.4 = 4.09t + 2.02t²
2.02t² + 4.09t - 23.4 = 0
solving the equation by using quadratic formula
Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :
a = 2.02
b = 4.09
c = -23.4
we get t=2.539 or t= -4.563
time cannot be negative so
t= 2.539 sec = 2.6 Sec is the answer
Can you put this image back together? Type the correct order of letters below.
Answers
Answer:
CABD
.....................
The image is cut and jumbled and placed into different places, so the correct order of the image that it can be back together is CABD.
What is an image?A specific piece of something is an image. It can provide information to the optical system in two dimensions, three components, or in another way. An item that mimics a subject, such as a picture or other two-dimensional image, might be considered an image. An image in the context of signal analysis is a dispersed color amplitude.
A graphical image need not make use of the complete visual system. A common illustration of this is a grayscale image, which does not employ color, but instead relies on the visual game's sensitivity to brightness across all wavelengths. Even if it doesn't make proper use of the visual system, a black-and-white and white visual depiction of something is nonetheless an image.
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Need help with this question!!
Answers
A beta particle or an electron is released during beta decay. The charges can be positive or negative. Co changes to Ni, Fe to Mn, Pb to Tl, and Pu to Am.
What is beta decay?Beta-decay is the radioactive decay that involves the release of the beta particle or the positron or the electron. The larger nucleus splits during nuclear fission and releases smaller nuclei.
The nuclear reactions are shown as:
⁶⁰Co₂₇ → ⁶⁰Ni₂₈ + ⁰e₋₁⁵⁶Fe₂₆ → ⁵⁶Mn₂₅ + ⁰e₋₁²¹⁰Pb₈₂ → ²¹⁰Tl₈₁ + ⁰e₋₁²⁴¹Pu₉₄ → ²⁴¹Am₉₅ + ⁰e₋₁Learn more about beta decay here:
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A sinewave has a period (duration of one cycle) of 460 μs. What is the corresponding frequency of this sinewave in kHz, expressed to three significant figures (see Study Note 7.1 in the Using numbers (maths) booklet)?
If the frequency of the sinewave is now reduced by a factor of 4.25, what will be the new period value? Express your answer in ms to three significant figures.
Note: in the second part of this question, to avoid rounding errors in your calculations, use the full frequency value from the first part, not the rounded value to three significant figures.
Answers
Answer:
See below
Explanation:
Frequency = 1 / period
= 1 / 460 X 10^-6
2173.91 Hz = ~ 2.17 Khz
2.17391 kHz / 4.25 = .511 kHz <====new frequency
period = 1/frequency
= 1955 microseconds =1.96 ms
Answer:
2.17 kHz1.96 msExplanation:
You want to know the frequency in kHz of a sine wave with a period of 460 μs, and its period if the frequency is reduced by a factor of 4.25.
UnitsWhen you are interested in the frequency in kHz, it is convenient to use milliseconds (ms) to express the period. Milliseconds and kilohertz are inverse units.
The period of the given sine wave can be expressed in milliseconds as ...
460 μs = 460×(0.001 ms) = 0.460 ms
FrequencyThe frequency of the wave is 1/(0.460 ms) ≈ 2.17 kHz.
PeriodBecause period and frequency are reciprocals of each other, reducing the frequency by a factor of 4.25 increases the period by the same factor.
The new period is 4.25×0.460 ms = 1.96 ms.
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A person lifts a ball of mass 1 kg through the
height of 1 m in 10 s. The average power
supplied by him is [Take g = 10 m/s²]
(2) 1 W
(4) 2 W
(1) 10 W
(3) 0.1 W
Answers
Answer:
1 W
Explanation:
Power is the ratio of the total work done by the body to the time it will takes to do the work.
Therefore Power = total work done ÷ timeTotal Work done = Potential energy (P.E) = mgh
Total work done = mgh
Total work done = 1kg × 10m/s² × 1m
Total work done = 10 J
Therefore Total Work done is 10J.
And then time is 10 s.
Power = Total Work done divided by time
Power = 10 J÷ 10 s = 1 W
Therefore Power is 1WWhich object has the least thermal energy?
Answers
Answer:
2kg brick
Explanation:
2kg brick at 25 degrees.
Answer: A 2kg brick at 20*C
Explanation:
The answer is not B.
Answers
Answer:
D
Explanation:
15m/ s
i hope this helps
Entropy is how quickly things get messy.
O A. True
OB. False
Answer : False
Answers
Answer:
false
Explanation:
it cant defined the messy and clean states
Which of the following causes a car's passenger to lean in the direction
opposite the direction in which the car is turning?
OA. Friction
OB. Inertia
OC. Centripetal acceleration
OD. Centripetal force
Answers
Explanation:
This case is inertia.
Mass is directly proportional to inertia
What work do you think is done when you carry a 20 N weight backpack for a 1000 m walk? will the work be positive, negative, or potentially zero?
Answers
The work done is positive and is equal to 20000 J
What is work done?Work done is defined as the product of force and the distance moved by the force.
Mathematically:
Work done = force * distanceThe work done by the force = 20 * 1000 = 20000J
The work done is positive and is equal to 20000 J
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You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, it accelerates because the engines are at full throttle and because there is a catapult that propels the jet forward. You begin to wonder how much force is supplied by the catapult. You look on the Web and find that the flight deck of an aircraft carrier is about 90.0 m long, that an F-14 has a mass of 24800 kg, that each of the two engines supplies 27000 lb of thrust, and that the takeoff speed of such a plane is about 158 mi/h. Estimate the average force on the jet due to the catapult.
Answers
You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, the average force on the jet is due to the catapult is mathematically given as
What is the average force on the jet is due to the catapult?
Generally, the equation for acceleration is mathematically given as
[tex]a=\frac{vf^2-vi^2}{2s}\\\\\Therefore\\\\a=\frac{69.29^2-0^2}{2(90}\\\\a=26.673m/s^2[/tex]
In conclusion, The force
F=m*a
F=15100*26.673
F=40272.3N
F 402 KN
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The cutoff frequency for a certain element is 1.22 x 10^15 Hz. What is its work function in eV?
Answers
The work function in eV for the given cutoff frequency is 5.05 eV.
What is cutoff frequency?The work function is related to the frequency as
Wo = h x fo
where, fo = cutoff frequency and h is the Planck's constant
Given is the cutoff frequency for a certain element is 1.22 x 10¹⁵ Hz
Wo = 6.626 x 10⁻³⁴ x 1.22 x 10¹⁵ Hz / 1.6 x 10⁻¹⁹
Wo = 5.05 eV
Thus, the work function is 5.05 eV
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Determine the amount of power
used in holding a 25 kg box, 1.5
meters above the floor, for 60
seconds.
[?] W
(answer is not 6.13)
Thank you in advance!
Answers
Here is your answer mate,
Question,
[tex]Determine\: the\: amount\\ \: of\: power\:used\: in\\\: holding\: a\: 25\: kg\: box\:\\ , \: 1.5\: meters \: above\: the\: floor\\\: for\: 60\: seconds[/tex]
Answer,
Power is equal to work done per unit timeWork is force × displacement SI UNIT OF WORK Newton meterSI UNIT OF POWER Watt[tex][/tex]
Solution,
[tex][/tex]
Given,
[tex]MASS \: IS\: 25\: KG\: \\ and \: HEIGHTIS\: 1.5m\: [/tex]
[tex][/tex]
WORK DONE (done against gravity) =
mass×acceleration due to gravity ×height
WORK = 25× 10× 1.5
[tex]\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: [/tex]= 375 Nm
[tex][/tex]
Now
POWER =
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{work}{time} [/tex]
POWER
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{375}{60} Watt [/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =6.25[/tex]
[tex]Therfore\: your \: answer\: is\: 6.25[/tex]
[tex][/tex]
Check this,
[tex]Acceleration\: due\: to \: gravity\\\: can\: be\: 9.8\: m/s²\: \\As\: nothing\: mentioned\\\: in\: question\: \\I \: took \: it \: as \: 10[/tex]
[tex][/tex]
Have a good day
Which picture correctly shows the path of refracted light rays given an object outside the focal point? Select one: a. A b. B c. C d. D
Answers
Answer:
Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor
Which force is reasonable for making fusion possible in the sun?
Answers
The force is reasonable for making fusion possible in the Sun is heat energy.
What is nuclear fission and fusion?When the slow moving neutrons are bombarded with the heavy radioactive nuclei, the product is the more number of neutrons are produced with the large amount of energy. This multiplying process is called nuclear fusion.
The amount of energy produced in such a reaction can be calculated using the equivalence of mass and energy relationship.
E = mc²
The same happens in nuclear fusion where large amount of energy is needed to make more heavy nuclei.
Thus, fusion requires heat energy to continue the reaction.
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Please help me!!
1. For objects like insulators (plastics), they can get charged by ___.
2. For metals, where there are more loosely bound electrons, they can get charged without contact by ___.
3. For metals, they can also be charged by direct contact by ___.
Answers
A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrofoam container. When thermal equilibrium is reached, the temperature of the alcohol-water solution is 12°C. What is the specific heat capacity of the alcohol? Assume the sealed container is an isolated system. The specific heat capacity of water is 4.19 kJ/kg · °C. 3.14 kJ/kg \xe2\x88\x99 °C 4.14 kJ/kg \xe2\x88\x99 °C 3.72 kJ/kg \xe2\x88\x99 °C 4.88 kJ/kg \xe2\x88\x99 °C
Answers
The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
What is the specific heat capacity?The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."
Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.
Given data;
Mass of liquid sample of Alcohol m₁ = 200-gram
The temperature of alcohol, T₁ = -6°C.
Mass of liquid sample of water m₂ = 400-gram
The temperature of the water, T₂= 20°C.
The specific heat capacity of the alcohol, S₁=?
The specific heat capacity of water is, S₂=4.19 kJ/kg.°C
As we know that;
[tex]\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C[/tex]
Hence the specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 100-kilogram weight 2-decimeters above the ground with an energy efficiency of 25%. How many repetitions can she do with the energy supplied from a single Oreo cookie? What happens to the number of repetitions that can be done if the efficiency increases?
Answers
Answer:
Approximately [tex]325[/tex] (rounded down,) assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].
The number of repetitions would increase if efficiency increases.
Explanation:
Ensure that all quantities involved are in standard units:
Energy from the cookie (should be in joules, [tex]{\rm J}[/tex]):
[tex]\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}[/tex].
Height of the weight (should be in meters, [tex]{\rm m}[/tex]):
[tex]\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}[/tex].
Energy required to lift the weight by [tex]\Delta h = 0.2\; {\rm m}[/tex] without acceleration:
[tex]\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}[/tex].
At an efficiency of [tex]0.25[/tex], the actual amount of energy required to raise this weight to that height would be:
[tex]\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}[/tex].
Divide [tex]2.551 \times 10^{5}\; {\rm J}[/tex] by [tex]784\; {\rm J}[/tex] to find the number of times this weight could be lifted up within that energy budget:
[tex]\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}[/tex].
Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.
