Answer:
The answer is C burning the wood
Two red blood cells each have a mass of 4.60×10−14 kg4.60×10−14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion arising from the excess charge prevents the cells from clumping together. Once cell carries −2.00 pC−2.00 pC of charge and the other −2.90 pC−2.90 pC , and each cell can be modeled as a sphere 8.20 μm8.20 μm in diameter. What minimum relative speed vv would the red blood cells need when very far away from each other to get close enough to just touch? Ignore viscous drag from the surrounding liquid.
Answer:
v = 5.26 10² m / s
Explanation:
We can solve this exercise using the concepts of conservation of mechanical energy, because there is no friction
starting point. Red blood cells too far away
Em₀ = K = ½ m v²
final point. Red blood cells touching r = 8.20 10⁻⁶ m
Em_f = U = k q₁ q₂ / r₁₂
Em₀ = Em_f
½ m v² = k q₁ q₂ / r₁₂
v = √ (2 k q₁ q₂ / m r₁₂)
we calculate
v =√ (2 9 10⁹ 2 10⁻¹² 2.9 10⁻¹² / (4.60 10⁻¹⁴ 8.20 10⁻⁶))
v = √ (0.276775 10⁶)
v = 0.526 10³ m / s
v = 5.26 10² m / s
a total journey of 170 miles, in 3 hours, what would be your average velocity? (in miles/hour) *
Answer:
Velocity = 56.67 mi/hr
Explanation:
Distance = 170 miles
Time = 3 hours
Velocity =?
[tex]Velocity = \frac{Distance}{Time} \\\\V = \frac{170}{3} \\\\V =56.67[/tex]
A young child cannot understand
another person's perspective because he is
Answer:
egocentric
Explanation:
Fill in the blank
A pot of water is placed on the cook top. For a while the temperature of the water increases,
indicating the____
is increasing.
Answer:
temperature
Explanation:
when you put a water on the stove the water will start to boil there for temperature
A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.55 m/s. If the roof is pitched at 22.0° below the horizon and the roof edge is 2.90 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground. HINT Solve for the time of flight using vy2 = v0y2 − 2gΔy and then vy = v0y − gt. Solve for the horizontal distance using the horizontal displacement equation. Click the hint button again to remove this hint. (a) the time the baseball spends in the air (in s) 0.434 Incorrect: Your answer is incorrect. s (b) the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m) m
Answer:
Explanation:
The time that the baseball spends in the air is known as time of flight and it is expressed as shown;
T = Using(theta)/g where;
U is the velocity of the baseball
g is the acceleration due to gravity.
Given parameters
U = 4.55m/s
theta = 22.0°
g = 9.81m/s²
Substituting the values in the formula;
T = 4.55sin22°/9.81
T = 4.55(0.3746)/9.81
T = 1.7045/9.81
T = 0.1738second
Hence the time flight of the baseball is 0.1738second
b) The horizontal distance covered by the ball is called the RANGE in projectile.
Range = U√2H/g
U = 4.55m/s
H is the maximum height = 2.90m
g = 9.81m/s²
Substitute the given parameters into the formula
Range = 4.55√2(2.90)/9.81
Range = 4.55√5.8/9.81
Range = 4.55√0.5912
Range = 4.55(0.7689)
Range = 3.4986m
Hence the horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.4986m
Can an object with constant acceleration reverse its direction of travel? Can it reverse its direction twice? Explain.
Answer:yes
Explanation:The constan acceleration means that it wont stop moving but if you kick it a different direction then it will change direction
1) A plane's velocity increases from 40 m/s to 100 m/s over a 10 second interval. What is the plane's average acceleration for this interval? *
Answer:
average acceleration = 6 [tex]\frac{m}{s^2}[/tex]
Explanation:
Recall that the average acceleration [tex](a)[/tex] is defined by the change in velocity from an initial velocity [tex](v_i)[/tex], to a final velocity [tex](v_f)[/tex] over the time (t) it took that change to happen. Then, in mathematical terms this is:
[tex]a=\frac{v_f-v_i}{t}[/tex]
with our information this becomes:
[tex]a=\frac{v_f-v_i}{t} = \frac{100-40}{10}=6\,\frac{m}{s^2}[/tex]
What is the control group used for
Answer:
What is a Control Group? The control group (sometimes called a comparison group) is used in an experiment as a way to ensure that your experiment actually works. It's a way to make sure that the treatment you are giving is causing the experimental results, and not something outside the experiment
Explanation:
A powerful missile reaches a speed of 5 kilometers per second in 10 seconds after its launch. What is the average acceleration of the missile during this period?
A.
0.2 meters/second2
B.
0.4 centimeters/second2
C.
0.5 meters/second2
D.
0.5 kilometers/second2
Answer:
D
Explanation:
5 Km=5000m
so Δv=5000 m/sec
a=Δv/Δt
=5000/10
a=500 m/sec² as 500÷1000=0.5 Km
a=0.5 km/sec²
so D is the right answer.
Two vectors A and B are at right angles to each other. The magnitude of A is 3.00. What should be the length of B, so that the magnitude of their vector sum is 5.00?
Answer: length of B =4.00
Explanation:
for the vectors A and B and the angle between them as x.
Magnitude of the sum of A and B is given as = √(A²+B²+2ABcosx
where
Magnitude of A = 3.00
Magnitude of the sum of A and B is 5.00
5.00=√(A²+B²+2ABcos90°
5.00= √3² +b² +0
5²= 3² +b²
25=9+b²
b²= 25-9
b² = 16
b= √16
b= 4
You carry a fire hose up a ladder to a height of 10 m above ground level and aim the nozzle at a burning roof that is 9 m high. You hold the hose horizontally and notice that the water strikes the roof at a horizontal distance of 7 m from where it exits the nozzle. The hose is connected to a large pressurized chamber in the fire truck 0.5 m above ground level. What is the pressure in the chamber
Answer:
The value is [tex]P_1 = 314645 \ Pa [/tex]
Explanation:
From the question we are told that
The height is [tex]h_2 = 10 m[/tex]
The height of the burning roof is [tex]k = 9 m[/tex]
The horizontal distance is [tex]d = 7 \ m[/tex]
The height of the truck is [tex]h_1 = 0.5 \ m[/tex]
Generally the time for the water to hit the roof from the hose is mathematically represented as
[tex]t = \sqrt{\frac{2 * (h_2 - k)}{g} }[/tex]
=> [tex]t = \sqrt{\frac{2 * (10 - 9)}{9.8} }[/tex]
=> [tex]t = 0.4518 \ s [/tex]
Generally the velocity of the water is mathematically evaluated as
[tex]v_2 = \frac{d}{t}[/tex]
[tex]v_2 = \frac{ 7}{0.4518}[/tex]
[tex]v_2 = 15.5 \ m/s [/tex]
Generally from Bernoulli's Equation we have that
[tex]P_1 + \frac{1}{2} v_1^2 * \rho + \rho *g *h_1 = P_2 + \frac{1}{2} v_2^2 * \rho + \rho *g *h_2[/tex]
Here [tex]P_1 [\tex] is pressure in the chamber which we are to calculate , [tex]P_2 [\tex] is the atmospheric pressure with value [tex]P_2 = 101325 \ Pa [\tex] , [tex]v_1 [\tex] is the velocity of the water before it starts flowing with value [tex]v_1 = 0 m/s [\tex] , [tex]\rho [\tex] is the density of water with value [tex]\rho = 1000 \ kg/m^3 [\tex]
So
[tex]P_1 + \frac{1}{2} 0^2 * 1000 + 1000 *9.81 *0.5 = 101325 + \frac{1}{2}* 15.5^2* 1000 + 1000 *9.81 *10[/tex]
[tex]P_1 = 314645 \ Pa [/tex]
A runner has an original velocity of 6 m/s and slows to a final velocity of 0 m/s. If the runner covers a
displacement of 12 m while slowing down, how long time) did it take the runner to stop?
Answer:
4 s
Explanation:
Given:
Δx = 12 m
v₀ = 6 m/s
v = 0 m/s
Find: t
Δx = ½ (v + v₀) t
12 m = ½ (0 m/s + 6 m/s) t
t = 4 s
A projectile is launched on earth at an angle θ, relative to horizontal direction. At half of its maximum height the speed of the projectile is 1.00 m/s, and at its maximum height the speed of the projectile is 0.50 m/s. What is the angle θ ?
Answer:
the angle is about 67.79 degrees
Explanation:
We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)
We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :
[tex]half\,\,max-height = \frac{v_{yi}^2}{4\,g}[/tex]
we can use this information to find the y component of the velocity at that height via the formula:
[tex]v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}[/tex]
Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:
[tex]1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\[/tex]
Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:
[tex]tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o[/tex]
what kind of energy is called mechanical energy?
Answer:
Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position).
Hope this helps! :)
Explanation:
Suppose the two carts have the same mass m. In the initial state, these two carts are moving toward each other with the same initial speed, vi, along a frictionless track (implying no net external forces acting on the two carts). These carts collide and the result is some final state. The three parts of this question are concerned with three different final states. A. Assume that the carts hit each other and stop (both carts are not moving). Draw a momentum chart for this situation: make a separate row for each cart. B) Assume that the carts bounce off each other so that the final state of the system has each cart moving oppositely to its initial motion but with the same speed. Draw a momentum chart for this situation. C) As in (B), assume that the carts bounce off each other with equal speeds and in opposite directions, but now assume that the final speeds are smaller than the initial speeds. Draw a momentum chart. D) For each case does the total momentum of the two cars change? How do the momentum charts tell you this? E) Is the total kinetic energy constant for all three cases? How do you know?
Answer:
in all three cases the total moment is zero
cases A and B the kinetic energy is conserved.
In case C the velocity decreases so the kinetic energy decreases
Explanation:
This is a momentum conservation exercise
p = mv
In order for the moment to be preserved, we must define a system formed by the two cars, so that the forces during coke have been internal.
Before crash
car 1 p₀₁ = m v₀
car 2 p₀₂ = - mv₀
pose us several situations, we analyze each one
A) After the crash the cars stop
[tex]p_{f}[/tex] = 0
p₀₁ m v₀
p₀₂ -m v₀
p_{f} 0
B) After the collision, each vehicle reverses its direction
p₀₁ m v₀
p₀₂ -m v₀
p_{f1} -m v₀
p_{f2} m v₀
C) In this case some of the kinetic energy is lost which is converted into internal energy, for example, deformation, heat, friction.
Consequently the speed of the cars is
v < v₀
p₀₁ m v₀
p₀₂ - m v₀
p_{f1} -m v
p_{f2} m v
D) in cases A and B the momentum is maintained, but in case C the total momentum is maintained, even when the speed of the cars decreases, this is pf_total = 0
In all cases the total impulse is zero
p₀ = p₀₁ + p₀₂ = m v₀ - mv₀
p₀_total = 0
in all three cases the total moment is zero
E) The total kinetic energy is the sum of the kinetic energy of each car
K_total = K₀₁ + K₀₂
K_total = ½ m v₀² + ½ m (-v₀)²
K_total = m v₀²
we see that because it is squared, the sign of the velocity does not matter, therefore in cases A and B the kinetic energy is conserved.
In case C the velocity decreases so the kinetic energy decreases
Kf_total < K₀_total
the missing energy is transformed into internal energy during sackcloth.
In the attachment we can see a vector diagram of the momentum in each case
An engineering team has a goal of installing new, more efficient solar panels
on the International Space Station. Which equation best describes a cost-
benefit analysis the team might perform?
O A. Improved efficiency - manufacturing expenses + fuel to reach
orbit = net benefit
O B. Improved efficiency + manufacturing expenses - fuel to reach
orbit = net benefit
C. Improved efficiency + manufacturing expenses + fuel to reach
orbit = net benefit
O D. Improved efficiency - manufacturing expenses - fuel to reach
orbit = net benefit
Answer:
Answer is D
Explain:
Just did it on Abex
The equation best describes a cost-benefit analysis for installing new, more efficient solar panels on the International Space Station which the team might perform is Improved efficiency - manufacturing expenses - fuel to reach orbit = net benefit.
What are solar panels?A panel designed to absorb the sun's rays as a source of energy for generating electricity or heating. Solar technologies convert sunlight into electrical energy either through photovoltaic (PV) panels or through mirrors that concentrate solar radiation. This energy can be used to generate electricity or be stored in batteries or thermal storage.
What are advantages of solar power?The advantages of solar power are:
Solar power is pollution-free and causes no greenhouse gases to be emitted after installation.Reduced dependence on foreign oil and fossil fuels.Renewable clean power that is available every day of the year, even cloudy days produce some power.What is meant by cost-benefit analysis?A cost-benefit analysis is the process of comparing the projected or estimated costs and benefits (or opportunities) associated with a project decision to determine whether it makes sense from a business perspective.
The Elements of Cost-Benefit Analysis is
Determine on-going staffing costs – an analysis of the operating costs (actual versus proposed).Estimated savings and expenses by user department areas (e.g., Manufacturing, Marketing, etc.).Itemized benefits – both tangible and intangible.Hence, Improved efficiency - manufacturing expenses - fuel to reach orbit = net benefit best describes a cost-benefit analysis the team might perform.
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One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is 2.08 km. They start at the west side of the lake and head due south to begin with. (a) What is the distance they travel? (b) What is the magnitude of the couple’s displacement? (c) What is the direction (relative to due east) of the couple’s displacement?
Answer:
(a). The distance is 9.80 km.
(b). The magnitude of the couple’s displacement is 2.94 km.
(c). The direction of the couple’s displacement is 45°.
Explanation:
Given that,
Radius of lake = 2.08 km
(a). We need to calculate the total distance
Using formula of distance
[tex]d=\dfrac{3}{4}(2\pi R)[/tex]
Put the value into the formula
[tex]d=\dfrac{3}{4}(2\pi\times2.08)[/tex]
[tex]d=9.80\ km[/tex]
(b). We need to calculate the magnitude of the couple’s displacement
Using formula of displacement
[tex]D=\sqrt{R^2+R^2}[/tex]
[tex]D=\sqrt{2R^2}[/tex]
[tex]D=\sqrt{2\times(2.08)^2}[/tex]
[tex]D=2.94\ km[/tex]
(c), The direction of the displacement is given by
Using formula of direction
[tex]\tan\theta=\dfrac{R}{R}[/tex]
[tex]\theta=\tan^{-1}(1)[/tex]
[tex]\theta=45^{\circ}[/tex]
Hence, (a). The distance is 9.80 km.
(b). The magnitude of the couple’s displacement is 2.94 km.
(c). The direction of the couple’s displacement is 45° relative to the east.
The distance they travel is approximately 6.54644 km. The magnitude of the couple's displacement is 4.16 km. The direction of their displacement is opposite to due east.
(a) The distance they travel is:
Distance = (3/4) × 2 × 3.14159 × 2.08
Distance = 3.14159 × 2.08 km
Distance = 6.54644 km
So, the distance they travel is approximately 6.54644 km.
(b) The magnitude of the couple's displacement is:
Displacement = 2 × 2.08
Displacement = 4.16 km
So, the magnitude of the couple's displacement is 4.16 km.
(c) The direction of their displacement is opposite to due east.
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A cube at 333 K contains two metals: 8.00 kg of solid Silver and 15.0 kg of solid Gold. It is placed in contact with a block of solid Iron at 1737 K. The system reaches equilibrium at 1337 K (all the silver and all the gold has melted) Find the mass of the iron. (cgold-liquid= 0.150 kJ/(kg K), csilver-liquid= 0.280 kJ/(kg K)
Answer:
Mass of Iron is 24.45 kg
Explanation:
Given that:
Mass of Silver, [tex]m_{S}[/tex] = 8.00 kg
Mass of Gold, [tex]m_{G}[/tex] = 15.0 kg
Initial temperature of Silver and Gold = 333 K
Initial temperature of Iron = 1737 K
Final temperature = 1337 K
Specific heat capacity of Gold-liquid, [tex]c_{G}[/tex] = 0.150 kJ/(kg K)
Specific heat of Silver-liquid, [tex]c_{S}[/tex] = 0.280 kJ/(kg K)
Known: Specific heat capacity of Iron, [tex]c_{I}[/tex] = 0.461 kJ/(kg K)
Therefore;
Heat lost by Iron = Heat gained by Silver + Heat gained by Gold
[tex]m_{I}[/tex] x [tex]c_{I}[/tex] x Δθ = ([tex]m_{S}[/tex] x [tex]c_{S}[/tex] + [tex]m_{G}[/tex] x [tex]c_{G}[/tex]) x Δθ
[tex]m_{I}[/tex] x 0.461 x (1737 - 1337) = (8 x 0.280 + 15 x 0.150) x (1337 - 333)
[tex]m_{I}[/tex] x 0.461 x 400 = (2.24 + 2.25) x 1004
[tex]m_{I}[/tex] x 184.4 = 4507.96
[tex]m_{I}[/tex] = [tex]\frac{4507.96}{184.4}[/tex]
[tex]m_{I}[/tex] = 24.4466
[tex]m_{I}[/tex] ≅ 24.45
The mass of Iron is 24.45 kg.
How does the mass of an object affect its motion through the air?
The motion of an object through the air does not affect by its mass. The rate of fall of objects does not depend upon the mass.
What are free fall and air resistance?Free fall is a motion of a body in which gravity is the only force acting upon it. An object moving upwards might not be considered to be falling. But if the object is under the effect of the force of gravity, it is said to be in free fall.
Free fall is a type of motion in which the force acting upon an object is only gravity. Objects are not encountering a significant force of air resistance as they are only falling under the sole influence of gravity. All objects under such conditions will fall with the same rate of acceleration, regardless of their masses.
As an object falls through the air, have gone through some degree of air resistance. Air resistance is the collisions of the object's leading surface with molecules present in the air. The two most common factors that have a direct effect on the amount of air resistance are the cross-sectional area of the object and the speed of the object.
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What are 1A, 3B, and 7A examples of on the periodic table?
groups.
numbers
periods
rows
Answer:
groups
Explanation:
I got a 100 on my quiz
Answer:
It's groups
Explanation:
2020 quiz
Grant sprints 50 meters to the right with an average velocity of 3.0 m/s.
How many seconds did Grant sprint?
Answer:
The Answer Is 17s
Explanation:
Displacement trianglex = 50m
Time t = ?
Average velocity v = 3.0 m/s
You can rearrange the equation v = trianglex/t to solve for time t.
v = tianglex/t
t = trianglex/v
= 50m/3.0m/s
= 17s
It's important to match your exercise shoes with the type of exercise in which you will be participating, Please select the best answer from the choices provided OT
Answer:
the answer is true
Explanation:
A train that is 268.7 m long undergoes constant acceleration the moment the last car (end of
the train) is outside of the station, how far is the front of the train from the station after 25.0 s
if its initial speed before acceleration is 4.48 m/s and its final speed is 27.4 meters per second?
Answer:
its 667m just took the test
Explanation:
Two forces are applied to a tree stump to pull it out of the ground. Force A has a magnitude of 2410 newtons (N) and points 36.0 ° south of east, while force B has a magnitude of 4470 N and points due south. Using the component method, find the (a) magnitude and (b) direction of the resultant force A + B that is applied to the stump. Specify the direction as a positive angle with respect to due east.
Answer:
The magnitude of the resultant force is 6201 N
The direction of the resultant force is 72°
Explanation:
Please find the attached file for explanation;
Which three statements make up the law of gravity
A) The strength of the gravitational force between two objects decreases as the distance between their centers increases.
B) The strength of the gravitational force between two objects increases as the total mass of the objects increases
C) matter attracts all other matter in the universe
D) no matter where you are in the universe you will always have the same weight.
Answer:
the answer is A, B, and C
Explanation:
D is wrong because your weight does change...it is your mass that doesn't change
which best describes the big bang theory ? A. matter was compressed into a single point and then exploded outward to form the universe. B. stars burn off their fuel, expand, and shed their shells. C. a black hole exploded, creating stars.
Answer:
A. matter was compressed into a single point and then exploded outward to form the universe
Explanation:
The big bang is how astronomers explain the way the universe began. It is the idea that the universe began as just a single point, then expanded and stretched to grow as large as it is right now (and it could still be stretching).
A. matter was compressed into a single point and then exploded outward to form the universe.
this option describes the big bang theory
What is the Big Bang theory?The Big Bang hypothesis states that all of the current and past matter in the Universe came into existence at the same time, roughly 13.8 billion years ago. At this time, all matter was compacted into a very small ball with infinite density and intense heat called Singularity.
How did the universe begin?The Big Bang was the moment 13.8 billion years ago when the universe began as a tiny, dense, fireball that exploded. Most astronomers use the Big Bang theory to explain how the universe began. But what caused this explosion in the first place is still a mystery.
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A reversible heat engine, operating in a cycle, withdraws thermal energy from a high-temperature reservoir (the temperature of which consequently decreases), performs work w, and rejects thermal energy into a low-temperature reservoir (the temperature of which consequently increases). The two reservoirs are, initially, at the temperatures T1 and T2 and have constant heat capacities C1 and C2, respectively. Calculate the final temperature of the system and the maximum amount of work which can be obtained from the engine.
Answer:
The final temperature is [tex]\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}[/tex]
and the maximum amount of workdone is [tex]C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}[/tex].
Explanation:
Assume that [tex]R_1[/tex] is the reservior having temperature [tex]T_1 K[/tex] and heat capicity [tex]C_1 \frac JK[/tex] and [tex]R_2[/tex] is the reservior having temperature [tex]T_2[/tex] and heat capicity [tex]C_2 \frac JK[/tex].
The work will be extracted till that both the reservior reach the thermal equilibrium. Let the final temperature of both the reservior is [tex]T_f[/tex].
Let total [tex]Q_1[/tex] heat is extracted by the heat engine from the reservior [tex]R_1[/tex] and its temperature decreases from [tex]T_1[/tex] to [tex]T_f[/tex] and [tex]Q_2[/tex] heat is rejected by the heat engine to the reservior [tex]R_2[/tex] and its temperature decreases from [tex]T_2[/tex] to [tex]T_f[/tex].
So, The maximum amount of work done,
[tex]w= Q_1 - Q_2\; \cdots (i)[/tex]
Now, as the heat engine is reversible, so change is entropy for the universe is 0, which means sum of change in entropy for the ststem as well as surrounding is 0.
As shown in figure, the system is the reversible engine, so, change is entropy for the system is 0. Hence, change in entropy for the the surrounding is 0.
As temperature of [tex]R_1[/tex] is changing fron [tex]T_1[/tex] to [tex]T_f[/tex], so, change in entropy of surrounding due to transfer of [tex]Q_1[/tex] is [tex]C_1 \ln \frac {T_f}{T_1}[/tex].
Similarly, change in entropy of surrounding due to transfer of [tex]Q_2[/tex] is [tex]C_2 \ln \frac {T_f}{T_2}[/tex].
As the net change in entropy of the surrounding is 0.
[tex]\Rightarrow C_1 \ln \frac {T_f}{T_1}+C_2 \ln \frac {T_f}{T_2}=0[/tex]
[tex]\Rightarrow \ln \left( \frac {T_f}{T_1} \right)^{C_1}+ \ln \left( \frac {T_f}{T_2}\right)^{C_2}=0[/tex]
[tex]\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^{C_1}=- \ln \left( \frac {T_f}{T_2}\right)^{C_2}[/tex]
[tex]\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^{C_1}= \ln \left( \frac {T_2}{T_f}\right)^{C_2}[/tex]
[tex]\Rightarrow \left( \frac {T_f}{T_1}\right)^{C_1}=\left( \frac {T_2}{T_f}\right)^{C_2}[/tex][taking anti-log both the sides]
[tex]\Rightarrow T_f^{(C_1 +C_2)}=T_1^{C_1}+T_2^{C_2}[/tex]
[tex]\Rightarrow T_f=\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}\; \cdots (ii)[/tex]
This is the required final temperature.
Now, from equarion (i), the maximum amount of work done is
[tex]w= Q_1 - Q_2[/tex]
As [tex]Q=C\Delta T[/tex]
[tex]\Rightarrow w=C_1(T_1-T_f)-C_2(T_f-T_2)[/tex]
[tex]\Rightarrow w=C_1T_1 + C_2T_2-(C_1+C_2)T_f[/tex]
From equation [tex](ii)[/tex],
[tex]w=C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}[/tex]
This is the required maximum workdone.
The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth is 1.5multiply1011 m. (a) Calculate the magnitude of the gravitational force exerted by the Sun on the Earth. N (b) Calculate the magnitude of the gravitational force exerted by the Earth on the Sun.
Answer:
a) 3.56 x 10^22 N
b) 3.56 x 10^22 N
Explanation:
Mass of the sun M = 2 x 10^30 kg
mass of the Earth m = 6 x 10^24 kg
Distance between the sun and the Earth R = 1.5 x 10^11 m
From Newton's law,
F = [tex]\frac{GMm}{R^2}[/tex]
where F is the gravitational force between the sun and the Earth
G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2
m is the mass of the Earth
M is the mass of the sun
R is the distance between the sun and the Earth.
Substituting values, we have
F = [tex]\frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2}[/tex] = 3.56 x 10^22 N
A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to 3.56 x 10^22 N
b) The force exerted by the Earth on the Sun = 3.56 x 10^22 N
Part A:
The magnitude of the gravitational force exerted by the Sun on the Earth is :
Mass of the sun M = 2 x 10^30 kgMass of the Earth m = 6 x 10^24 kgDistance between the sun and the Earth R = 1.5 x 10^11 mFrom Newton's law,
F = GmM/F
G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2m is the mass of the EarthM is the mass of the sunR is the distance between the sun and the Earth.Substituting values,
F = = 3.56 x 10^22 N
The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to 3.56 x 10^22 N.
Part B:
The force exerted by the Earth on the Sun = 3.56 x 10^22 N.
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What is the effect on the gas’ pressure when compressing a gas to 1/3 of its volume? Explain
Explanation:
It's pressure become three times larger because according to Boyles Law the pressure of fixed mass of gas is inversely proportional to it's volume provided that temp remains constant. That means a reduction in volume, will result in an increase in pressure and vice versa.
Mary throws a baseball straight upward. We can ignore air resistance.
Answer:
Acceleration Remains Constant and Velocity decreases.
Explanation:
