Answer:
Approximately [tex]261\; \rm K[/tex], if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.
Explanation:
Let [tex]P_1[/tex] and [tex]P_2[/tex] denote the pressure of this gas before and after the changes.
Let [tex]V_1[/tex] and [tex]V_2[/tex] denote the volume of this gas before and after the changes.
Let [tex]T_1[/tex] and [tex]T_2[/tex] denote the temperature (in degrees Kelvins) of this gas before and after the changes.
Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the quantity (number of moles of gas particles) in this gas before and after the changes.
Assume that this gas is an ideal gas. By the ideal gas law, the ratios [tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex] and [tex]\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex] should both be equal to the ideal gas constant, [tex]R[/tex].
In other words:
[tex]R = \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex].
[tex]R =\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].
Combine the two equations (equate the right-hand side) to obtain:
[tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} = \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].
Rearrange this equation for an expression for [tex]T_2[/tex], the temperature of this gas after the changes:
[tex]\displaystyle T_2 = \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2} \cdot T_1[/tex].
Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: [tex]n_2 = n_1[/tex], [tex](n_2 / n_1) = 1[/tex].
[tex]\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}[/tex].
Can someone tell me anything useful about energy management in the human body?
Answer:
The human body carries out its main functions by consuming food and turning it into usable energy. Immediate energy is supplied to the body in the form of adenosine triphosphate (ATP). Since ATP is the primary source of energy for every body function, other stored
Explanation:
this what teacher explain to us
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
Each minute, the disk completes 100 revolutions, so a point on the rim traverses a distance of 100 times the circumference of the disk and would have a linear speed of
100 rev/min
= (100 rev/min) × (2π × 50 cm/rev) × (1/100 m/cm) × (1/60 min/s)
= 5π/3 m/s ≈ 5.236 m/s
Then after 30 s of rotation, the point would have traveled a distance of
(5π/3 m/s) × (30 s) = 50π m ≈ 157.08 m
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 41 mm , while nonathletes' stretch only 33 mm .
Hello. Your question is incomplete. However, I managed to find it completely on the internet and I realized that you forgot to mention that the question asks you for the maximum energy difference between velovistas and non-athletes, considering that the spring constant for the tendon of the two groups is equal to 33n/mm.
To make this calculation you will need to use Hooke's law, using the formula: ¹/2*K*x², where "K" will be the value of the spring constant for the tendon and "X" will be the value of the sprinter and non-athlete terms.
So for the sprinter we will have the calculation:
¹/2*33*41² -------> 0,5*33*1681 = 27736. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
For the non-athlete we will have the calculation:
¹/2*33*33² -------> 0,5*33*1089 = 17968. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
Now, to reach the final result, you only need to subtract the two values presented by the sprinter and the non-athlete.
27736.5 - 17968.5 = 9768 Nmm
A +0.0129 C charge feels a 4110 N
force from a -0.00707 C charge. How
far apart are they?
[?] m
Answer:
r = 14.13 m
Explanation:
Given that,
Charge 1, q₁ = +0.0129 C
Charge 2, q₂ = -0.00707 C
The force between charges, F = 4110 N
We need to find the distance between charges. The formula for the force between charges is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Where
r is the distance between charges
So,
[tex]r=\sqrt{\dfrac{kq_1q_2}{F}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 0.0129 \times 0.00707 }{4110 }} \\\\r=14.13\ m[/tex]
So, the distance between charges is equal to 14.13 m.
Answer:
14.13 m
Explanation:
acellus
Two students are sitting 1.50 m apart. One student has a mass of 70.0 kg and
the other has a mass of 52.0 kg. What is the gravitational force between them?
A. 8.01 x 10-9
B. 1.08 x 10-2
C. 2.28 x 10-8
Answer:
B
Explanation:
I need help with this review question.
Answer:
The acceleration of the football is greatest
Explanation:
The more mass the more acceleration
Will give brainliest!
Describe how heat is moving in the image and label each as Radiation, Conduction, or Convection.
Radiation / Conduction / Convection
Answer:
well in the pot there is conventional heat, the pot itself is giving off conductable heat, and the radiational heat is coming from the stove.
2) Which of the following is not a quality of a healthy relationship?
A) respect
B) trust
C) closed communication
D) honesty
One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 42.9o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.
Answer:
x = 0.455 L
Explanation:
For this exercise we must use the rotational equilibrium condition
Σ τ = 0
it has two forces, the first is perpendicular to the rod, so its stub is
τ₁ = F₁ L
the second force is applied with an angle, so we can use trigonometry to find its components
sin θ = F_parallel / F₂
cos θ = F_perpendicular / F₂
F_parallel = F₂ sin θ
F _perpendicular = F₂ cos θ
torque is
τ₂ = F_perpendicular x + F_parallel 0
the parallel force is on the rod therefore its distance is zero
we apply the equilibrium equation
τ₁ - τ₂ = 0
F₁ L = F₂ cos θ x
x = [tex]\frac{L}{cos \theta} \ \frac{F_1}{F_2}[/tex]
let's calculate
x = [tex]\frac{L}{cos \ 42.9} \ \frac{2.00}{6.00}[/tex]
x = 0.455 L
What is the order of the events for the water cycle on a typical warm day?
А
rain, snow, sleet
B
precipitation, evaporation, rain
с
evaporation, condensation, precipitation
D
condensation, evaporation, precipitation
true or false
The Total electric potential due to two or more charges is equal to the algebraic sum of the potentials due to the individual charges.
Answer:
i guess the answer is false
. Indiana Jones needs to ascend a 10-m-high building. There is a large hose filled with pressurized water hanging down from the building top. He builds a square platform and mounts four 4-cm-diameter nozzles pointing down at each corner. By connecting hose branches, a water jet with a velocity, u, can be produced from each nozzle. Jones, the platform, and the nozzles have a combined mass of 150 kg. Determine the minimum water jet velocity, u (m/s), needed to raise the system.
Answer:
u = 14 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. On the floor just when u speeding out
Em₀ = K = ½ m v²
final point. When on top of the building, no speed
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ m v² = m g h
v = u
u = [tex]\sqrt{2 g h}[/tex]
u= [tex]\sqrt{2 \ 9.8 \ 10}[/tex]
u = 14 m / s
A wave has a wavelength of 1.5 meters and period of 0.083s. What is the waves speed?
When jeremiah stands in a swimming pool and looks at hid feet, his legs appear to be bent. Which is the term for this phenomenon?
A. Diffraction
B. Dispersion
C. Reflection
D. Refraction
Pls help will mark Brainliest
Which of the following happens to
density as air pressure decreases?
С C
A. Density increases.
B. Density stays the same.
C. Density decreases.
D. There is no correlation between air pressure and
density.
Explanation:
As pressure increases, with temperature constant, density increases. Conversely when temperature increases, with pressure constant, density decreases. Air density will decrease by about 1% for a decrease of 10 hPa in pressure or 3 °C increase in temperature.
Convert (a) 50 oF, (b) 80 oF, (c) 95 oF to Celsius
the higher the objects " ? ", the more kinetic energy
The liquid emerges into a vertical jet as it drains from the container, with the velocity profile in the jet remaining uniform. The outlet of the container is located 2.0 m above ground, and the radius of the emerging liquid jet changes with vertical distance from the bottom of the container as it accelerates under the action of gravity. Neglecting viscous losses and surface tension effects in the liquid jet, what is the velocity of the water jet as it strikes the ground when the container begins to drain
Answer:
6.26 m/s
Explanation:
Since we are neglecting viscous losses and surface tension effects in the liquid jet, by conservation of energy, the potential energy loss of the jet = kinetic energy gain of the jet
So, mgh = 1/2mv² where m = mass of water in jet, g = acceleration due to gravity = 9.8 m/s², h = height of outlet = 2.0 mand v = velocity of liquid jet
So, mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 2.0 m)
v = √(39.2 m²/s²)
v = 6.26 m/s
What is the unit of measurement of mass and weight?
Answer:
kilogram
In the International System of Units (SI), the kilogram is the basic unit of mass, and the newton is the basic unit of force. The non-SI kilogram-force is also a unit of force typically used in the measure of weight.
hhhep faaast plssssss
Answer:
false
Explanation:
I am in need of points sorry
why is potassium and sodium considered as reactive metals?
Answer:
because they are found freely in nature uncombined so they are highly reactive with other elements
What voltage would be measured across the 15 ohm resistor?
A)
2.5 volts
B)
5.0 volts
C)
7.5 volts
D)
10 volts
Answer:
7.5 volts
Explanation:
I did it on USA Testprep
As altitude decreases, what happens to
air pressure?
A. increases
B. decreases
C. stays the same
D. not enough information to tell
Answer:
A. Increases
Explanation:
As altitude decreases, the amount of gas molecules in the air increases - the air becomes less dense. As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense.
Answer:
It decreases so it is B
Explanation:
As altitude rises, air pressure drops.
Please help me with this review question.
Answer:
28.7%
Explanation:
efficiency = work output /work input × 100
What is the term used to describe the
force pushing on an area or surface?
A. density
B. pressure
C. surface tension
D. volume
The kinetic energy and the potential energy of the cannonball is constantly ________ as it travels through the air.
A. Changing
B. Increasing
C. Constant
D. Decreasing
Answer:
C. Constant
Explanation:
The total energy of the cannonball remains constant as it travels through the air.
Answer:
Explanation:
hi my name is Ava
The moment of inertia of the club head is a design consideration for a driver in golf. A larger moment of inertia about the vertical axis parallel to the club face provides more resistance to twisting of the club face for off-center hits. The mass of one club head is 200 g and its moment of inertia is 5000 g cm2 . What is the radius of gyration of this club head
Answer:
Explanation:
Moment of inertia I = M k² , where M is mass and k is radius of gyration .
Putting the given values in the equation
5000 = 200 x k²
k² = 25
k = 5 cm .
Radius of gyration is 5 cm .
A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 20
kg. The carriage has
energy. Calculate it
Answer:
Energy in carriage (Potential energy) = 4,116 J
Explanation:
Given:
Mass of baby = 20 kg
Height = 21 m
Find:
Energy in carriage (Potential energy)
Computation:
The energy accumulated in an object as a result of its location relative to a neutral level is known as potential energy.
In carriage accumulated energy is potential energy.
Energy in carriage (Potential energy) = mgh
Energy in carriage (Potential energy) = (20)(9.8)(21)
Energy in carriage (Potential energy) = 4,116 J
Fairly easy question I’ll give extra points help.
1. third law
2. first law
3. third law
4. second law