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].
Review please help.
Answer:
1 and 3
Explanation:
because they are going up from 0
What is the importance of using locally available resources in creating art?
Answer:
please give me brainlist and follow
Explanation:
Using locally available resources for art help in the preservation of environment. A significant and practical aspects of art is material significance. The items used by artists while making an art piece affects both the form and the material. Every material delivers something special in the creative process.
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
light of wavelength 485 nm passes through a single slit of width 8.32 *10^-6m. what is the single between the first (m=1) and second (m=2) interference minima?
Answer:
3.35
Explanation:
Got it on Acellus
The light of wavelength 485 nm passes through a single slit. The single between the first (m=1) and second (m=2) interference minima is 3.36°.
What is diffraction?Diffraction is the phenomenon of bending of waves through obstacles.
Given is the wavelength λ= 485 nm, silt width d = 8.32 *10⁻⁶ m, then the angle θ will be
d sinθ =mλ
for m=1, sin θ₁ = λ/d
for m=2, sin θ₂ = 2λ/d
Substitute the values into both expressions to find the angles,
sin θ₁ = 485 x 10⁻⁹ / 8.32 *10⁻⁶
θ₁ = 3.34°
and sin θ₂ = (2 x 485 x 10⁻⁹ )/ 8.32 *10⁻⁶
θ₂ = 6.7°
The angle between m =1 and m=2 will be
θ₂ -θ₁ = 6.7° - 3.34° =3.36°
Thus, angle between the first (m=1) and second (m=2) interference minima is 3.36°.
Learn more about diffraction.
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A 2.0-kg cart is rolling along a frictionless, horizontal track towards a 1.8-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is 5.9 m/s, and the second cart's velocity is -2.7 m/s. (a) What is the total momentum of the system of the two carts at this instant
Answer:
the total momentum of the system before collision is 6.94 kgm/s
Explanation:
Given;
mass of the first cart, m₁ = 2.0 kg
mass of the second cart, m₂ = 1.8 kg
velocity of the first cart before collision, u₁ = 5.9 m/s
velocity of the second cart before collision, u₂ = -2.7 m/s
The total momentum of the system before collision is calculated as follows;
[tex]P_t = P_1 + P_2 \\\\P_t = m_1u_1 + m_2u_2\\\\P_t = (2\times 5.9) + (1.8 \times -2.7)\\\\P_t = 11.8 - 4.86\\\\P_t = 6.94 \ kgm/s[/tex]
Therefore, the total momentum of the system before collision is 6.94 kgm/s
I’ll mark you as brinlist please help.
Answer:
245 divided by 5.14=47.6653696 or 47.66
Explanation:
An object undergoing simple harmonic motion takes 0.15 s to travel from one point of zero velocity to the next such point. The distance between those points is 30 cm. (a) Calculate the period of the motion. s (b) Calculate the frequency of the motion. Hz (c) Calculate the amplitude of the motion. cm
Answer:
Explanation:
Point of zero velocity are extreme points situated on either side of equilibrium position .
a )
Time taken to travel between these two points is .15 s
time for half the oscillation = .15 s
Time for full one oscillation = .30 s .
Time period of oscillation = .30 s
b)
frequency = 1 / time period
= 1 / .30s = 3.33 oscillation per second.
c )
Distance between these two point is equal to two times amplitude
2 x amplitude = 30 cm
amplitude = 15 cm
Which of the following best defines
weather?
A. the expanding or contracting of the atmosphere
B. the measurement of the amount of water vapor in the
atmosphere
C. the condition of the atmosphere at a certain time and
place
Help Resources
D. the average air temperature of a specific region
Answer:
I'd say D
Explanation:
because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)
A water balloon weighing 4.5 N rests on a table. The balloon has an area of 2.6 x 10-3
m² in contact with the table. What pressure does the balloon exert on the table?
Answer:
the pressure the balloon exerts on the table is 1,730.77 N/m²
Explanation:
Given;
weight of the water balloon, F = 4.5 N
area of the balloon, A = 2.6 x 10⁻³ m²
The pressure the balloon exerts on the table is calculated as follows;
[tex]P = \frac{F}{A}[/tex]
substitute the given values and solve for pressure, P;
[tex]P = \frac{4.5}{2.6 \times 10^{-3}} \\\\P = 1,730.77 \ N/m^2[/tex]
Therefore, the pressure the balloon exerts on the table is 1,730.77 N/m²
Which of the
following
DECREASES
as you go UP a
mountain?
A. climate
B. altitude
C. amount of oxygen
D. buoyancy
Answer:
C. Amount of oxygen
Explanation:
Options A and D are invalid as they aren't affecting factors.
Option B is false as the altitude increases as you go up a mountain.
Option C is true as the air pressure (atmospheric pressure) is inversely proportional to the height/altitude of the mountain.
What kind of energy is in a moving skateboard
Answer:
I guess it is kinetic energy
Answer:
kinetic energy because my dog told me
A 1.65-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 16.0 N. (a) What are the frequencies of the first three allowed modes of vibration
Answer:
Explanation:
mass per unit length ρ = .100 / 1.65 = .0606 . kg /m
length of wire L = 1.65 m
For fundamental frequency , the expression is as follows
n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
L = 1.65 , T = 16 n and m = .0606
n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]
= 4.9 /s .
This is fundamental frequency .
other mode of vibration ( first three ) will be as follows
4.9 x 2 = 9.8 /s ,
4.9 x 3 = 14.7 /s .
if the density of a napthalene ball is 0.02kg.what is the mass of the napthalene ball if it has a volume of 100m³
Please solve this and please don’t put in a link.
Answer:
synthesis
Explanation:
It is a combination (synthesis) reaction.
A+B→AB
Answer:
i think it is a synthesis
Explanation:
synthesis reaction occurs when two substances combine and form a compound...
hope this helps!!
A student takes the elevator up to the fourth floor to see her favorite physics instructor. She stands on the floor of the elevator, which is horizontal. Both the student and the elevator are solid objects, and they both accelerate upward at 5.80 m/s2. This acceleration only occurs briefly at the beginning of the ride up. Her mass is 87.0 kg. What is the normal force exerted by the floor of the elevator on the student during her brief acceleration? (Assume the j direction is upward.)
The normal force exerted by the floor of the elevator on the student during her brief acceleration is 1357.2 N.
To determine the normal force exerted by the floor of the elevator on the student during the brief acceleration, we need to analyze the forces acting on the student.
The force of gravity acting on the student can be calculated using the equation F_gravity = m * g, where m is the mass of the student and g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, F_gravity = 87.0 kg * 9.8 m/s² = 852.6 N.
Since the elevator and the student are accelerating upward, there is an additional upward force acting on the student. This force is equal to the mass of the student multiplied by the acceleration of the elevator, which is given as 5.80 m/s². So, the upward force is F_upward = m * a = 87.0 kg * 5.80 m/s² = 504.6 N.
Now, the normal force exerted by the floor of the elevator on the student is equal to the sum of the gravitational force and the upward force. In this case, the normal force is given by F_normal = F_gravity + F_upward = 852.6 N + 504.6 N = 1357.2 N.
For such more questions on acceleration
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What do thermal energy and electrical energy have in common
Answer:
you can write some points its an explanation
and similarities. or common
Explanation:
Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy. A whole branch of physics, thermodynamics, deals with how heat is transferred between different systems and how work is done in the process (see the 1ˢᵗ law of thermodynamics).
The faster the atoms or molecules move, the more heat or thermal energy they have. ... A hair straightener turns the electrical energy from a wall outlet into heat (thermal energy). 4. As electricity runs through the filaments in a space heater, the electrical energy is converted into heat (thermal energy).
Please help me easy question just get mixed up.
Answer:
The mechanical energy of the ball is 112.667 J
Explanation:
Mechanical energy = kinetic energy + potential energy
Where:
kinetic energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]
potential energy = mgh
So that for the given ball, its mechanical energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] + mgh
= m([tex]\frac{1}{2}[/tex][tex]v^{2}[/tex] + gh)
But, m = 400 g = 0.4 kg, v = 23.0 m/s, g = 9.81 m/[tex]s^{2}[/tex] and h = 1.75 m.
Mechanical energy of the ball = 0.4([tex]\frac{1}{2}[/tex]([tex]23^{2}[/tex]) + 1.75*9.81)
= 0.4(264.5 + 17.1675)
= 0.4 x 281.6675
= 112.667
The mechanical energy of the ball is 112.667 J.
What is a transfer of energy called?
A. Displacement
B. Acceleration
C. Work
D. Torque
Select the correct answer Which object is an insulator
A. iron
b. cooper
c. plastic
d. salt water
HELP URGENT PLEASE!!!!!!!
Answer:
I think c I dont know sorry if I'm wrong
_____is the nitrogen base found only in DNA.
Answer:
thymine
Explanation:
The nitrogen base found only in DNA is known as thymine which is also called 5-methyl uracil as thymine is a derivative of uracil.
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.
What is Newton's scientific view?
Answer:
Newton's first law of motion concerns any object that has no force applied to it.
Explanation:
three laws of motion and the law of universal gravitation.
Please answer this for 15 points please don’t put in a link.
Answer:
c. Double Replacement
Explanation:
As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.
Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2
and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.
A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.
Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm
Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t[tex]_{min[/tex] = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t[tex]_{min[/tex] = 750 / 2(1.20)
t[tex]_{min[/tex] = 750 / 2.4
t[tex]_{min[/tex] = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t[tex]_{min[/tex] = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t[tex]_{min[/tex] = 750 / 4(1.50)
t[tex]_{min[/tex] = 750 / 6
t[tex]_{min[/tex] = 125 nm
Therefore, the minimum thickness be now will be 125 nm
Which of these cubes absorb the most light?
Answera black cube or dark colors cause dark colors suck in heat
Pls help will mark Brainliest
Balance the equation by choosing the correct coefficient numbers in the drop down menus.
[Select]
SO2 +
[Select]
VH₂ →
[Select]
S +
[ Select]
H20
It is suggested you write this on scratch paper and balance it before choosing your answers :)
Answer:
SO₂ + 2H₂ —> S + 2H₂O
The coefficients are: 1, 2, 1, 2
Explanation:
SO₂ + H₂ —> S + H₂O
The above equation can be balance as follow:
SO₂ + H₂ —> S + H₂O
There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by writing 2 before H₂O as shown below:
SO₂ + H₂ —> S + 2H₂O
There are 2 atoms of H on the left side and 4 atoms the right side. It can be balance by writing 2 before H₂ as shown below:
SO₂ + 2H₂ —> S + 2H₂O
Now, the equation is balanced.
The coefficients are: 1, 2, 1, 2
An object is projected from a height of 100m above the ground at an angle of 300to the
horizontal with a velocity of 100m/s.
Calculate
(4)
(4)
The maximum height reached above the ground
Time of flight
The velocity and the direction of the object 1 sec before it hit the ground
(4)
Answer:
a) y = 127.6 m, b) 11.9s, c) v = 103.6 m / s, θ’= 326.7º
Explanation:
This is a missile throwing exercise
let's start by breaking down the initial velocity
sin 30 = [tex]\frac{v_{oy} }{v_o}[/tex]
cos 30 = v₀ₓ / v₀
v_{oy} = v₀ go sin 30
v₀ₓ = v₀ cos 30
v_{oy} = 100 sin 30 = 50 m / s
v₀ₓ = 100 cos 30 = 86.6 m / s
a) the maximum height is requested.
At this point the vertical velocity is zero (v_y = 0)
v_y² = [tex]v_{oy}^2[/tex] - 2 g y
0 = v_{oy}^2 - 2g y
y = [tex]\frac{v_{oy}^2 }{2g}[/tex]
y = 50² / (2 9.8)
y = 127.6 m
b) Flight time
this is the time it takes to reach the ground, the reference system for this movement is taken on the ground this is a height of y = 0 m and the body is at an initial height of i = 100m
y = y₀ + v₀ t - ½ g t²
0 = 100 + 50 t - ½ 9.8 t²
we solve the quadratic equation
4.9 t² - 50 t - 100 = 0
t = [tex]\frac{50 \pm \sqrt{50^2 + 4 \ 4.9 \ 100} }{2 \ 4.9}[/tex]
t = [tex]\frac{50 \ \pm \ 66.8}{9.8}[/tex]
t₁ = 11.9 s
t₂ = -8.4 s
flight time is 11.9s
c) The time 1 s before hitting the ground is
t1 = 11.9 -1
t1 = 10.9 s
let's find the vertical speed
v_y =[tex]v_{oy}[/tex] - g t
v_y = 50 - 9.8 10.9
v_y = -56.8 m / s
the negative sign indicates that the direction of the velocity is downward.
On the x-axis there is no acceleration therefore the speed is constant.
Let's use the Pythagorean theorem
v = [tex]\sqrt{v_x^2+v_y^2}[/tex]
v = [tex]\sqrt { 86.6^2 + 56.8^2}[/tex]
v = 103.6 m / s
let's use trigonometry
tan θ = [tex]\frac{v_y}{v_x}[/tex]
θ = tan⁻¹ \frac{v_y}{v_x}
θ = tan⁻¹ (-56.8 / 86.60)
θ = -33.3º
the negative sign indicates that it is measured clockwise from the x-axis
for a counterclockwise measurement
θ’= 360 - θ
θ' = 360 - 33.3
θ’= 326.7º
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