Answer:2000 cm³
Here, pressure remains constant.
So, b the gas law
V/V' = T/ T'
1000 / V' = 300 / 600
V' = 2000 cm³
Explanation:also pls mark brainliest
A car has a mass of 2000 kg and accelerates at 2 meters per second per second. What is the magnitude of the net force exerted on the car?
Hello!
[tex]\large\boxed{4000 N}[/tex]
Use the following equation to solve for the net force (N):
∑F = m × a
Plug in the given mass (kg) and speed (m)
∑F = 2000 * 2
Simplify:
∑F = 4000 N
Which is a valid velocity reading for an object?
45 m/s
45 m/s north
O 0 m/s south
O 0 m/s
Answer:45 m/s north
Explanation:
The superheroine Xanaxa, who has a mass of 65.1 kg , is pursuing the 78.7 kg archvillain Lexlax. She leaps from the ground to the top of a 153 m high building then dives off it and comes to rest at the bottom of a 17.5 m deep excavation where she finds Lexlax and neutralizes him. Does all this bring about a net gain or a net loss of gravitational potential energy
Answer:
There is net loss of gravitational energy .
Explanation:
When Xanaxa is on the ground , her potential energy is assumed to be zero . When she leaps to a height of 153 m , she gains gravitational energy . When she dives and reaches the surface , she loses potential energy and on reaching the ground her potential energy becomes zero . When she further goes down inside ground to a depth of 17.5 m , she loses potential energy further . Her potential energy becomes less than zero or negative .
Ultimately her potential energy changes from zero to negative in the whole process . So there is net loss of potential energy .
Did I hear correctly that the speed of light is different in deep space observation?
Answer:
Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.
5.0 L/s water flows through a horizontal pipe that narrows smoothly from 10.0 cm diameter to 5.0 cm diameter. A pressure gauge in the narrow section reads 50 kPa. What is the reading of the pressure gauge in the wide section
Solution :
The volume rate of flow is given by : R = 5.0 L/s
[tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]
The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]
∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]
= 0.637 meter per second
Then the speed of the water at wider section,
[tex]$R=A_1v_1$[/tex]
Similarly, the speed of water at narrow pipe.
The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m
[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]
= 2.55 meter per sec
Now from Bernoulli's theorem,
[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]
[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]
[tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]
= 50 kPa + 3.05 kPa
= 53.05 kPa
or 53000 Pa
This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.
The pressure reading in the wide section is "53.05 KPa".
First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.
[tex]V = A_1v_1[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m
A₁ = Area of narrow section = πr₁² = π(0.025 m)²
v₁ = velocity at narrow section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]
Similarly,
[tex]V = A_2v_2[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m
A₂ = Area of wide section = πr₁² = π(0.05 m)²
v₂ = velocity at wide section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]
Now, we will use Bernoulli's Theorem to find out the pressure wide section.
[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]
where,
[tex]\rho[/tex] = density of water = 1000 kg/m³
P₁ = pressure in narrow section = 50 KPa = 50000 Pa
P₂ = pressure in wide section = ?
Therefore,
[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]
P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa
P₂ = 53046.45 Pa = 53.05 KPa
Learn more about Bernoulli's Theorem here:
https://brainly.com/question/13098748?referrer=searchResults
The attached picture shows Bernoulli's Theorem.
Help me out please. It’d be greatly appreciated
Answer:
Option D
2Na + Cl₂ —> 2NaCl
Explanation:
We'll begin by stating the law of conservation of matter.
The law of conservation of matter states that matter can neither be created nor destroyed during a chemical reaction but can be transferred from one form to another.
For an equation to comply with the law of conservation of matter, the number of atoms of each element must be the same on both side of the equation. This simply means that the equation must be balanced!
NOTE: An unbalanced equation simply means matter has been created or destroyed.
Now, we shall determine which equation is balanced. This can be obtained as follow:
For Option A
Na + Cl₂ —> 2NaCl
Reactant:
Na = 1
Cl = 2
Product:
Na = 2
Cl = 2
Thus, the equation is not balanced!
For Option B
2Na + 2Cl₂ —> 2NaCl
Reactant:
Na = 2
Cl = 4
Product:
Na = 2
Cl = 2
Thus, the equation is not balanced!
For Option C
2Na + Cl₂ —> NaCl
Reactant:
Na = 2
Cl = 2
Product:
Na = 1
Cl = 1
Thus, the equation is not balanced!
For Option D
2Na + Cl₂ —> 2NaCl
Reactant:
Na = 2
Cl = 2
Product:
Na = 2
Cl = 2
Thus, the equation is balanced!
From the above illustrations, only option D has a balanced equation. Thus, option D illustrate the law of conservation of matter.
what is the mathematical definition of momentum? what is a more conceptual or descriptive definition of momentum?
Answer:
Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion.
Explanation:
The driver of a 3000 lb. car, coasting down a hill, sees a red light at the bottom, and must stop. His speed when he applies the brakes is 60 mph, and he is 100 feet (vertically) above the bottom of the hill. (a)How much energy as heat must be dissipated by the brakes if we neglect wind resistance and other frictional effects
Answer:
Explanation:
60 mph = 60 x 1760 x 3 / (60 x 60) ft /s
speed of car , v = 88 ft /s
kinetic energy of car = 1/2 m v²
= .5 x 3000 x 88²
= 11616 x 10³ poundal - foot
Potential energy = mgh
= 3000 x 32 x 100
= 9600 x 10³ poundal - foot
Total energy = potential energy + kinetic energy
= ( 11616 + 9600 )x 10³
= 21216 x 10³ poundal - foot .
This energy is dissipated as heat when brakes are applied on the car to stop the car .
In July 2015, Oregon State University, the National Oceanic and Atmospheric Administration, and the Coast Guard cooperated to send a hydrophone into Challenger Deep, the deepest part of the Mariana Trench. The titanium shelled recording device withstood the pressure 10,994 meters (nearly 7 miles!) under the ocean's surface. The hydrophone recorded 23 days of audio from the deepest part of the ocean floor. If the spherical hydrophone has a radius of 10 cm, what is the total force exerted on the titanium shell by the ocean water
Answer:
Explanation:
Pressure due to water column as deep as 10994 meters can be given by the following expression
Pressure = h d g , where h is depth of water , d is density of water and g is acceleration due to gravity .
Pressure = 10994 x 10³ x 9.8
= 10.77 x 10⁷ N / m²
Pressure will act on curved surface of the spherical shell , the effective surface area will be π R² where R is radius of the surface .
Effective surface = 3.14 x 0.1²
= .0314 m²
Total force = pressure due to water column x effective surface
= 10.77 x 10⁷ x .0314 N.
= 33.82 x 10⁵ N .
Question 18 of 25
Which type of reaction is shown in this energy diagram?
Energy
Products
Activation
Energy
Reoctants
to
ti
Time
A. Endothermic, because the products are lower in energy
B. Exothermic, because the reactants are lower in energy
C. Endothermic, because the reactants are lower in energy
D. Exothermic, because the products are lower in energy
Answer:
Endothermic, because the reactants are lower in energy (C)
Explanation:
From the graph, you can see the energy of the products is higher than the energy of the reactants. If you recall that when the enthalpy change Eproducts is gretater than Ereactants, the reaction is said to be endothermic.
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 1 m/s2 for 4 seconds. It then continues at a constant speed for 12.9 seconds, before getting tired and slowing down with constant acceleration coming to rest 66 meters from where it started. The tortoise accelerates uniformly for the entire distance, finally catching the hare just as the hare comes to a stop. 1)How fast is the hare going 1.6 seconds after it starts
Answer:
v = 4 m/s
Explanation:
Given that,
Initial speed of hare, u = 0
The hare accelerates uniformly at a rate of 1 m/s² for 4 seconds.
We need to find how fast is the hare going 1.6 seconds after it starts. Let the speed be v. So,
v = u+at
Substitute all the values,
v = 0+1×4
v = 4 m/s
So, the required speed of the hare is 4 m/s after it starts.
A heat pump is used to heat a building. The external temperature is lower than the internal temperature. The pump's coefficient of performance is 3.70, and the heat pump delivers 7.27 MJ as heat to the building each hour. If the heat pump is a Carnot engine working in reverse, at what rate must work be done to run it
Answer:
Heat pump needs 1.965 megajoules each hour to run.
Explanation:
The Coefficient of Performance ([tex]COP[/tex]), no unit, of a Carnot's heat pump is:
[tex]COP = \frac{Q_{H}}{W}[/tex] (1)
Where:
[tex]Q_{H}[/tex] - Heat received by the building, measured in megajoules.
[tex]W[/tex] - Work needed to run the heat pump, measured in megajoules.
If heat pump is a Carnot engine working in reverse, then the amount of work needed to run the heat pump is the least possible work. If we know that [tex]Q_{H} = 7.27\,MJ[/tex] and [tex]COP = 3.70[/tex], then the amount needed by the heat pump each hour is:
[tex]W = \frac{Q_{H}}{COP}[/tex]
[tex]W = \frac{7.27\,MJ}{3.70}[/tex]
[tex]W = 1.965\,MJ[/tex]
Heat pump needs 1.965 megajoules each hour to run.
An electron moves from point i to point f, in the direction of a uniform electric field. During this motion:Group of answer choicesthe work done by the field is positive and the potential energy of the electron-field system increasesthe work done by the field is negative and the potential energy of the electron-field system increasesthe work done by the field is positive and the potential energy of the electron-field system decreasesthe work done by the field is negative and the potential energy of the electron-field system decreasesthe work done by the field is positive and the potential energy of the electron-field system does not change
Answer:
the work done by the field is positive and the potential energy of the electron field system decreases
Explanation:
This exercise asks to find the work and the potential energy of an electron in an electric field.
Work is defined by
W = F .d = F d cos θ
the electric force is
F_e = q E
W = q E d cos θ
since the charge of the electron is negative the force is in the opposite direction to the electric field
W = - e E d
we select the direction to the right is positive, point i is to the left of point f,
therefore the work moving from point i to point F has two possibilities
* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated
* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too. The electron moves away from this point, this is in accordance with the indicated movement
In the latter case the electric field lines go from f to i point, therefore the Work is positive
Now let's examine the potential energy
ΔU = - q E .d
so we see that this definition is related to work,
ΔU = -W
Therefore, as the work is positive, the power energy must decrease
When reviewing the different answers, the correct ones are:
the work done by the field is positive and the potential energy of the electron field system decreases
The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.
An electron moves from point i to point f, in the direction of a uniform electric field, then the potential energy of the electron can be calculated s given below.
[tex]\Delta V=-qEd[/tex]
Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.
The work done by the electron in the uniform electric field can be calculated as,
[tex]W = F\times d \times cos\theta[/tex]
Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and for uniform electric field, the value of [tex]\theta[/tex] is zero.
Hence [tex]W=F\times d\times 1\\W=F \times d[/tex]
Electric force [tex]F = q E[/tex]
By substituting the value of electric force on the above formula,
[tex]W = qEd[/tex]
Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.
[tex]W = -\Delta V[/tex]
The work done by the electron is negative and the potential energy of the electron increases.
For more information, follow the link given below.
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