Please help, I keep trying a bunch of things but keep getting them wrong. I don't know where I am going wrong here.
1. Boyle's Law states the volume and pressure of a gas are inversely proportional.
Name the three units of the constant of proportionality between pressure and volume in alphabetical order. (**I have the first two)
2. The ideal gas law can be written as (PV/nT=R). Name the units for R.

Answer:

equal

Explanation:

Answer:

23

Explanation:

3x-4

Answer:

Explanation:

d = ½(2.0)4² - ½(2.0)3² = 7 m

. Power

Patintero

, also known as harangang-taga or tubigan, (Intl. Translate: Escape from the hell or Block the runner) is a traditional Filipino children's game. Along with tumbang preso, it is one of the most popular outdoor games played by children in the Philippines.[1]

is an ultramarathon trail race held in Frozen Head State Park near Wartburg, Tennessee. If runners complete 60 miles (97 km) this is known as a "fun run." The full course is about 100 miles (160 km). The race is limited to a 60-hour period and takes place in late March or early April of each year.

We have that for the Question "find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A."

Answers:

From the question we are told

The concentration of [tex]CCl_4[/tex] at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s

A) the mass flow rate of CCI_4 as it passes point A is the same as the mass flow rate at which CCI_4 enters the left end of the tube

Therefore, the mass flow rate of CCI_4 at point A

=  [tex]5.86*10^{-13} kg/s[/tex]

B) From Fick's law

[tex]\deltaC = \frac{mL}{DAt}\\\\ Assume L = 5*10^{-3}, A = 3*10^{-4}\\\\\deltaC = \frac{5.86*10^{-13} * 5*10^{-3}}{21.9*10^{-10} * 3*10^{-4}}\\\\\deltaC = 4.46*10^{-3}kg/m^3[/tex]

Then,

[tex]Concentration = 1.71*10^{-2} - 4.46*10^{-3}\\\\= 12.6*10^{-3}kg/m^3[/tex]

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Answer:

This happens when the mass of one of the objects is very large and does not move.

Explanation:

Answer:

A straight wire in a magnetic field experiences a force of 0.022 N when the current in the wire is 1.1 A. The current in the wire is changed ...

1 answer

·

Top answer:

The magnetic force that acts on the wire in the first case can be written as

Explanation:

Answer:

The box arrives first.

Explanation:

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According to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

Friction may be defined as the resistance that is offered by the surfaces that are in contact when they move past each other. It is a type of force that opposes the motion of a solid object over another.

There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. According to the context of this question, the sphere possesses less friction as compared to the box. This is because the box has an irregular surface that possesses high friction over the inclined surface.

Therefore, according to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

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Answer:

Explanation:

θ = arctan(56.0/76.0) = 36.4° West of North

average velocity is √(56.0² + 76.0²) / (8 + 5) = 94.4/13 = 7.26 m/s

The formula of force is F = ma.

Force: This can be defined as the product of mass and acceleration. The S.I unit of force is Newton (N).  Force is a vector quantity

Types of force

⇒ The formula of force is

⇒ Where:

Hence, the formula of force is F = ma.

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Before looking at the formula, Lets understand what is force?

The formula used to find force is:

[tex] \sf \nrightarrow \: F=m×a[/tex]

Where,

Equation in words:- Force applied on an object is directly proportional or equal to the product of the mass of object and the acceleration of object.

Answer:

Answer: At its lowest point, the kinetic energy of a watermelon just as it touches the ground is zero if it does not touch anything on its way down.

Explanation:

This is because that upon having been dropped from a height, an object no longer has any kinetic energy at all. Kinetic energy transforms to gravitational potential energy during the fall and there's nothing left over for kinetic once you stop accelerating anymore. Fortunately, things don't stay still until they land very often! For example, if a person catches the fruit with his hands after some air resistance slows him down - making him more similar in speed to the lag of the trajectory - then he'll be able to share some of his saved up gravitational potential with that watermelon and do some

The distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m

If we consider the potential energy (U) between the pions, then (U) can be expressed as:

[tex]\mathbf{U = \dfrac{kq^2}{r} ---- (1)}[/tex]

Given that at some instance, the potential energy becomes negligible compared to the final K.E.

As such the conservation of the total energy in the system can be given as:

Again, if we consider the ratio of the potential energy to the kinetic energy to be about 0.01, then:

[tex]\mathbf{\dfrac{U}{K}= 0.01} \\ \\ \mathbf{U = 0.01 K----(2)}[/tex]

∴

Equating both equations (1) and (2) together, we have:

[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 K}[/tex]

[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }[/tex]

[tex]\mathbf{r =\dfrac{kq^2}{ 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }}[/tex]

where:

Replacing their values in the  above equation, the distance (r) between the pions is calculated as:

[tex]\mathbf{r =\dfrac{(9\times 10^9 \ N.m^2/C^2) (1.6022 \times 10^{-19} \ C)^2}{ 0.01 \Bigg [ (2.5 \times 10^{-28\ kg } )\times (3\times 10^8 \ m/s)^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{(2.97 \times 10^8 \ m/s)^2}{(3 \times 10^8 \ m/s)^2} }} \Big ] \Bigg] }}[/tex]

distance (r) = 1.45 × 10⁻¹⁶ m

Therefore, we can conclude that the distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m

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-cant travel through space since there's no molecules to travel through

-sound travels 4.3 times faster in water than air

-sounds are waves that pass through our ears via vibrations and travel by vibrations of molecules

-different types of sound like audible, inaudible, infrasonic, ultrasonic,

-sounds waves are either longitudinal, mechanical and pressure waves

-sound travels at 767 miles per hour

Answer:

What is the average force of collision?

Explanation:

The velocity of the combined mass after impact is found by conservation of momentum

0.480(75) + 0.240(0) = (0.480 + 0.240)v

v = 50 m/s

An impulse results in a change of momentum

FΔt = mΔv

F = mΔv/Δt

for the pigeon

F = 0.240(50 - 0)/0.015

F = 800 N

for the falcon

F = 0.480(50 - 75)/0.015

F = -800 N

The final speed(v) of the Peregrine falcons and pigeon is 50 m/s

The objective of this question is to determine the final speed(v) of the Peregrine falcons and pigeon

From the information given:

According to the conservation of momentum, we can say that the totality of momentum before and after the collision is the same. As such;

[tex]\mathbf{m_fv_f = (m_f+m_p) v}[/tex]

[tex]\mathbf{480 g \times 75 m/s = (480 + 240) \ g \times v}[/tex]

[tex]\mathbf{v = \dfrac{480 g \times 75 m/s}{(480 + 240) \ g}}[/tex]

[tex]\mathbf{v = \dfrac{36000 \ m/s}{ 720}}[/tex]

v = 50 m/s

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Answer:D: the velocity is zero

Explanation:

Answer:

Explanation:

½(75)v² = ½(0.008)390²

        v² = (0.008)390²/75

        v² = 16.224

         v = 4.027...

         v = 4.0 m/s

Answer:

Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.

Explanation:

Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.

Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.

The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.

Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].

The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].

The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:

[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].

The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:

[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:

[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:

[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].

[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.

[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].

[tex]9995\, \Delta v = 25000[/tex].

[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].

Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.

Answer:

17 J

Explanation:

work done= force X distance moved

Answer:

Explanation:

s = s₀ + v₀t + ½at²

s = 0 + 5.4(12) + ½(0)12²

s = 64.8 m

Answer: Yes, the " Cosmology, philosophy and physics" is the most famous book of the philosopher, Alexis karpouzos. But and the other books are important. For example, the " The self-criticism of science", the "Universal conscilusness" and the "Non-duality".

Explanation:

Answer:

What does Newton's third law describe?

Answer:

Energy consumed is 0.00033 Joules.

Explanation:

the formula of Energy is:

Energy = power/ time.

From the diagram, the angular speed will increase clockwise, the sign of the angular acceleration will be negative and the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative. The correct answer is option B

Given that two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Counterclockwise is considered the positive rotational direction.

If m1 is 24 kilograms, m2 is 12 kilograms, and mbar is 10 kilograms, The moment of object m1 will be equal to the moment of object m2 without the Mbar

Let assume that the length L of the seesaw is 9 cm.

Anticlockwise moment = 24 x 9/3 = 72Nm

Clockwise moment = 12 x 2(9/3) = 72 Nm

With the consideration of mass of the bar Mbar, this will add to clockwise moment of the seesaw.

Therefore, the direction of rotation will be clockwise direction.

Angular acceleration is positive when object is speeding up and negative when slowing down. Also, angular acceleration is positive when speed increases in an anticlockwise direction and negative when speed increases in the clockwise direction.

From the diagram, since the angular speed increase clockwise, the sign of the angular acceleration will be negative.

We can conclude that the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative.

The correct answer is option B

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Hi there!

We can use the equation:

v = at, where in this instance:

v = velocity (m/s)

a = acceleration due to gravity (m/s²)

t = time (s)

g ≈ 9.8 m/s², so:

v = 9.8(3) = 29.4 m/s

Answer:

how many feet?

Explanation:

Answer:

Its acceleration will increase.

Explanation:

For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity

This is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves smaller radii and for lager radii acceleration will be less.

acceleration, a = v²/r

For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.

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Answer:

D

Explanation: