What is the average momentum of Cart 1 during the entire time shown before
kg-m
the collision, in units .?
$
Note that the collision appears to take place somewhere between 0.5 s and 0.6
s (perhaps at 0.6 s exactly, but we cannot be sure), so we can safely say that
data up to and including 0.5 s is "before the collision."

The temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.

The given parameters;

The mean velocity of the gas molecules is calculated as follows;

[tex]\tau = \frac{\lambda _{rms}}{V_{rms}} \\\\V_{rms} = \frac{\lambda _{rms}}{\tau} \\\\V_{rms} = \frac{4.32 \times 10^{-8} }{3 \times 10^{-10}} \\\\V_{rms} = 144 \ m/s[/tex]

The temperature  of the gas molecules is calculated as follows;

[tex]V_{rms} = \sqrt{\frac{3kT}{M} } \\\\V_{rms}^2 = \frac{3kT}{M} \\\\T = \frac{V_{rms} ^2 M}{3k}[/tex]

where;

[tex]T = \frac{V_{rms} ^2 M}{3k} \\\\T = \frac{(144)^2 \times (6.0 \times 10^{-25})}{3 \times 1.38 \times 10^{-23}} \\\\T = 300.5 \ K[/tex]

The number of gas molecules per unit volume is calculated as follows;

[tex]\lambda = \frac{1}{4\pi \sqrt{2} \ r^2 n} \\\\n = \frac{1}{\lambda 4\pi \sqrt{2} \ r^2} \\\\n = \frac{1}{(4.32 \times 10^{-8}) \times 4 \pi \times \sqrt{2} \ \times (1\times 10^{-10})^2} \\\\n = 1.303 \times 10^{26} \ molcules/m^3[/tex]

The pressure of the gas molecule is calculated as follows;

[tex]n = \frac{P}{kT} \\\\P = nkT\\\\P = (1.303 \times 10^{26} ) \times (1.38 \times 10^{-23}) \times (300.5)\\\\P = 540,341.07 \ Pa\\\\P = 5.33 \ atm[/tex]

Thus, the temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.

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Answer:

When the input to an inverter is high (1) the output is low (0); and when the input is low, the output is high.

Answer:

Explanation:

A wonderful (NOT) mixed unit problem

convert mph to m/s

v = 10 mi/hr(5280 ft/mi)(12 in/ft)(2.54 cm/in) / (100 cm/m) / (3600 s/hr)

v = 4.4704 m/s

KE = ½(0.006)4.4704²

KE = 0.05995342848 J

KE = 0.06 J

Answer:

Hey mate

Explanation:

Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height.

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Question

what is the force exerted on a selectively permeable membrane beacuse water has moved from an area of higher concentration to an area of Lower concentration of water?

Answer

OSMOSIS

Explanation

Osmosis: This is the hydrostatic force acting to equalize the concentration of water on both sides of the membrane that is impermeable to substances dissolved in that water. Water will move along its concentration gradient.

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[tex]\large\huge\green{\sf{Yes}}[/tex]

Answer:

I believe The Knowledge that we apply everyday or based on these methods and discoveries may represent us

Explanation:

Answer:

Explanation:

vA / vB = √(TA/(m/L)) / √(TB/(m/L))

The lengths are the same, so the L divides out to 1

The material is identical so the mass will be directly proportional to the cross sectional area of the string

vA / vB = √(TA/(πdA²/4)) / √(TB/(πdB²/4))

π/4 is common so divides out to 1

vA / vB = √(TA/dA²) / √(TB/dB²)

vA / vB = √(410/0.489²) / √(809/1.27²)

vA / vB = 41.407 / 23.396

vA / vB = 1.8488961...

vA / vB = 1.85

Based on the information given, it can be noted that the bat was either shorter or longer than what he expected.

From the information given, it was stated that Raghu wants a cricket bat to be made by a carpenter and he tells the carpenter that the length of the bat should be 7 hand spans.

Since he got disappointed when he collected the bat, the reason for this will be because the bat was either shorter or longer than what he requested.

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Answer:

The removal of heat energy slows the speed of particles

Explanation:

When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.

Answer:

B

Explanation:

Simply take all forces pointing to the right of the box as positive and all of the forces pointing to the left of the box as negative and add all values.

ΣF = 7 + 18 + (-20) = 5N to the right

Answer:What is the half life of the element Lokium?

The half-live of the element Lokium is 4.

Explanation:

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Answer:

The half-life of the element Lokium is 4

What is half-life?

Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.

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Hi there!

1.

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

2.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

3.

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

4.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

Answer:

Explanation:

Objects want to continue doing what they're already doing because they are "lazy." This is called inertia.

Answer:

Decant it.

Explanation:

Pour the water/sugar solution off the sand. When the sand wants to start coming out as well, Stop and add fresh water to the beaker, stir to rinse the remaining solution into a less concentrated solution and decant again.

Repeat the dilution process until the mix is essentially sand and water, then drive the remaining water from the sand by drying.

Answer:

35m/s

Explanation:

Answer:

35m/s

Explanation:

Simply multiply 5 and 7.

Answer:

Water erosion can slowly wash away dirt and rocks, wearing away landforms and sometimes forming rivers that can even carve out stone over time.

Answer:

Electrons

Explanation:

Electrons are negatively charged sub-atomic particles, therefore when a body's negatively charged, it means that there's more electrons than protons.

Answer:

Explanation:

if you are pulling in the positive direction and Upward is also positive.

F = -0.96i + 0.58(9.8)j

F = -0.96i + 5.7j

F = √(-0.96² + 5.7²) = 5.8 N

You are pulling a sled with constant velocity by using a rope then the surface force the ground exerts on the sled will be 5.8 N.

The pulling force conveyed axially by a string, cable, chain, or another analogous object, or by either end of a rod, truss member, or another such three-dimensional object, is referred to as tension in physics. Another way to think of tension is as the action-reaction pair of forces acting at each end of the previous elements. Perhaps tension is the polar opposite of compression.

A restoring force may cause what is also referred to as tension when atoms or molecules are torn apart from one another at the atomic level and build up potential energy.

If you are moving upward and pushing in a positive way,

F = -0.96 i + 0.58(9.8) j

F = -0.96 i + 5.7 j

F = [tex]\sqrt{(-0.96^2 + 5.7^2)}[/tex]

F= 5.8 N

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Answer:

[tex]\huge\color{skyblue}\boxed{\colorbox{black}{Answer ☘}}[/tex]

[tex]total \: resistance \: ( R_{t} ) = (3 + 1)ohm \\ (since \: the \: resistors \: are \: connected \\ in \: series)[/tex]

[tex]current \: flowing \: through \: circuit(I) = 2A \\ \\ now... \: by \: ohms \: law \\ V = IR\\ V = (2)(4) \\ = > 8v[/tex]

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Answer:

B. Increase the electric force by rubbing the balloon for a longer period of time.

Explanation:

The relative kinetic energy of molecules in the soda is least energy and above the soda in the glass is greatest energy.

The relative kinetic energy of gas molecules increases with increase in the mean distance between the gas molecules.

Also, relative kinetic energy of gas molecules increases with in the temperature of the gas molecules and decreases with a decrease in the temperature of the of the gas molecules;

ΔK.E ∝ T

The ice in the soda lowers the temperature of the gas molecules, thereby reducing their average speed which in turn reduces the average kinetic energy of the gas molecules in the soda.

Above the soda in the glass, the concentration of the gas molecules is less and their mean distance is greatest when compared to inside the soda. This results to an increase in the speed of the gas molecules which increases their average kinetic energy.

Thus,  the relative kinetic energy of molecules in the soda is least energy and above the soda in the glass is greatest energy.

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Answer:

3. years

Explanation:

The independent variable is the x axis of graph

Answer:

Explanation:

The word "sheet" implies that the copper is quite thin.

Copper is also a very good conductor of heat.

Therefore, with a very short heat flow distance to cover and a high rate of heat transmission, temperature differences on either side of the sheet are almost instantaneously eliminated by heat flow.

Answer:

Explanation:

Ignoring friction, the initial kinetic energy will convert to maximum potential energy at its highest point.

PE = KE

mgh = ½mv²

    h = v²/2g

    h = 36.4²/ (2(9.81))

    h = 67.53109...

    h = 67.53 m

The relationship between the potential and the electric field allows to find the results for the value of the electric field as a function of the distance is:

Electric potential is defined by the change in potential energy of a test charge between two points, between the value of the test charge.

          dV = - E . ds

          E = [tex]- \frac{dV}{ds} \ \hat s[/tex]  

Where the bold letters indicate vectors, V is the potential difference, E the electric field and s the path.

Let's apply this expression for each section of the given graph:

1) section from x₀ = 0 to x_f = 2 m, the potential is V₀ = 2 V is constant.

  The derivative of a constant is zero.

        E = 0

2) Section between x₀ = 2 and x_f = 4 m, the potential varies linearly from V₀ = 2 v to V_f = -2 V.

We look for the equation of the line.

       V-V₀ = m (x- x₀)

We carry out the derivative.

      E = - m i ^

The slope (m) is:

       [tex]m= \frac{V_f - V_o}{x_f- x_o}[/tex]  

Let's calculate.

       [tex]m= \frac{-2 -2}{4-2} = \ -2 \ V/m[/tex]  

Let's substitute.

       E =  [tex]2 \hat i \ V/m[/tex]  

3) From x₀ = 4 to x_f = 4.5 m, the potential varies from V₀ = -2 to V_f = 0.

We look for the equation of the line and we derive.

      E = - m i ^

Let's  substitute.

      [tex]m = \frac{0-(-2)}{4.5-4} = \ 4 V/m[/tex]  

    E = - 4 [tex]\hat i[/tex] V / m

4) From x₀ = 4.5 m to x_f = 6m.  The potential is constant and the derivative of a constant is zero.

      E = 0

5) From x₀ = 6m to x_f = 8 m, the potential changes linearly from v₀ = 0 to V_f = 1 V

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{1-0}{8-6} = \ 0.5 \ V/m[/tex]  

      E = - 0.5 [tex]\hat i[/tex] V/m

6) From x₀ = 8m to x_f = 9m, the potential changes linearly from V₀ = 1 V to V_f = -1.

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-1-1}{9-8} = \ -2 \ V/m[/tex]

Let's substitute.

       E = 2 [tex]\hat i[/tex] V/m

7) From x₀ = 9m to x_f = 10 m, the potential changes linearly from V₀ = -1 V to V_f = -2 V

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-2+1}{10-9} = \ -1 \ V/m[/tex]

Let's substitute.

       E = 1 [tex]\hat i[/tex]  V/m

In the attachment we can see these Electric fields as a function of distance.

In conclusion, the relationship between the potential and the electric field we can find the results for the value of the electric field as a function of the distance is:

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The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

[tex]\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}[/tex]

[tex]\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}[/tex]

[tex]\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}[/tex]

[tex]\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }[/tex]

[tex]\mathbf{\omega^2=\dfrac{39.24 }{2}}[/tex]

[tex]\mathbf{\omega=\sqrt{19.62 } \ rad/sec}[/tex]

[tex]\mathbf{\omega=4.429 \ rad/sec}[/tex]

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

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The angular velocity of the wheel depends on the mass, radius and the

mode of rotation of the wheel (with or without slipping).

Reasons:

The given parameters are;

Radius of the wheel, r = 2.0 m

Height of the incline, h = 8.0 m

Required:

Angular velocity of the wheel at the bottom of the incline.

Solution:

The potential energy of the wheel at the top of the hill, P.E. = m·g·h

[tex]Sum \ of \ the \ kinetic \ energy \ of \ the \ wheel, \ K.E. = \mathbf{\displaystyle \frac{1}{2} \cdot m \cdot v^2 + \frac{1}{2} \cdot I \cdot \omega ^2}[/tex]

Where;

v = The translational velocity of the wheel = ω·r

I = The moment of inertia of the wheel = m·r²

Therefore'

[tex]Sum \ of \ K.E. = \displaystyle \frac{1}{2} \cdot m \cdot (\omega \cdot r)^2 + \frac{1}{2} \cdot m \cdot r^2 \cdot \omega ^2 = \mathbf{m \cdot r^2 \cdot \omega^2}[/tex]

At the bottom of the hill, the potential energy is converted to kinetic energy

Therefore;

P.E. = Sum of K.E.

m·g·h = m·r²·ω²

g·h = r²·ω²

[tex]\displaystyle \omega = \sqrt{ \frac{g \cdot h}{r^2} } = \mathbf{ \frac{\sqrt{g \cdot h} }{r}}[/tex]

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Therefore;

[tex]\displaystyle \omega = \frac{\sqrt{9.81 \times 8} }{2} \approx \mathbf{ 4.43}[/tex]

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