Assume ideal behavior. How many moles of are needed to fill a 2000L weather balloon at 210K and 1atm pressure? Round your answer to the nearest mole (use R = 0.082057 L atm mol-1K-1)

To make the boiler more efficient, we would replace the resistor with another double the length.

We know that the resistance of the resistor in the boiler is given by R = ρl/A where

Also, the power generated by the resistor is P = i²R where

If i is constant, we see that P ∝ R.

So, to increase the efficiency, the value of the resistance has to increase.

Also, since R = ρl/A and ρ and A are constant, we see that R ∝ l.

Since P ∝ R and R ∝ l ⇒ P ∝ l.

So, to increase the efficiency, we would double the length of the resistor.

So, to make the boiler more efficient, we would replace the resistor with another double the length.

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Answer:

because of the gravity of the earth

Force = mass x acceleration:

Force = 13.2 kg x 9.4 m/s

Force = 124.08

Force = 124 Newtons

The force will be  = 124 N              

Newton's second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

according to newton's second law of motion

Force = mass * acceleration

given

mass = 13.2 kg

acceleration = 9.4 m/s

force = 13.2 * 9.4

        = 124.08 N

Rounding off answer to  the appropriate number of significant figures

force = 124 N

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D)

Explanation:

Coz the position changes with time but the position change is not constant throughout time So, D) does not have constant velocity

[tex]\\ \sf\Rrightarrow F_{net}=F_1+F_2+F_3+F_4[/tex]

[tex]\\ \sf\Rrightarrow F_{net}=20N+20N+10N-10N[/tex]

[tex]\\ \sf\Rrightarrow F_{net}=40N[/tex]

Answer:

We know that ΔK = Kf - Ki = 1/2 m Vf^2 - 1/2 m Vi^2 = 1/2m(Vf^2-Vi^2) = 1/2 m ΔV^2.

The mass remains the same, just calculate the difference of squared velocities and multiply it by half of the mass.

Answer:

density can be written as:

d = F / V  (weight density) or M / V (mass density)

Since F in the first equation can be written as

B = d V where B is the buoyant force and is proportional to the

weight of the liquid displaced

For a given volume the weight of the liquid displaced is proportional to the density and hence the buoyant force

I am familiar with Russian Mountaineering

Answer:

V = 4/3 R^3

V2 / V1 = (R2 / R1)^3 = 2^3 = 8

8.00

The volume of water in the larger balloon is 8 times greater than in the smaller balloon

We'll begin by calculating the volume of each balloon.

V = 4/3 πr³

V = 4/3 × 3.14 × 3³

V = 4/3 πr³

V = 4/3 × 3.14 × 6³

Finally, we shall determine how much greater the larger balloon is to the smaller balloon

Greatness => large / small

Large / small = 904.32 / 113.04

Large / small = 8

Cross multiply

Therefore, the larger balloon is 8 times greater than the smaller balloon.

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Here are all the steps.

When you are pulling a sled with constant velocity with a rope then the ground exerts a surface force on the sled will be equal to 573.56 N.

The pulling force delivered axially by a cable, string, chain, or by either end of a rod, truss member, or another such 3-D object is known as tension in physics. Tension can be described as the action-reaction pair of forces acting at each end while tension is a polar opposite of compression.

A restoring force causes tension when atoms or molecules are torn apart at the atomic level and build up potential energy.

When you are trying to move upward and push in a positive way,

F = - 112 i + 57.4 (9.8) j

F = -112 i + 562.52 j

[tex]F = \sqrt{(-112)^2+ (562.52)^2}[/tex]

[tex]F =\sqrt{12544 + 316428.7}[/tex]

F = 573.56 N

Therefore, the surface force the ground exerts on the sled is 572.52 N.

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Sorry your tired, ill try to make it worth reading.

One day I walked outside, I looked around. THEN BAM a wasp lands right on my face. Its hideously sharp stinger impaled my flesh. I could almost hear it going deeper and deeper. I screamed and ran around in circles until it finely flew away. As I started walking to the door I could feel my heart pules where the thing had stung. I opened the door and went into the bathroom. Right in the middle of my face just above my nose was a red dot, with a slightly less red ring around it, like an ateroid impact. I sat down and went over the entire event five times. I decided to go back outside, as I walked around the corner, another wasp landed on me.

Answer:

the slope of velocity-time graph represent an object acceleration

Answer:

d and c are the answers :)

Answer:

Plan B.

Because flexibility is best improved by stretching.

Explanation:

Improving and increasing flexibility is done by having stretching sessions daily which maintains and widens the range of motion in the joints and stretches muscles.

Explanation:

For the first question look at the picture. It should be pretty clear. Remember that cos45° = sqrt2 / 2.

2) We calculate it thanks to the conservation of mechanical energy:

Em1 = Em2

U1 + K1 = U2 + K2

K1 = 0J

U2 = 0j

U1 = K2

3) mgh = 1/2mv^2

v = sqrt(2gh)

Answer:

The value of [tex]\theta=63.43^0[/tex]

Explanation:

Range of projectile [tex]\mathrm{R}=\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}[/tex]

Height of projectile [tex]\mathrm{h}=\frac{\mathrm{u}^{2} \cdot \sin ^{2} \theta}{2 \cdot \mathrm{g}}[/tex]

Here we have

[tex]R=2h\\\\\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}=2\frac{\mathrm{u}^{2} \cdot \sin ^{2} \theta}{2 \cdot \mathrm{g}}\\\\\sin 2 \theta=\sin ^{2} \theta\\\\2\sin \theta\cos \theta=\sin ^{2} \theta\\\\\tan \theta=2\\\\\theta=63.43^0[/tex]

Range be R and height be h

[tex]\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}[/tex]

[tex]\boxed{\sf h=\dfrac{u^2sin^2\theta}{2g}}[/tex]

ATQ

[tex]\\ \sf\longmapsto R=2h[/tex]

[tex]\\ \sf\longmapsto \dfrac{u^2sin2\theta}{g}=\dfrac{2u^2sin^2\theta}{2g}[/tex]

[tex]\\ \sf\longmapsto sin^2\theta=sin2\theta[/tex]

[tex]\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta [/tex]

[tex]\\ \sf\longmapsto \dfrac{sin^2\theta}{sin\theta cos\theta=2[/tex]

[tex]\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2[/tex]

[tex]\\ \sf\longmapsto tan\theta=2[/tex]

[tex]\\ \sf\longmapsto \theta=tan^{-1}(2)[/tex]

[tex]\\ \sf\longmapsto \theta=63.4°[/tex]

Answer:

Inside the film the wavelength will be λ/n

For constructive interference to occur the film must be λf/4 thick where λf is the wavelength of the light in the film - there will be a 180 degree phase shift at the water/film interface since the index of refraction of the film is greater than that of water - and the light has to travel λ/2 inside the film for constructive interference to occur

280 nm / 1.6 * 4 = 700 nm is the greatest wavelength allowed

Note that 700 nm is also the upper wavelength of the visible spectrum

So, your science teacher has given your class the classic "mousetrap car" assignment: to make, design, and build a small vehicle powered by the snapping action of a mousetrap to make your car travel as far as possible. If you want to come out ahead of all the other students in your class, you'll need to make your car as efficient as possible so you can squeeze every last inch out of your "car". With the right approach, it's possible to streamline your car's design for maximum distance using only common home materials. You could also buy a mousetrap car kit from any craft store and skip wondering if it will work.

Use large rear wheels. Large wheels have greater rotational inertia than small wheels. In practice, this means that once they start rolling, they're harder to stop rolling. This makes large wheels perfect for distance-based contests — theoretically, they'll accelerate less quickly than smaller wheels, but they'll roll much longer and they'll travel a greater distance overall. So, for maximum distance, make the wheels on the drive axle (the one the mousetrap is tied to, which is usually the rear one) very large. The front wheel is a little less important — it can be large or small. For a classic drag racer look, you'll want big wheels in the back and smaller ones in front.

Use thin, light wheels. Thinner wheels have less friction and may go farther if the distance is what you want or need with your mousetrap racer. It's also important to take the weight of the wheels themselves into account — any unneeded weight will ultimately slow your car down or lead to added friction. In addition, it's worth noting that wide wheels can even have a small negative effect on the car's drag due to air resistance. For these reasons, you'll want to use the thinnest, lightest wheels available for your car.

Old CDs or DVDs work fairly well for this purpose — they're large, thin, and extremely light. In this case, a plumbing washer may be used to reduce the hole size in the middle of the CD (to fit the axle better).

If you have access to old vinyl, these also work extremely well, though they may be too heavy for the smallest mousetraps.

Use a narrow rear axle. Assuming your car is a rear-wheel-drive car, each time your rear axle turns, the rear wheels turn. If your rear axle is extremely skinny, your mousetrap car will be able to turn it more times for the same length of string than it would if it were wider. This translates to turning your rear wheels more times, meaning greater distance! For this reason, it's a wise idea to make your axle out of the skinniest material available that can still support the weight of the frame and wheels.

Narrow wooden dowel rods are a great, easily-accessible choice here. If you have access to thin metal rods, these are even better — when lubricated, they usually have less friction.

Create traction by giving the edges of the friction of the wheels. If the wheels slip against the ground when the trap is sprung, energy is wasted — the mousetrap works to make the wheels turn, but you don't get any extra distance. If this happens with your car, adding a friction-inducing material to the rear wheels may reduce their slippage. To keep your weight requirements down, use only as much as is necessary to give the tips of the wheels some grip and no extra. Some suitable materials are:[1]

Electrical tape

Rubber bands

Additionally, placing a piece of sandpaper under the rear wheels at the start line can reduce slippage as the car begins to move (when it is most likely)

Answer:

C.Compacting the waste and burying it in a shallow landfill.

Explanation:

It is buried in a shallow landfill because it is less reactive.

Answer:

Reaction force = Weight

the force exerted by A on B is equal.

M1a = M2a  

Explanation: I think the answer is right if wrong message me.

Answer:

You should try your best to answer the question.

Explanation:

hope this may help you

have a nice day

To Find The Acceleration of  A Car Slowing Down:

The change in speed is the final speed minus the initial speed. To find acceleration, divide the change in velocity by the length of time during which the velocity changed.

Answer:

I think you ask for another way of writing the joule (J). It can be expressed as the product between a force applied to a body and the deployment of the body (not necessarily caused by that force applied), so N*m (newton * meter).

The quantity work—which has units of energy—was defined, another unit for energy, equivalent to the joule (j), is Newton ·meter (N· m).

The quantity with a constant magnitude that is used to measure the magnitudes of other quantities of the same type is referred to as a unit in physics.

Unit of work is that of energy, that is, Joule (J).

Again, work = force × displacement

Unit of work = Unit of force × Unit of displacement = Newton . meter (N.m).

Hence, the quantity work—which has units of energy—was defined, another unit for energy, equivalent to the joule (j), is Newton ·meter (N· m).

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Answer:

I put the answer

Explanation:

Can you see it

10Hz is the frequency of the wave.

To calculate the frequency of a wave, divide the velocity of the wave by the wavelength. Write your answer in Hertz or Hz

Given,

F = 2

λ=2m

v=20m/s

By applying the formula, we get

V = Fλ

F = V / λ

F = 20 / 2

=10Hz

Wave frequency is the number of waves that pass a fixed point in a given amount of time.

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Developing Psychological Theories: Mastery Test

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Hope this helps :)

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The distance traveled by object 1 is greater than the distance traveled by object 2.

The shortest distance between two points is a straight line. More distance is covered when the path of motion is curved.

The distance traveled by each object at different time is compared as follows;

[tex]at \ t_ A , \ \ \ d_1>d_2 \ \ \ (d_2 = 0)\\\\at \ t_B, \ \ d_1 > d_2 \\\\at \ t_C, \ \ \ d_2 > d_1 \\\\at \ t_D \ \ \ d_1 = d_2\\\\at \ t_f \ \ \ d_2 > d_1 \ \ \ (d_1 = 0)[/tex]

The major distinguishing distance traveled by each occurs at [tex]t_B[/tex] and [tex]t_C[/tex].

[tex]d_{(t_B)} > d_{(t_C)} \\\\d_1 > d_2[/tex]

Thus, we can conclude that the distance traveled by object 1 is greater than the distance traveled by object 2.

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Answer:

The earth travels around the sun in a (parabolic) path known as the ecliptic.

The axis of the earth's rotation is tilted about 23 deg to the ecliptic.

At the winter solstice (around Dec. 21), the sun will appear to be at its farthest south in its orbit. (This marks the beginning of winter for observers in the northern hemisphere and the beginning of summer for an observer in the southern hemisphere)

The tilt of the axis of the earth's rotation causes the change in the daylight hours.