When it rises, air has more weight above it, and the higher pressure allows the air to expand.A.TrueB.False

Which of the following best characterizes an electron?

barxatty [35]

A, electrons are negatively charged and do orbit around the nucleons

4 0

3 years ago

To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster that

noname [10]

Answer:

a. 2v₀/a b. 2v₀/a

Explanation:

a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

Since the dragster starts from rest with an acceleration, a, using

s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster

s' = 0t + 1/2at²

s' = 1/2at²

Since the distance moved by me and the dragster must be the same,

s = s'

v₀t. = 1/2at²

v₀t. - 1/2at² = 0

t(v₀ - 1/2at) = 0

t= 0 or v₀ - 1/2at = 0

t= 0 or v₀ = 1/2at

t= 0 or t = 2v₀/a

So the maximum time tmax = 2v₀/a

b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)

= 2v₀/a

4 0

3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

3 years ago

The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and

Zolol [24]

Answer:

gas is dioatomic

T_f = 330.0 K

$\eta = 7.07 mole$

Explanation:

Part 1

below equation is used to determine the type Gas by determining $\gamma$ value

$\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}$

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

$\gamma = \frac{ln(P_f/P_i)}{ln(V_i/V_f)}$

$\gamma = \frac{ln(3.61/1.50)}{ln(151/80.6}$

\gamma = 1.38

therefore gas is dioatomic

Part 2

final temperature in adiabatic process is given as

$T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)$

substituing value to get final temperature

$T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}$

T_f = 330.0 K

Part 3

determine number of moles by using following formula

$\eta =\frac{PV}{RT}$

$\eta =\frac{1.013*10^{5}*0.151}{8.314*260}$

$\eta = 7.07 mole$

4 0

3 years ago

Study the analogy, and then answer the question that follows.

nordsb [41]

What kind of analogy is this?
A. synonyms
B. part to whole
C. degrees of intensity
D. cause to effect
This is because a synonym refers to things like "normal" or "regular" like your calling someone something.

6 0

3 years ago

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