Describe your acceleration if you ride your bike up a hill,then ride down the other side.

Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho

uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

$\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

$\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

${\frac{M_{B}}{M_{A}}} = 4$

${\frac{M_{B}}{4}} = M_{A}$

7 0

3 years ago

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri

Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but here is an explanation of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t >0$ ), the displacement y of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement y not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x>0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0

3 years ago

If a coworker repeatedly asks you out for a date even though she knows you are not interested, what is she guilty of?

Over [174]

D sexual harassment

5 0

3 years ago

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Which of the following represents a chemical change to matter? A. digestion of food B. melting an ice cube C. shredding paper D.

guajiro [1.7K]

The answer is A. digestion of food. The chemical change to matter means that the molecule structure has been destroyed. For B, C and D. the matter are still ice, paper and ice. So they are physical change.

8 0

3 years ago

Inside a television picture tube there is a build-up of electrons (charge of 1.602 × 10^–19 C) with an average spacing of 38.0 n

Brut [27]

The magnitude of electric field is produced by the electrons at a certain distance.

E = kQ/r²

where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9 N. m2 / C2

Given are the following:
Q = 1.602 × 10^–19 C
r = 38 x 10^-9 m

Substitue the given:
E = $\frac{( 1.602 x 10^{-19} )( 9.0x10_{9} )}{(38x10^{-9}) ^{2} }$

E = 998.476 kN/C

8 0

3 years ago

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