Which statement about subatomic particles is NOT true?

The period of a sound wave is 0.002 seconds. The speed of a sound is 344 m/s. Find the frequency and wavelength of the sound wav

ankoles [38]

Answer:

I think that's the solution

5 0

2 years ago

For 1983 through 1989, the per capita consumption

Tresset [83]

Answer:

y = 43.55 + 2.15t

Explanation:

We were told that in 1983, the per capita consumption was 37.1 pounds, and in 1989 it was 50 pounds.

If we assume t = 0 corresponds to year 1980. Then, for 1983 it will be t = 3 and for 1989,it will be t = 9.

Thus, expressing the information as ordered pairs, we have; (3,37. 1) and (9,50).

Let us now find slope of the linear function:

m1 = (y2 - y1)/(t2 - t1)

m1 = (50 - 37.1)/(9 - 3)

m1 = 2.15

So, we can find the linear equation as;

y - 37.1 = 2.15(t - 3)

y = 37.1 + 2.15t - 6.45

y = 43.55 + 2.15t

8 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

Conservation of linear momentum:

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

Conservation of kinetic energy:

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

2 years ago

Which steps are important when designing and conducting a scientific experiment?

Pepsi [2]

Make an Observation. Scientists are naturally curious about the world. ... Form a Question. After making an interesting observation, a scientific mind itches to find out more about it. ... Form a Hypothesis. ... Conduct an Experiment. ... Analyze the Data and Draw a Conclusion.

8 0

2 years ago

Read 2 more answers

a car travels between the 100 meter and the 250 meter highway marker in 10 seconds calculate the velocity of the car during this

kozerog [31]

$Given:\\\Delta s= 250m-100m=150m\\t=10s\\\\Find:\\v=?\\\\Roz.\\\\v=\frac{\Delta s}{t} \\\\v=\frac{150m}{10s} =15\frac{m}{s}$

4 0

2 years ago