What pressure will be exerted by a 20 N box that has equal sides of 1 meter

A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.

Answer:

The phase at the second point is $\phi _2 = 1.57 \ rad$

Explanation:

From the question we are told that

The wavelength of the spherical wave is $\lambda = 2.0 \ mm = \frac{2}{1000} = 0.002 \ m$

The first radius is $r_1 = 4.0 \ mm = \frac{4}{1000} = 0.004 \ m$

The phase at that instant is $\phi _1 = \pi \ rad$

The second radius is $r_2 = 3.5 \ mm = \frac{3.5}{1000} = 0.0035 \ m$

Generally the phase difference is mathematically represented as

$\Delta \phi = \phi _2 - \phi _1$

this can also be expressed as

$\Delta \phi = \frac{2 \pi }{\lambda } (r_2 - r_1 )$

So we have that

$\phi _2 - \phi _1 = \frac{2 \pi }{\lambda } (r_2 - r_1 )$

substituting values

$\phi _2 - \pi = \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 )$

$\phi _2 = \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 ) + 3.142$

$\phi _2 = 1.57 \ rad$

7 0

3 years ago

A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to

grandymaker [24]

P1V1/T1 = P2V2/T2

(1.6atm)(168cm^3)/(255K) = (1.3atm)V2/(285K)

Final volume = 231cm^3

6 0

4 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago