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3 years ago

A wave travels a distance of 60cm in 3s. The distance blw successive crests of thd wave is 4cm. What is the frequency?

1 answer:

6 0

Data:
ΔS = 60cm
Δt = 3s
Vm = ?

Vm = ΔS/Δt
$Vm = \frac{60}{3}$
$Vm = 20cm/s$

We have:

Vm = v
ΔS = λ
Δt = T

$v = \frac{\lambda}{T}$
$20 = \frac{\lambda}{4}$
$\lambda = 20*4$
$\lambda = 80cm$

Soon:

$v = \lambda*f$
$20 = 80*f$
$80f = 20$
$f = \frac{20}{80} simplify( \frac{\div20}{\div20} )= \frac{1}{4}$
$\boxed{f = 0,25Hz}$

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20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu

Anna71 [15]

Answer:

B. $V_{f}= 10\,cubic\,inches$

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

$PV=nRT$

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

$P_{i}V_{i}=n_{i}RT_{i}$ (1)

$P_{f}V_{f}=n_{f}RT_{f}$ (2)

Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because $T_{i}=T_{f}$ , $n_{i}=n_{f}$ and R is an universal constant:

$P_{i}V_{i}= P_{f}V_{f}$ , solving for $V_{f}$

$V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}$

$V_{f}= 10 cubic\,inches$

6 0

3 years ago

A light beam has speed c in vacuum and speed v in a certain plastic. The index of refraction n of this plastic is

Mnenie [13.5K]

Answer:

$n = \frac{c}{v}$

Explanation:

Refractive Index: It is a measure to find how fast the light travels through a medium. It is ration of the speed of light in vacuum to speed of light in the medium. Speed of light is not constant and varies depending on the density of the medium.

In vacuum the speed of light is 300000 km/s and is denoted by c. When the light beam enters any medium the speed will decrease. Here it is given that the speed in plastic is v. Thus the refractive index(n) is given as:

$n = \frac{c}{v}$

It is a dimensionless no.

3 0

3 years ago

!!HELP HELP!!

borishaifa [10]

Answer:

7.5s

Explanation:

Given parameters:

Velocity = 30m/s

Deceleration = 4m/s²

Unknown:

Time it takes for the car to come to complete rest = ?

Solution:

To solve this problem, we use the kinematics expression below:

v = u + at

Since this is a deceleration

v = u - at

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

v - u = -at

0 - 30 = -4 x t

-30 = -4t

t = 7.5s

6 0

2 years ago

Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards Vernonville, and another car leaves Verno

iragen [17]

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is $v_{A} =\frac{x_{A} }{t_{A} }$ (Equation A) and Average speed of the car B is $v_{B} =\frac{x_{B} }{t_{B} }$ (Equation B), where $x_{A}$ and $x_{B}$ are the distances and $t_{A}$ and $t_{B}$ are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

$96 minutes . \frac{1 hour}{60 minutes} = 1.6 h$

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

$v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h}$ (Equation C)

$v_{B} = \frac{208-x}{1.6h}$ (Equation D)

From Equation A and C, we have:

$\frac{208-x}{1.6}+10 = \frac{x}{1.6}$

208-x+16 = x

208 + 16 = 2x

$x = \frac{224}{2}$

x = 112 miles

Replacing x in Equation A:

$v_{A} = \frac{112miles}{1.6h}$

$v_{A} = 70 miles per hour$

Replacing x in Equation B:

$v_{B} = \frac{208miles-112miles}{1.6h}$

$v_{B} = \frac{96miles}{1.6h}$

$v_{B} = 60 miles per hour$

3 0

3 years ago

Ax = 44.4 m and Ay = 25.1 m<br> Find the magnitude of the<br> vector.

max2010maxim [7]

Answer:

The magnitude of the vector A is <u>51 m.</u>

Explanation:

Given:

The horizontal component of a vector A is given as:

$A_x=44.4\ m$

The vertical component of a vector A is given as:

$A_y=25.1\ m$

Now, we know that, a vector A can be resolved into two mutually perpendicular components; one along the x axis and the other along the y axis. The magnitude of the vector A can be written as the square root of the sum of the squares of each component.

Therefore, the magnitude of vector A is given as:

$|\overrightarrow A|=\sqrt{A_{x}^2+A_{y}^2}$

Now, plug in 44.4 for $A_x$ , 25.1 for $A_y$ and solve for the magnitude of A. This gives,

$|\overrightarrow A|=\sqrt{(44.4)^2+(25.1)^2}\\|\overrightarrow A|=\sqrt{1971.36+630.01}\\|\overrightarrow A|=\sqrt{2601.37}\\|\overrightarrow A|=51\ m$

Therefore, the magnitude of the vector A is 51 m.

6 0

3 years ago