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2 years ago

The period (T) of an oscillating wave is 1/5s. What happens to the frequency (f) of the wave if T increases to 1/2s

1 answer:

6 0

Frequency = 1/T
as the 5 is reduced, frequency is increase.
as 1 whole wave travels through a point in a lesser time now

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The weather conditions of an area may be expected to change over a(n) _________ period of time.

jenyasd209 [6]

Answer:

Your answer should be 2. Short

Explanation:

4 0

3 years ago

A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid.

Vinil7 [7]

Answer:

Refractive index of unknown liquid = 1.56

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 65.0° )

${\theta_r}$ is the angle of refraction ( 53.0° )

${n_r}$ is the refractive index of the refraction medium (unknown liquid, n=?)

${n_i}$ is the refractive index of the incidence medium (oil, n=1.38)

Hence,

$1.38\times {sin65.0^0}={n_r}\times{sin53.0^0}$

Solving for ${n_r}$ ,

Refractive index of unknown liquid = 1.56

4 0

3 years ago

I need help please, thank you!!!

stiks02 [169]

Answer:

135 mph

because 7+9=16

8 is left

if 7=40

9=50

then 8=45

add 40+50+45mph

=135

4 0

2 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

3 0

3 years ago

Read 2 more answers

PLEASE HURRY 20 PTS

Crank

Answer:

A. The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

Explanation:

The electric force exerted on a charge by an electric field is given by:

where

F is the force

q is the charge

E is the electric field

We see that if the charge is negative, q contains a negative sign, so the force F and the electric field E will have opposite signs (which means they have opposite directions). This is due to the fact that the direction of the lines of an electric field shows the direction of the electric force experienced by a positive charge in that electric field: therefore, a negative charge will experience a force into opposite direction.

5 0

3 years ago