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4 years ago

7

Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short

wavelength. It has a long wavelength. It has a high intensity. It has a low intensity.

1 answer:

Alik [6]4 years ago

7 0

Answer:

Longest wavelength, lowest intensity

Explanation:

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A stationary speed gun emits a microwave beam at 2.10x10^10 Hz. It reflects off a car and returns 638 Hz lower. What is the spee

stiv31 [10]

Answer:

The answer is 4.55 m/s!

Explanation:

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3 years ago

Is your increase in gravitational potential energy the same in both cases? When Climbing a mountain on a zigzag path and on a st

baherus [9]

Answer:

The increase in gravitational potential energy is the same in both cases

Explanation:

It is easier to climb a mountain in a zigzag way rather than climbing on a straight line but since the distance is the same ( vertical height ) , mass and gravity is the same. Hence the increase in gravitational potential energy is the same in both cases.

gravitational potential energy = mgh ( same in both cases )

m = mass

g = acceleration due to gravity

h = distance ( vertical height )

7 0

3 years ago

A string vibrates at a frequency of 20Hz. what is its period?

Andreas93 [3]

We know, Period = 1/frequency = 1/20 sec.

So, your final answer is 0.05 sec

Hope this helps!

7 0

3 years ago

The resistance created by waves on a 120-m-long ship is tested in a channel using a model that is 4 m long Y Part A If the ship

DanielleElmas [232]

Answer:

$V_m = 12.78 km/hr$

Explanation:

given,

length of the ship = 120 m

length of model of the ship = 4 m

Speed at which the ship travels = 70 km/h

speed of model = ?

by using froude's law

$F_r = \dfrac{V}{\sqrt{L g}}$

for dynamic similarities

$(\dfrac{V}{\sqrt{L g}})_P = (\dfrac{V}{\sqrt{L g}})_{model}$

$(\dfrac{V_p}{\sqrt{L_p}}) = (\dfrac{V_m}{\sqrt{L_m}})$

$(\dfrac{70}{\sqrt{120}}) = (\dfrac{V_m}{\sqrt{4}})$

$V_m = 12.78 km/hr$

hence, the velocity of model will be 12.78 km/h

6 0

3 years ago

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

mixer [17]

The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

where

B is the bulk modulus

$\rho_0$ is the density at surface

$\Delta p$ is the variation of pressure

$\Delta \rho$ is the variation of density

In this problem, we have:

$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

$\rho_0 =1100 kg/m^3$

$\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa$ is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

$\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3$

Learn more about pressure in a fluid:

brainly.com/question/9805263

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7 0

3 years ago