1. Calculate the height of tree, 250 m away that produces

An electromagnetic wave has a frequency of 6.0 x 10^18 Hz. What is the

jonny [76]

Answer:

Wavelength = 5 * 10^{-11} meters

Explanation:

Given the following data;

Frequency = 6.0 x 10^18 Hz

Speed = 3 * 10⁸ m/s

To find the wavelength of the wave;

Mathematically, the wavelength of a wave is given by the formula;

$Wavelength = \frac {speed}{frequency}$

Substituting into the formula, we have;

$Wavelength = \frac {3 * 10^{8}}{6.0 x 10^{18}}$

Wavelength = 5 * 10^{-11} meters

8 0

3 years ago

A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre

diamong [38]

Answer:

A) 1568.60 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

$f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\$

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

$f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz$

6 0

3 years ago

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago

In which item is energy stored in the form of gravitational potential energy? A. A slice of bread B. A compressed spring C. An a

marysya [2.9K]

The things that determine the amount of an object's gravitational potential energy are ...

-- mass of the object

-- gravitational acceleration in the place where the object is

-- height of the object above the ground or the floor

A). a slice of bread; No. It's still a slice of bread even if it's on the ground.

B. A compressed spring; No. It's still a compressed spring even if it's on the ground.

C. An apple on a tree; Yes ! It can't be an apple on a tree if it's on the ground.

D. A stretched bow string; No. It's still a stretched bowstring even if it's on the ground.

3 0

3 years ago

Read 2 more answers

Consider the roller coaster on the track. If you want to increase the kinetic energy of the roller coaster as it proceeds throug

a_sh-v [17]

Answer:

The answer is A and D but if you want one anwer i think the anwer is A

Explanation:

5 0

4 years ago