Which type of radiation has the highest penetrating power?

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

Alik [6]

Answer:

Longest wavelength, lowest intensity

Explanation:

7 0

3 years ago

Ipaliwanag ang paggalang

cestrela7 [59]

paraan ito ng pagrerespeto

5 0

2 years ago

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

a. The frequency with which the waves strike the hill is 242.61 Hz

b. The frequency of the reflected sound wave is 254.23 Hz

c. The beat frequency produced by the direct and reflected sound is

11.62 Hz

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

2 years ago

A cow wanders 30 m North, turns 22 degrees right of its original path, and wanders another 40 m. Find its total displacement.

scoundrel [369]

Answer:OB=58.3m

Explanation:

So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.

now take the starting point as a origin such that cow moves in x-y co-ordinate axis.

As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.

So the final displacement is the length of cow from the origin that is length OB.

now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]

now displacement of cow= length of OB

= $\sqrt{[37.084]^{2}+[44.984]^{2} }$

= $\sqrt{3398.78}$

OB = $58.3$

7 0

11 months ago

MARKING BRAINLIST | Which situation below would have the STRONGEST gravitational force between them?

maks197457 [2]

Case d) has the strongest gravitational force

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

a) For this pair of objects:

m1 = 10 kg

m2 = 2 kg

r = 30 km = 30,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(2)}{30000^2}=1.48\cdot 10^{-18}N$

b) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 30 km = 30,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(10)}{30000^2}=7.41\cdot 10^{-18}N$

c) For this pair of objects:

m1 = 2 kg

m2 = 2 kg

r = 10 km = 10,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(2)(2)}{10000^2}=1.33\cdot 10^{-17}N$

d) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 10 km = 10,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(10)}{10000^2}=6.67\cdot 10^{-17}N$

Therefore, the strongest gravitational force is in case d).

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago