PLS HELP ME I'LL GIVE BRAINLIEST

Question 7 (2 points)

Arlecino [84]

Answer:

Tire

Explanation:

3 0

3 years ago

A climber using bottled oxygen accidentally drops the oxygen bottle from an altitude of 4500 m. If the bottle fell straight down

Trava [24]

Answer:

v= 300 m/s

Explanation:

Given that

altitude ,h= 4500 m

The mass ,m = 3 kg

Lets take acceleration due to gravity , g= 10 m/s²

The speed before impact at sea level = v

Initial speed ,u = 0 m/s

We know that

v²=u²+2 g h

v=final speed

u=initial speed

h=height

Now by putting the values in the above equation

v² = 0²+ 2 x 10 x 4500

v²=90000

v= 300 m/s

Therefore the speed at sea level will be 300 m/s.

3 0

3 years ago

What kind of front occurs when a cold air mass replaces a warm air mass?

Andreyy89

Answer:

Explained below:

Explanation:

When a cold air mass meets a warm air mass, a front is formed and if the cold air is replacing the warm air, it is known as a cold front. Cold fronts frequently cause thunderstorms or rain showers because they pressure the air in a steep upward direction at the front's edge. They are also responsible to bring the changes in atmospheric pressure and wind direction.

7 0

4 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

3 years ago

The same net force is applied to two bodies. The first body acquires half of the acceleration of the second body. What is the re

Yanka [14]

The mass of the first body is double the mass of the second body.

This all revolves around Newtons 2nd law of motion that states F=m*a.

If we understand that body one has the same force applied but half of the acceleration than body two, we can easily understand that it has to do something with the objects mass. This is because mass and acceleration in newtons 2nd law are inversely proportional. This means if and objects force stays the same, but it’s mass doubles, then the acceleration must be half its original value. And if the acceleration doubles and the force stays constant than the mass must have halved from its original value. It sounds quite complicated but it’s very simple. Hopefully this helps!!!

4 0

3 years ago