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3 years ago

PLEASE ANSWER BOTH IF YOU CAN, I NEED THESE ASAP!!

1 answer:

8 0

Solar eclipse: when the sun, moon, and earth align
Lunar eclipse: when the moon blocks the sun's light

1, there is no water there, so plants cant grow
2, there are no plants, hence, there would be no oxygen.
3, (not sure) the temperature is different from earth and could possible not sustain life

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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:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is </em>

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 mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl

frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ = 34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

W = 292.9686 J

W ≅ 292.97 J

Hence, the work done by tension before the block goes up the incline = 292.97 J

8 0

2 years ago

View the chart on the characteristics of electromagnetic and mechanical waves and answer the question.

Elza [17]

I think the answer is c

4 0

2 years ago

Read 2 more answers

A 20-kg child is tossed up into the air by her parent. The child is 2 meters off the ground traveling 5 m/s.

nignag [31]

The kinetic energy (KE) is 250 J and the gravitational potential energy (GPE) is 392 J

4 0

3 years ago

A player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

____ [38]

At the highest point in its trajectory, the ball's acceleration is zero but its velocity is not zero.

<h3>What's the velocity of the ball at the highest point of the trajectory?</h3>

At the highest point, the ball doesn't go more high. So its vertical velocity is zero.
However, the ball moves horizontal, so its horizontal component of velocity is non - zero i.e. u×cosθ.
u= initial velocity, θ= angle of projection

<h3>What's the acceleration of the ball at the highest point of projectile?</h3>

During the whole projectile motion, the earth exerts the gravitational force with a acceleration of gravity along vertical direction.
But as there's no acceleration along vertical direction, so the acceleration along vertical direction is zero.

Thus, we can conclude that the acceleration is zero and velocity is non-zero at the highest point projectile motion.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

A- neither the ball's velocity nor its acceleration are zero.

B- the ball's acceleration points upward.

C- the ball's acceleration is zero but its velocity is not zero.

D- the ball's velocity points downward.

Learn more about the projectile motion here:

brainly.com/question/24216590

#SPJ1

7 0

2 years ago