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3 years ago

12

The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o

f 0.6 meters?

1 answer:

N76 [4]3 years ago

3 0

Frequency = (speed) / (wavelength)

Frequency = (331 m/s) / (0.6 m) = 551.7 Hz

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All of the following are true about scientific evidence except: *

goldenfox [79]

Answer:

Your answer is D It does not need to be repeatable.

the reason for this is because C is correct.

You need to be able to experiment on something multiple times so that you can gather further data to imbedded your evidence in facts.

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3 years ago

Ben walks 500 meters from his house to the corner store. He then walks back toward his house but stops to talk to a neighbor whe

spin [16.1K]

Average velocity =

(displacement) / (time for the displacement)
and
(direction of the displacement) .

Displacement =

(distance from the start-point to the end-point)
and
    (direction from the start-point to the end-point) .

When Ben is 200 meters from the corner store,
he is (500 - 200) = 300 meters from his house.

His displacement is

   300 meters in the direction
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His average velocity is

   (300/910) = 0.33 meters per second, in the
   direction from his house to the neighbor .

7 0

3 years ago

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An astronaut holds a rock 100m above the surface of Planet X. The rock is then thrown upward with a speed of 15m/s, as shown in

Harlamova29_29 [7]

The acceleration due to gravity of the planet X is 1 m/s².

The given parameters;

height above the ground, h = 100 m
initial velocity of the rock, u = 15 m/s
time of motion of the rock, t = 10 s

The acceleration due to gravity is calculated as follows;

$h = ut - \frac{1}{2} gt^2\\\\100 = 15(10) - (0.5\times 10^2)g\\\\100 = 150 - 50g\\\\50g = 150-100\\\\50g = 50\\\\g = 1 \ m/s^2$

Thus, the acceleration due to gravity of the planet X is 1 m/s²

Learn more here: brainly.com/question/24564606

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2 years ago

A ski jumper travels down a slope and

AleksandrR [38]

Answer:

304.86 metres

Explanation:

The x and y cordinates are $dcos\theta$ and $dsin\theta$ respectively

The horizontal distance travelled, $x=v_{ox}t=dcos\theta$

Making t the subject, $t=\frac{dcos\theta}{v_{ox}}$

Since $y=0.5gt^2=dsin\theta$ , we substitute t with the above and obtain

$0.5g(\frac{dcos\theta}{v_{ox}})^2=dsin\theta$

Making d the subject we obtain

$d=\frac{2v_{ox}^2sin\theta}{gcos^2\theta}$

$d=\frac{2*30^2sin48}{9.8cos^248}$

d=304.8584

d=304.86m

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3 years ago

75 POINTS AND BRAINLY

kramer

Answer:

B=work=power/time

Explanation:

8 0

3 years ago

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