5. Convert 0.984 miles per hour to meters per minute

The distance from one Crest to the next Crest is the blank

Luda [366]

The distance from one crest to the next crest in a set of waves is called the wavelength. The distance from the crest of one wave to the equilibrium point is called frequency.
Hope that helped! =)

8 0

3 years ago

Read 2 more answers

Use the information from the graph to answer the question.

galina1969 [7]

The displacement of the object as determined from the velocity-time graph is 562.5 m.

<h3>What is a velocity-time graph?</h3>

A velocity-time graph is a graph of the velocity of an object plotted in the vertical or y-axis of the graph against the time taken on the horizontal or x-axis.

The displacement of an object can be obtained from its velocity-time graph by calculating the total area under the graph.

The total area under the graph = area of triangle + area of rectangle

Area of triangle = b*h/2 =

Area of triangle = 25 * (35 - 10)/2 = 312.5 m

Area of rectangle = l * b

Area of rectangle = 10 * 25 = 250 m

Total area = (312.5 + 250) m

Total area = 562.5 m

Therefore, the displacement of the object is 562.5 m

In conclusion, the total area of a velocity-time graph gives the displacement.

Learn more about velocity-time graph at: brainly.com/question/28064297

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5 0

1 year ago

Tires of a Bigfoot truck has a diameter of 2.2 m. If it rotates 60 revolutions find distance travel on the road.

Firlakuza [10]

Answer:

$s = 414.7 m\\$

Explanation:

The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:

s = rθ

where,

s = distance traveled on road = ?

r radius of tires = diameter/2 = 2.2 m/2 = 1.1 m

θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad

Therefore,

$s = (1.1 m)(377 rad)\\s = 414.7 m$

8 0

2 years ago

Suppose that a star has a spectrum that includes red, blue, and violet lines spaced in the pattern of the lines from hydrogen bu

ladessa [460]

Answer:

It can be concluded that the star is moving away from the observer.

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest for this case is 434 nm and 410 nm ( $\lambda_{0} = 434nm$ , $\lambda_{0} = 410nm$ )

$Redshift: \lambda_{measured} > \lambda_{0}$

$Blueshift: \lambda_{measured} < \lambda_{0}$

Since, $\lambda_{measured}$ (444nm) is greater than $\lambda_{0}$ (434 nm) and $\lambda_{measured}$ (420nm) is greater than $\lambda_{0}$ (410 nm), it can be concluded that the star is moving away from the observer

6 0

3 years ago

A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe

joja [24]

Answer:

The Doppler Effect is given by the following relation;

$f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f$

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

$v_o$ = The velocity of the observer

$v_s$ = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(ii) When the source is receding from the observer, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

(1) Given that (v + $v_s$ ) > v, therefore, v < (v + $v_s$ ), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(1) Given that (v - $v_s$ ) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

7 0

3 years ago