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2 years ago

Please help!!

Physics

1 answer:

sukhopar [10]2 years ago

3 0

Answer:

: WHY DIDN'T THE POD DOCK LIKE IT WAS SUPPOSED TO DO? 

ANSWER;

The force exerted by the thrusters caused the pod to change direction.

WHAT NEW THEORIES DO YOU HAVE?

ANSWER;

This pod moved differently because it was more massive.

CARRYONLEARNING:)

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The position of an object that is oscillating on a spring is given by the equation x = (17.4 cm) cos[(5.46 s-1)t]. what is the a

nekit [7.7K]

the angular frequency of this motion is 5.46rad /sec

The formula for the angular frequency is = 2π/T. Radians per second are used to express angular frequency. The frequency, f = 1/T, is the period's inverse. The motion's frequency, f = 1/T = ω/2π, determines how many complete oscillations occur in a given amount of time so in this case the It is measured in units of Hertz, (1 Hz = 1/s).

herex

(17.4cm)cos[(5.46s− 1)t]

is written in the general form

where we can identify: A=17.4cm and ω

=5.46rad /sec

To learn more about angular frequency :

brainly.com/question/12446100

#SPJ4

8 0

1 year ago

A car travels eastwards at 60km/h for 2h, then travels northwards at 20km/h for 8h. Find,

sveta [45]

Answer:

$a) Average\ Speed=28\ km/h\\b) Average\ Velocity= 20\ km/h$

Explanation:

$We\ are\ given\ that,\\Velocity\ of\ the\ car\ Eastwards=60\ km/h\\Time\ taken\ by\ the\ car\ Eastwards=2\ h\\Velocity\ of\ the\ car\ Northwards=20\ km/h\\Time\ taken\ by\ the\ car\ Northwards=8\ h\\Hence,\\As\ we\ know\ that,\\Speed=\frac{Distance}{Time}\\Distance= Speed* Time\\Now,\ lets\ find\ the\ distance\ covered\ by\ the\ car\ in\ both\ the\ cases.$

$Hence,\\Distance\ Covered\ During\ its\ Eastward\ Journey=60*2=120\ km\\Distance\ Covered\ During\ its\ Northward\ Journey=20*8=160\ km\\Now,\\As\ we\ know\ that\ Average\ Speed=\frac{Total\ Distance}{Total\ Time} \\Here,\\Total\ distance\ of\ the\ car=Distance\ Covered\ During\ its\ Northward\ Journey+Distance\ Covered\ During\ its\ Eastward\ Journey\\Hence,\\Total\ distance\ of\ the\ car=120+160=280\ km\\Total\ time\ taken\ by\ the\ car=8+2=10\ hours\\Hence,\\$ $Average\ Speed\ Of\ the\ Car\ throughout\ its\ journey=\frac{280}{10}=28\ km/h$

$Now,\\For\ Average\ Velocity\ we\ need\ to\ consider\ displacement\ as:\\Average\ Velocity\ =\frac{Total\ Displacement}{Total\ Time} \\Now,\\As\ we\ already\ know\ that\ displacement\ is\ the\ shortest\ distance\\ from\ the\ initial\ to\ the\ final\ point.\\We\ observe\ that, \\The\ car\ forms\ a\ right\ triangle\ during\ its\ complete\ journey.\\Hence,\\As\ we\ already\ know\ that,\\Distance\ travelled\ Eastwards= 120\ km\\Distance\ travelled\ Northwards= 160\ km\\Hence,\\$ $We\ may\ apply\ Pythagoras\ Property\ to\ find\ the\ net\ displacement.\\Hence,\\a^2+b^2=c^2\\120^2+160^2=c^2\\14400+25600=c^2\\40000=c^2\\c=\sqrt{40000}\\c=200\\Hence,\\Total\ displacement=200\ km\\Total\ Time\ taken=2+8=10\ hours\\Hence,\\Average\ Velocity\ Of\ the\ Car=\frac{200}{10}=20\ km/h$

7 0

3 years ago

a rocket initially at rest on the ground lifts off vertically with a constant acceleration of 2.0 x 10^1 meters per second^2. Ho

xenn [34]

Here's the formula for the distance covered by an accelerating body in some amount of time ' T '. This formula is incredibly simple but incredibly useful. It pops up so often in Physics that you really should memorize it:

D = 1/2 a T²

Distance = (1/2)·(acceleration)·(time²)

This question gives us the acceleration and the distance, and we want to find the time.

(9,000 m) = (1/2) (20 m/s²) (time²)

(9,000 m) = (10 m/s²) (time²)

Divide each side by 10 m/s²:

(9,000 m) / (10 m/s²) = (time²)

900 s² = time²

Square root each side:

T = 30 seconds

7 0

3 years ago

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea

Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$