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3 years ago

Crest : trough :: compression : _____ A. frequency B. amplitude C. rarefaction D. wavelength

1 answer:

3 0

Rarefraction.

Crest- tallest spot on transverse wave.

Trough- shortest point on transverse wave.

Compression - spot on a compressional wave where the waves are closer together.

Rarefraction - spot on a compressional wave where the waves are farther apart.

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How do I solve such problem???

pashok25 [27]

As far as I'm concerned, this is a bogus question, or at least a severely corrupted one.

The three numbers given can NOT all be true on Earth.

-- It rolled off the table at 7.6 m/s . By golly, there you are! Its initial horizontal velocity is 7.6 m/s, and it has no vertical velocity until it leaves the table.

-- There are no horizontal forces that we're aware of acting on the object. So it maintains the same horizontal velocity for the rest of the story. It's 10.5m away from the table in (10.5 m) / (7.8 m/s) = 1.35 second .

-- Vertically, it's just an object dropped from 17.6m off the floor. Shockingly, the distance it falls in time 'T' is (1/2 g) T². In 1.35 second, that's 8.88 meters ! . . . only about halfway to the floor !

-- In order to fall 17.6 m to the floor, it would need 1.89 seconds. In <u>that</u> length of time, however, it would travel (7.8 m/s) x (1.89 s) = 14.78 m away from the base of the table.

So you see, either . . .

-- the table is NOT 17.6m tall, or

-- the object does NOT roll off of the table at 7.8 m/s, or

-- it does NOT land 10.5 m away from the base of the table.

OR . . .

-- the table is not on Earth, and gravity is not 9.8 m/s² !

We often see questions posted on Brainly with not enough given information, OR with some information given that's not needed because it's not involved the answer.

THIS one is different, and it's unusual. In this one, we have<em> too much</em> given information, we can't ignore any of it because it's all related, but it's inconsistent and it CAN't all be true.

(Unless the whole story takes place on a mystery planet that is not Earth. Which I'm not going to take the time and effort right now to figure out what the acceleration of gravity has to be in order to make all of the given information compatible.)

7 0

3 years ago

A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r

sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana $m_{b}$ = 8.97 kg

at distance r = 4/5 { radius of platform }

mass of monkey $m_{m}$ = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + $m_{b}$ ( $\frac{4}{5}$ R)² + $m_{m}$ R² ]w

m/2 × ω₀ = [ m/2 + $m_{b}$ ( $\frac{4}{5}$ )² + $m_{m}$ ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0

3 years ago

Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

arsen [322]

Answer:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates to find the horizontal distance traveled by the electron when it hits the plate.

acceleration a=qE/m= $1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}$ = $7*10^{14}$ m/s

now we find the horizontal distance traveled by electrons hit the plates

horizontal distance

$X=u[2y/a]^{1/2}$

= $4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}$

= $3*10^{-2}$ = 3 cm

5 0

3 years ago

ALL OF MY POINTS FOR THIS!

scoray [572]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

   = (1,127 kg) x (6 m/s² forward)

   = (1,127 x 6) newtons forward

   = 6,762 newtons forward
______________________________

   Momentum = (mass) x (speed)

   =   (69 kg) x (6 m/s)

   = 414 kg-m/s

3 0

3 years ago

If one object is 103 km away and a second object is 106 km away, one could say that the second object is _____ times further awa

vaieri [72.5K]

Answer:

$1.03$

Explanation:

$\frac{object_{second}}{object_{first}} = \frac{106}{103} = 1.02912621359$

Round to three significant digits

$1.03$

7 0

3 years ago