If the_____of a wave increases, its frequency must decrease.a. period
b. energy
c. amplitude
d. velocity

Answers

Answer 1

Answer:

Answer:wrong. Jts not velocity. Its period.

Explanation:

Took the test

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A businessperson took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours, and each way the trip was 200 miles. What was the speed of the wind that affected the plane, which was flying at a speed of 120mph? Round your answer to the nearest whole number.

Answers

Answer:

Speed of the wind is 48.989 mph

Explanation:

We have given each trip is of 200 miles

So total distance = 200 +200 = 400 miles

Speed of the airplane = 120 mph

Let the speed of the wind = x mph

So the speed of the airplane with wind = 120+x

So time taken by airplane with wind = $(200)/(120+x)$

Speed of the airplane against the wind = 120 - x

So time taken by the airplane against the wind $=(200)/(120-x)$

Total time is given as t= 4 hour

So $(200)/(120+x)+(200)/(120-x)=4$

$(200(120-x)+200(120+x))/((120+x)(120-x))=4$

$48000=57600-4x^2$

$4x^2=9600$

x = 48.989 mph

Answer:

Explanation:

Type Distance Rate Time

Headwind 200 120-r 200/120-r

Tailwind 200 120 - r 200/120 - r

We know the times add to 4, so we write the equation:

200/120−r + 200/120 + r = 4

We multiply both sides by the LCD and simplify to get:

(120−r)(120+r) ((200/120 -r ) + 200/120+r) = 4(120 -r) (120 +r)

200(120−r)+200(120+r)=4(120−r)(120+r)

Factor the 200 and simplify inside the parentheses to find:

200(120−r+120+r)=4(1202−r2)

200(240)=4(1202−r2)

200(60)=120^2−r^2

12,000=14,400−r^2

−2,400= −r^2

49 ≈ r

The speed of the wind is 49mph.

The planet uranus is tilted nearly on its side so that its axis or rotation is only 8 degress abway from its orbit plane. if you lived at latitude 45 degrees on uranus for what fraction of the uranian year would answer?

Answers

The rotation of Uranus, like that of Venus, is retrograde and its axis of rotation is inclined almost ninety degrees above the plane of its orbit. During its orbital period of 84 years one of the poles is permanently illuminated by the Sun while the other remains in the shade. Exactly its rotation period is equivalent to 17 hours and 14 Earth minutes and its translation period is equivalent to 84 years, 7 days and 9 Earth hours.

Only a narrow band around the equator experiences a rapid cycle of day and night, but with the Sun very low on the horizon as in the polar regions of the Earth. On the other side of the orbit of Uranus, the orientation of the poles in the direction of the Sun is inverse. Each pole receives about 42 years of uninterrupted sunlight, followed by 42 years of darkness. Therefore an observer at latitude of 45 degrees in Uranus will probably experience a long winter night that is equivalent to one third of the year uranium.

Two ice skaters, Lilly and John, face each other while at rest, and then push against each other's hands. The mass of John is twice that of Lilly. How do their speeds compare after they push off? Lilly's speed is one-fourth of John's speed. Lilly's speed is the same as John's speed. Lilly's speed is two times John's speed. Lilly's speed is four times John's speed. Lilly's speed is one-half of John's speed.

Answers

Answer:

Lilly's speed is two times John's speed.

Explanation:

m = Mass

a = Acceleration

t = Time taken

u = Initial velocity

v = Final velocity

The force they apply on each other will be equal

$F=ma\n\Rightarrow a_l=(F)/(m_l)$

$F=ma\n\Rightarrow a_j=(F)/(2m_l)\n\Rightarrow a_j=(1)/(2)a_l$

$v=u+at\n\Rightarrow v_l=0+(F)/(m_l)* t\n\Rightarrow v_l=a_lt$

$v=u+at\n\Rightarrow v_l=0+(F)/(2m_l)* t\n\Rightarrow v_j=(1)/(2)a_lt\n\Rightarrow v_j=(1)/(2)v_l\n\Rightarrow v_l=2v_j$

Hence, Lilly's speed is two times John's speed.

Answer:

Lilly's speed is 2 times Johns speed

Explanation:

Hearing the siren of an approaching fire truck, you pull over to side of the road and stop. As the truck approaches, you hear a tone of 460 Hz; as the truck recedes, you hear a tone of 410 Hz. How much time will it take to jet from your position to the fire 5.00 km away, assuming it maintains a constant speed?

Answers

Answer:

The truck will reach there in 250 seconds.

Explanation:

The frequency due to doppler effect, when the observer is stationary and the source is moving towards it is

$f_(obv)$ = $(v)/(v-v_(s) ) f$

where v= velocity of sound in air

$v_(s)$ = velocity of source of sound

f= frequency of sound and

$f_(obv)$ = frequency oberved due to Doppler effect

$(v)/(v-v_(0) ) f$ = 460------------------------------------------( 1 )

The frequency due to doppler effect, when the observer is stationary and the source is moving away from it

$f_(obv)$ = $(v)/(v+v_(s) ) f$

where v= velocity of sound in air

$v_(s)$ = velocity of source of sound

f= frequency of sound and

$f_(obv)$ = frequency oberved due to Doppler effect

$(v)/(v+v_(0) ) f$ = 410-------------------------------------------( 2 )

Dividing ( 1 ) by ( 2 )

$(v+v_(s) )/(v-v_(s) ) =(460)/(410)$

$(v+v_(s) )/(v-v_(s) ) =(46)/(41)$

41v + 41 $v_(s)$ = 46v - 46 $v_(s)$

87 $v_(s)$ = 5v

$v_(s)$ = $(5)/(87)v$

Velocity of Sound (v)= 348 m/s

$v_(s)$ =20 m/s

Therefore, the truck is moving at 20 m/s.

$Time=(Distance)/(Time)$

Distance= 5000 m

Time= $(5000)/(20)$

Time= 250 s

Time = 4 min 10 sec

What is gravity at north pole, South pole and at different point on the equatorial regions. Give reasoning for your answers why do you think it is different or same. Can you imagine same concept for the electric charge, yes or No

Answers

Answer:

The gravity at Equator is 9.780 m/s2 and the gravity at poles is 9.832 m/s2. The gravity at poles are bigger than at equator, principally because the Earth is not totally round. The gravity is inversely proportional to the square of the radius, that is the reason for the difference of gravity (The radius at Poles are smaller than at Equator).

If Earths would have a net charge Q. The Electric field of Earth would be inversely proportional to the square of the radius of Earth (Electric field definition for a charge), the same case as for gravity. So there would be a difference between the electric field at poles and equator, too.

You received a shipment 20 days ago of 13 I for treatment of hyperthyroidism. What fraction of the original shipment would you still have with a half-life of 8.040 days for 31?