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Fantom [35]
3 years ago
13

Wat is the volume of a marble in mL?

Physics
1 answer:
rewona [7]3 years ago
6 0
4/3 pi (radius)^3
If the radius is in cm the volume will be in mL
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Item 4 Which conditions produce the smallest and largest ocean waves? Choose the two correct answers.
Darya [45]

Answer:

strong winds that blow for a long time over a great distance

weak winds that blow for short periods of time with a short fetch

Explanation:

When the winds are weak and blow for short periods, we experience the smallest ocean waves but when there are strong winds over a longer duration, the largest ocean waves are seen. Therefore, the conditions to produce the smallest and largest ocean waves are strong winds that blow for a long time over a great distance and weak winds that blow for short periods of time with a short fetch.

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3 years ago
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A sound system is designed to produce a 1.0W/m2 sound intensity over the surface of a hemisphere 20 m in radius. What acoustic p
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Answer:

2513.6 W

Explanation:

Acoustic power = sound intensity × area of hemisphere

Sound intensity = 1 W/m^2

Area of hemisphere = 2πr^2 = 2×3.142×20^2 = 2513.6 m^2

Acoustic power = 1 W/m^2 × 2513.6 m^2 = 2513.6 W

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4 years ago
3. A skater moves across the ice a distance of 12 m before a constant frictional force of 15 N cause him to stop. His initial sp
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Explanation:

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3 years ago
Which is not a good not a function of proteins
Yuki888 [10]

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3 years ago
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
3 years ago
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