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lidiya [134]
2 years ago
9

2. Below what depth would a submarine have to submerge so that it would not be swayed by surface waves with a wavelength of 24 m

eters?
Physics
1 answer:
Travka [436]2 years ago
4 0

<u>Answer:</u> Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

<u>Explanation:</u>

To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.

This wave base is equal to half of the wavelength. The equation becomes:

Wave base = \frac{\text{Wavelength}}{2}

We are given:

Wavelength = 24 m

Putting values in above equation, we get:

Wave base = \frac{24m}{2}=12m

Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

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A train travels 4200 km in 90 hours. What is its speed?
denis23 [38]

Answer: I think its 120

Explanation: thx for the free points :)

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2 years ago
Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi
Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

   P₁ = m v

   P₁ = 0.0575 x 26.7

   P₁ = 1.535 kg.m/s

b) momentum of ball after collision

   P₂ = m v'

   P₂ = 0.0575 x (-29.5)

   P₂ = -1.696 kg.m/s

c) change in momentum

    Δ P = P₂ - P₁

    Δ P = -1.696 -1.535

    Δ P = -3.231 kg.m/s

d) using conservation of momentum

  initial speed of racket = 0 m/s

  M u + m v = Mu' + m v

  M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

  0.331 u' = 3.232

     u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

5 0
3 years ago
A force of 100N is applied to move an object a horizontal distance of 20m to the right. The work done by this force on the objec
horsena [70]
WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
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5 0
3 years ago
You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock
3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

y = 29.77 ft

The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

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chemical reaction brainly.com/question/3953793

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8 0
3 years ago
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