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Virty [35]
3 years ago
13

A rogue wave is a random monstrous wave that occurs unexpectedly in the ocean.

Physics
2 answers:
Tpy6a [65]3 years ago
8 0

A wave with a large amplitude

a wave check

natulia [17]3 years ago
8 0

Answer:

give the other man brainliest

Explanation:

he deserves it

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If the person has less villi than normal in their small intestine, then the surface area will not be as large meaning there is less area for absorption to occur across so less soluble molecules will be absorbed.
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I need help with this Physics problem. I've been stuck forever:
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6 0
3 years ago
Due Ma<br> duart<br> ded<br> out<br> 25 N<br> 35 N<br> 1-03
Anuta_ua [19.1K]

Answer:

what's that all about

hehehwhe

Explanation:

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8 0
3 years ago
In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable
pav-90 [236]

Answer:

Approximately 3.98\; \rm m \cdot s^{-2}.

Assumption: air resistance on the rocket is negligible. Take g = \rm 9.81\; m \cdot s^{-2}.

Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}.

Note that in this case, the uppercase letter \rm M in the units stands for "mega-", which is the same as 10^6 times the unit that follows. For example, \rm 1\; Mg = 10^6\; g, while \rm 1\; MN = 10^6\; N.

Convert the mass of the rocket and the thrust of its engines to SI standard units:

  • The standard unit for mass is kilograms: \displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g}  = 552 \times 10^3 \; kg.
  • The standard for forces (including thrust) is Newtons: \text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N.

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

  • Thrust (which is supposed to go upwards), and
  • Weight (downwards due to gravity.)

The thrust on the rocket is already known to be \rm 7.61 \times 10^6\; N. Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately 9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}. Hence, the weight on the rocket would be approximately 9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N.

The magnitude of the net force on the rocket would be

\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}.

Apply the formula \displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}} to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg}  \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}.

6 0
3 years ago
Review Conceptual Example 7 before starting this problem. A uniform plank of length 5.0 m and weight 225 N rests horizontally on
Andreas93 [3]

Answer:

0.82 m

Explanation:

We are given that

Weight,w_1=225 N

Length,l=5 m

d=1.1 m

We have to find the distance x  for which person weighs 385 N walk on the overhanging part of the plank before it just begins to tip.

Half length=\frac{l}{2}=\frac{5}{2}=2.5 m

r=\frac{l}{2}-d=2.5-1.1=1.4 m

w_2=385 N

When system in equilibrium then

w_1r=w_2 x

225\times 1.4=385\times x

x=\frac{225\times 1.4}{385}

x=0.82 m

6 0
4 years ago
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