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3 years ago

Be-36 what hull type is best for use on ponds, small lakes and calm rivers?

2 answers:

8 0

The hull type that is best for use on ponds, small lakes and calm rivers is Flat Bottom Hull.
A flat bottomed boat is a boat with a flat bottomed, two-chined hull, which allows it to be used in shallow bodies of water, such as rivers, because it is less likely to ground. The flat hull also makes the boat more stable in calm water.

Pachacha [2.7K]3 years ago

8 0

Answer:

Flat bottomed two chined hull

Explanation:

For a boat to be used in ponds, small lakes and calm rivers, flat bottomed and two chined hull are perfect. As this type of hull allows them to be used in pond's shallow water. Since the depth is not that much having such type of hull prevents grounding of the boat. Also, this type of hull makes the boat stable in calm water in comparison to a boat with v-shaped or rounded hull.

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A stone is dropped at t = 0. A second stone, with 6 times the mass of the first, is dropped from the same point at t = 59 ms. (a

xxTIMURxx [149]

Answer: $y_{com}=0.707 m$

$v_{com}=3.713 m/s$

Explanation:

Given

first stone mass is m

second stone mass is 6 m

distance traveled by first stone in 430 ms

$y_1=ut+\frac{at^2}{2}$

$y_1=0+\frac{g(0.43)^2}{2}$

$y_1=0.9069 m$

Distance traveled by stone 2 in t=430-59=371 ms

$y_2=ut+\frac{at^2}{2}$

$y_2=0+\frac{g(0.0.371)^2}{2}$

$y_2=0.674 m$

velocity of first stone after t=0.43 s

$v_1=u+at$

$v_1=0+9.8\times 0.43=4.214 m/s$

velocity of second stone after t=0.371 s

$v_2=u+at$

$v_2=0+9.8\times 0.371=3.63 m/s$

Position of Center of mass of system

$y=\frac{y_1m_1+y_2m_2}{m_1+m_2}$

$y=\frac{0.9069\times m+0.674\times 6m}{m+6m}$

$y=\frac{4.95m}{7m}=0.707 m$

Velocity of COM

$v_{com}=\frac{v_1m_1+v_2m_2}{m_1+m_2}$

$v_{com}=\frac{4.214\times m+3.63\times 6m}{m+6m}$

$v_{com}=3.713 m/s$

6 0

4 years ago

He first step in the problem-solving process is

german

Answer:

To identify the problem

Explanation:

The first step in the problem-solving process is to identify the problem. It is not as simple as it sounds as different people may have different ideas of what the "problem" is. Clearly stating the problem and getting everyone involved to agree is an important first step.

5 0

3 years ago

4. Solid ‘X’ turns into a liquid at 80 0C and into a gas at 140 0C. Describe the changes in the

Paladinen [302]

Answer:

Explanation:

The melting point of the solid is 80°C

Vapor point of the liquid is 140°C

What happens to particles of X when heated from 70°C to 85°C?

Firstly, there would be a phase change from solid to liquid.
Below the melting point, a substance will exist as a solid.
With increase in thermal energy inputted by heat, as the temperature climbs above the melting point, it changes to the liquid.
When the solid begins to heat up, the particles of X starts vibrating about their fixed point.
At they melting point, they break lose and flow to form a liquid.
The particles will have more kinetic energy.

5 0

3 years ago

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Neporo4naja [7]

Answer:

Explanation:

Let the charge on proton be q .

energy gain by proton in a field having potential difference of V₀

= V₀ q

Due to gain of energy , its kinetic energy becomes 1/2 m v₀²

where m is mass and v₀ is velocity of proton

V₀ q = 1/2 m v₀²

In the second case , gain of energy in electrical field

= 2 V₀q , if v be the velocity gained in the second case

2 V₀q = 1/2 m v²

1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²

mv² = 2 m v₀²

v = √2 v₀

6 0

3 years ago

You push a piano 5 meters across the floor with a force of 400 n. how much work have you done?

kobusy [5.1K]

2,000 j(apex)

i hope this was right

3 0

3 years ago