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storchak [24]
4 years ago
14

A catarmaran boat is made of a pair of hulls taht are idealized as parallelpiped bodies. The boat is to carry a heavy slab of ma

terial with weight W. Determine the min hull dimensions so that the draft doesn't exceed the mas___________.
Physics
1 answer:
max2010maxim [7]4 years ago
3 0

Answer:

Explanation:

<em><u>How to determine the min hull dimensions of a catamaran</u></em>

                                      Lenght/beam ratio

<h3>                                 catamaran-LBRC</h3><h3>                                  LBRC = LH/Bcb</h3><h3>if we set Lbrc = 2.2 the longitudinal and tranversal stability will come very near  to the same value. we can design a sailing cataraman wider or narrower , if we like wider contsruction makes its heavier , narrower means that it will carries less sails.</h3>
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Two forces, F1 and F2, act at a point. F1 has a magnitude of 9.80 N and is directed at an angle of α = 62.0° above the negative
oksano4ka [1.4K]

Answer:

Explanation:

Fx = F₁ cos62 + F₂cos 53.7

= -(9.8cos 62 + 5.2 cos 53.7 )

= - ( 4.6 + 3.07

= - 7.67 N

Fy = F₁ sin 62 -  F₂sin  53.7

= 9.8 sin 62 -  5.2 sin  53.7

= 8.65 - 4.19

= 4.46 N

c )

magnitude of resultant F

= √ ( Fx² + Fy²)

= √ ( 7.67² + 4.46²)

= √ ( 58.83  + 19.89)

= 8.87 N  

d )

Tanγ = Fy / Fx

= 4.46 / 7.67

.58

γ  = 30° approx .

6 0
3 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
Weather balloons appear only partially inflated when launched into the air. Why would scientists do this?
zvonat [6]

The reason weather balloons are only partially inflated is to allow room for the helium in the balloon to expand.

7 0
3 years ago
What is the relationship between dipole moment and boiling point?
Lelu [443]
The dipole moment of a molecule and its boiling point are directly varied. If the dipole moment is large, the attraction between the positive and negative charges of a molecule is strong thus, requiring stronger forces to break them. Hence, they will have higher boiling points. 
3 0
3 years ago
It is 6.00 km from your home to the physics lab. as part of your physical fitness program, you could run that distance at 10.0 k
kirill115 [55]
<span>1.) It is 6.00km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0km/hr (which uses up energy at the rate of 700W ), or you could walk it leisurely at 3.00km/hr (which uses energy at 290 W). A.)Which choice would burn up more energy? running or walking? b.)How much energy (in joules) would it burn? c.)Why is it that the more intense exercise actually burns up less energy than the less intense one? Follow 2 answers Report Abuse Answers billrussell42 Best Answer: running, at 10 km/hour for 6 km is 6 km / 10 km/hour = 0.6 hour or 36 min energy used is 700 watts or 700 joules/s x 36 min x 60s/min = 1.512e6 joules or 1.5 MJ walking, at 3 km/hour for 6 km 6 km / 3 km/hour = 2 hour or 120 min energy used is 290 watts or 290 joules/s x 120 min x 60s/min = 1.872e6 joules or 1.8 MJ C) should be obvious PS, this has nothing to do with potential energy. billrussell42 · 5 years ago 0 Thumbs up 1 Thumbs down Report Abuse Comment Simon van Dijk I assume the watt consumption is per hour. Then running 6km at 10.0 km/h results in 700*6/10 = 420 w.h and walking in 290*6/3 = 580 w.h So walking would burn up more energy (kwh) b) 1 kilowatt hour = 3 600 000 joules so 420 wh = 0.42 kwh = 1.51.10^6 joule c) when you put more effort in making the distance your energy is used more efficient. Simon van Dijk · 5 years ago 0 Thumbs up 2 Thumbs down Report Abuse Comment</span>
7 0
3 years ago
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