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2 years ago

Your boat capsizes but remains floating upside down. what should you do?

2 answers:

8 0

On the chance that you capsize or swamp your boat, or if you have fallen over the edge and can't get back in, stay with the boat if possible.

Further Explanation:

Try to shake a flag:

Your overwhelmed vessel is simpler to see and will flag that you are in a difficult situation. Additionally signal for assistance utilizing different gadgets accessible (visual trouble signals, whistle, reflect).

Wear a PFD:

In the event that you committed the error of not wearing a PFD, discover one and put it on. In the event that you can't put it on, clutch it. Have your travelers do likewise.

Head check:

Take a head check. Reach, toss, push, or go, if necessary.

On the top of boat

On the off chance that your boat stays above water, attempt to reboard or climb onto it so as to get however much of your body out of the virus water as could be expected. Stepping water will make you lose body heat quicker, so attempt to utilize the pontoon for help.

Answer Details:

Subject: Physics

Level: High School.

Key Words:

Try to shake a flag:

Wear a PFD:

Head check:

On the top of boat

For further Evaluation :

brainly.com/question/10872128

brainly.com/question/4379740

Ludmilka [50]2 years ago

5 0

You should stay on the top of the boat, this will help rescuers find you because a boat is easier to find than a small body in a large ocean. Never swim away from the boat you could catch a cold and/ or get hypothermia or drift far away from where you were last seen. This will make the search much harder.

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When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t

antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0

2 years ago

An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

kolbaska11 [484]

Answer:

$mg=200.4 N$ .

Explanation:

This problem can be solved using Newton's law of universal gravitation: $F=G\frac{m_{1}m_{2}}{r^{2}}$ ,

where F is the gravitational force between two masses $m_{1}$ and $m_{2}$ , $r$ is the distance between the masses (their center of mass), and $G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2})$ is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting $w_{s}$ be the weight on the surface:

$w_{s}=mg$ ,

$mg=8*10^{2}$ ,

$m=(8*10^{2})/g$ ,

since we now that $g=9.8m/s^{2}$ we get that the mass is

$m=81.6kg$ .

Now we can use Newton's law of universal gravitation

$F=G\frac{Mm}{r^{2}}$ ,

where $m$ is the mass of the astronaut and $M$ is the mass of the earth. From Newton's second law we know that

$F=ma$ ,

in this case the acceleration is the gravity so

$F=mg$ , (becarefull, gravity at this point is no longer $9.8m/s^{2}$ because we are not in the surface anymore)

and this get us to

$mg=G\frac{Mm}{r^{2}}$ , where $mg$ is his new weight.

We need to remember that the mass of the earth is $M=5.972*10^{24}kg$ and its radius is $6.37*10^{6}m$ .

The total distance between the astronaut and the earth is

$r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6}$ meters.

Now we can compute his weigh:

$mg=G\frac{Mm}{r^{2}}$ ,

$mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}}$ ,

$mg=200.4 N$ .

5 0

2 years ago

Determine the change in thermal energy of 100 g of copper (M = 63,5, Debye 348K) if it is cooled from

Setler [38]

Answer:

given,

mass of copper = 100 g

latent heat of liquid (He) = 2700 J/l

a) change in energy

Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (300 - 4)

Q = 11153.63 J

He required

Q = m L

11153.63 = m × 2700

m = 4.13 kg

b) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (78 - 4)

Q = 2788.41 J

He required

Q = m L

2788.41 = m × 2700

m = 1.033 kg

c) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (20 - 4)

Q = 602.90 J

He required

Q = m L

602.9 = m × 2700

m =0.23 kg

8 0

2 years ago

Materials have unique properties because each one is made up of different kinds of which particle?

inysia [295]

D. Atoms.

Explanation:

All the matter is made of elementary particles called "atoms".

Further, an atom is made of electrons, protons and neutrons. The electrons & protons are again made of the fundamental sub-particles, electrons (leptons) and the protons(quarks).

The classification of particles is shown in the figure attached

7 0

2 years ago

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A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu

Illusion [34]

Starting from the angular velocity, we can calculate the tangential velocity of the stone:
$v=\omega r= (12 rad/s)(0.5 m)= 6 m/s$

Then we can calculate the angular momentum of the stone about the center of the circle, given by
$L=mvr$
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
$L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}$

3 0

2 years ago