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jonny [76]
3 years ago
13

What do I have to do to figure out " What % of an object’s mass is above the water line if the object’s density is 0.82g/ml " wi

th just that information?
Physics
1 answer:
blsea [12.9K]3 years ago
4 0

Answer:

18%

Explanation:

There are two equal and opposite forces on a floating object: weight and buoyancy.

W = B

The weight of an object is its mass times gravity: W = mg

Buoyancy is the weight of the displaced fluid: W = mf g

Plugging in:

mg = mf g

m = mf

Mass is density times volume:

ρV = ρf Vf

Solving for the ratio of Vf / V:

Vf / V = ρ / ρf

Given that ρ = 0.82 g/mL and ρf = 1.00 g/mL:

Vf / V = 0.82

That means 82% of the object's volume (and therefore, 82% of its mass, assuming uniform density) is submerged.  Which means that 18% is above the water line.

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Find the value of F1 + F2 + F3.<br>​
Dovator [93]

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

<u>For F₁</u>

<u />F_{y}=2[N]<u />

<u>For F₂</u>

F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]

<u>For F₃</u>

<u />F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]<u />

Now we can sum each one of the forces in the given axes:

F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]

Now using the Pythagorean theorem we can find the total force.

F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]

8 0
3 years ago
A passenger on your boat falls overboard. what should you do first?
yarga [219]
The very first thing that you should do when a passenger on your boat falls overboard is to throw a PFD or also known as a Personal Flotation Device. This would include anything that can help the passenger to float. But this step would differ, only if the passenger is not wearing a lifevest. 
5 0
4 years ago
A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill
LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s

(ii) The linear speed of the small metal object is calculated as;

v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s

6 0
3 years ago
A capacitor is charged to a potential difference of 3 volt it delivers 30% store energy to lamp what is the final potential diff
nexus9112 [7]

Answer:

3.98V

Explanation:

Given

Pontential difference V as 3v

Energy delivered is 30%,

Recall that Enery E=1/2cv^2 from this E=V^2(since Current C is not provided we can assume a value 2)

So E=V^2

E=3^2=9

At full charge E=9,30%of 9,0.3*9=2.7 energy in capacitor is 9-2.7=6.3

But E=V^2

✓E=V

✓6.3=3.98V

4 0
3 years ago
A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic
barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

(b) Change in momentum of the truck, \Delta p=m(v-u)

\Delta p=2000\ kg\times (16.11-9.44)

\Delta p=13340\ kg-m/s

Hence, this is the required solution.

6 0
4 years ago
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