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allochka39001 [22]
3 years ago
13

What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?

Physics
1 answer:
Sladkaya [172]3 years ago
4 0

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

\rho_w =1 g/mL \rightarrow \text{Water Density}

\rho_o = 0.82g/mL \rightarrow \text{Object density}

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

\sum F= 0

F_w-F_o = 0

F_w = F_o

m_w g = m_og

\rho_w V_w g = \rho_o V_o g

The value of gravity is canceled because it is a constant

\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}

\frac{V_w}{V_o} = \frac{0.82}{1}

\frac{V_w}{V_o} = 0.82

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

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Kay [80]

Answer:

333.3 m

Explanation:

Given

m =100g\ =\  0.1kg\\v = 100 m/s\\g = 10 m/s ^2

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We know that

Potential energy=mgh

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Now From the Equation(1)

mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\  gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\  m

3 0
3 years ago
A 2320 pound roller coaster starts from rest and is launched such that it creates a 110 ft high hill with a speed of 65 mph. The
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Answer: E = 941738.537J

Explanation:

to begin,

given that the mass = 2320 pound = 1052.334 kg

Δh = 110 ft = 33.528 m

given that  distance (d) = 1283 ft = 391.058 m

also the speed (v) is 65 mph = 29.058 m/s

force (F) = 87 pounds = 386.995 N

from our knowledge in work energy theory;

E = Fd + 1/2mv² + mgh

E = (386.995 × 391.058) + (1/2×1052.334×29.058²) + (1052.334×9.81×33.528)

E = 151337.491 + 444278.2 + 346122.84

E = 941738.537J

i hope this helps, cheers.

3 0
3 years ago
A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin
BARSIC [14]

Answer:

U_k = 0.113

Explanation:

using the law of the conservation of energy:

E_i -E_f=W_f

\frac{1}{2}Kx^2=NU_kd

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N = 32.8635

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\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)

solving for U_k:

U_k = 0.113

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