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dezoksy [38]
3 years ago
13

A block of density pb = 9.50 times 10^2 kg/m^3 floats face down in a fluid of density pt = 1.30 times 10^3 kg/m^3. The block has

height H = 8.00 cm. By what depth ft is the block submerged? If the block is held fully submerged and then released, what is the magnitude of its acceleration?
Physics
1 answer:
Nookie1986 [14]3 years ago
4 0

Answer:

The depth and acceleration are 0.1919291 ft and 3.61 m/s².

Explanation:

Given that,

Density of block \rho_{b} =9.50\times10^2\ kg/m^3

Density of fluid \rho_{t} =1.30\times10^3\ kg/m^3

We need to calculate the depth

Using balance equation

mg=\rho g V....(I)

We know that,

The density is

\rho=\dfrac{m}{V}

m=\rh0\times V

Put the value of m in equation (I)

\rho_{b}\times V\times g=\rho_{t}\times g\times V

\rho_{b}\times A\times H\times g=\rho_{t}\times g\times A\times h

h=\dfrac{\rho_{b}H}{\rho_{t}}

Put the value into the formula

h=\dfrac{9.50\times10^2\times8.00\times10^{-2}}{1.30\times10^3}

h= 5.85\ cm

h=0.1919291\ ft

We need to calculate the acceleration

Using formula of net force

F_{net}=\rho_{t}\times g\times V- \rho_{b}\times g\times V

ma=\rho_{t}\times g\times V- \rho_{b}\times g\times V

\rho_{b}\times V\times a=\rho_{t}\times g\times V- \rho_{b}\times g\times V

a=(\dfrac{\rho_{t}}{\rho_{b}}-1)g....(II)

Put the value in the equation (II)

a=(\dfrac{1.30\times10^3}{9.50\times10^2}-1)\times9.8

a=3.61\ m/s^2

Hence, The depth and acceleration are 0.1919291 ft and 3.61 m/s².

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Answer:

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3 years ago
A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r
sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0
3 years ago
An unknown material has a mass
atroni [7]

Answer: 1896.55J/kg°C

Explanation:

The quantity of Heat Energy (Q) required to heat a material depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = 1320 joules

Mass of material = 5.61kg

C = ? (let unknown value be Z)

Φ = 0.124°C

Then, Q = MCΦ

1320J = 5.61kg x Z x 0.124°C

1320J = 0.696kg°C x Z

Z = (1320J / 0.696kg°C)

Z = 1896.55 J/kg°C

Thus, the specific heat of the material is 1896.55J/kg°C

4 0
3 years ago
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Kisachek [45]

Answer:

0.2 T

Explanation:

Magnetic field is inversely proportional to the distance from wire since the distance is halved therefore magnetic field will be doubled.

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Elza [17]
The solution that would most likely be a strongest conductor of electricity is the solution that is most saturated or concentrated. This is because the atoms that are found within the aqueous solutions have become positively charged resulting to the attraction of negatively charged ions that are found in electricity. On the other hand, the least conductive from the aqueous solutions would be the most unsaturated one because of less conductive ions present.
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