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omeli [17]
3 years ago
15

the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is

40 kg, calculate the mass of another body. (And:119.9 kg)​
Physics
1 answer:
cluponka [151]3 years ago
3 0

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

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Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th
RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2} (1)

Where:

P_{1}, P_{2} - Pressures of the water on the first and second floors, measured in pascals.

\rho - Density of water, measured in kilograms per cubic meter.

v_{1}, v_{2} - Speed of the water on the first and second floors, measured in meters per second.

z_{1}, z_{2} - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]

\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]

v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})

v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) } (2)

If we know that P_{1}-P_{2} = 0\,Pa, \rho=1000\,\frac{kg}{m^{3}}, v_{1} = 15.5\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and z_{1}-z_{2} = -3.5\,m, then the speed of the water flow in the pipe on the second floor is:

v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}

v_{2} \approx 13.100\,\frac{m}{s}

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0
3 years ago
A block of ice at 0°C is added to a 150g aluminum calorimeter cup which holds 210 g of water at 12°C. If all but 2.0 g of ice me
Over [174]

Answer:

original mass of the block of ice is 38.34 gram

Explanation:

Given data

cup mass = 150 g

ice temperature = 0°C

water mass = 210 g

water temperature = 12°C

ice melt = 2 gram

to find out

solution

we know here

specific heat of aluminum is c = 0.900 joule/gram °C

Specific heat of water C =  4.186 joule/gram °C

so here temperature difference is dt =  12- 0 = 12°C

so here heat lost by water and cup are given by

heat lost  = cup mass × c  × dt + water  mass × C × dt

heat lost  = 150 × 0.900  × 12 + 210 × 4.186 × 12

heat lost  = 12168.72 J

so

mass of ice melt here = heat lost / latent heat of fusion

here we know latent heat of fusion = 334.88 joule/gram

so

mass of ice melt  =  12168.72 / 334.88

mass of ice melt  is 36.337554 gram

so mass of ice is here = mass of ice melt + ice melt

mass of ice  =  36.337554 + 2

mass of ice  =  38.337554 gram

so original mass of the block of ice is 38.34 gram

7 0
3 years ago
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