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Vaselesa [24]
2 years ago
11

when a metal sphere is dropped in to a tall cylinder containing liquid its acceleration is g÷2 (gravity over 2) show that : dens

ity of metal =2density of liquid​
Physics
1 answer:
Anastaziya [24]2 years ago
5 0

The density of the metal sphere is 2 times the density of the liquid as proved.

<h3>Net upward force acting on the metal sphere</h3>

The net upward force acting on the sphere as it is dropped into the liquid is calculated as follows;

F =  σVg - ρVg

ma =  σVg - ρVg  

where;

  • ρ is density of the liquid
  • σ is the density of the metal
  • a is acceleration of the metal

σV(a) =  σVg  - ρVg

σ(a) = σg  - ρg

σ(g/2) = σg -  ρg  

g(σ/2) = g(σ -  ρ)

σ/2 = σ -  ρ

σ/2 - σ = -  ρ

-σ/2 = -  ρ

σ = 2ρ --- proved

Thus, the density of the metal sphere is 2 times the density of the liquid as proved.

Learn more about density here: brainly.com/question/1354972

#SPJ1

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3 0
3 years ago
Read 2 more answers
An iron anchor of density 7890.00 kg/m3 appears 299 N lighter in water than in air. (a) What is the volume of the anchor? (b) Ho
PtichkaEL [24]

Answer:

weigh is 2353.13 N

Explanation:

Given data

density = 7890.00 kg/m3

lighter =  299 N

to find out

the volume of the anchor and weigh in air

solution

from question we can say that

apparent weight = actual weight - buoyant force

we know weight = mg and buoyant force = water density × g

so volume of anchor is = actual weight - apparent weight / buoyant force

volume of anchor is = 299 / 1000 × 9.81

volume of anchor is = 0.0304791 m³

and

weight of anchor is mg

here mass m = density Fe g

density Fe = 7870 from table 14-1

so weight = 7870 × 0.0304791  × 9.81

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7 0
4 years ago
1.45 L of 16°C water is placed in a refrigerator. The refrigerator's motor must supply an extra 10.7 W power to chill the water
Vika [28.1K]

Answer:

The coefficient of performance of the refrigerator is 2.251.

Explanation:

In this case, the coefficient of performance of the refrigerator (COP), no unit, is equal to the ratio of the heat rate received from the water to the power needed to work, that is:

COP = \frac{\dot Q_{L}}{\dot W} (1)

COP = \frac{\rho\cdot V\cdot c_{w}\cdot \Delta T}{\dot W \cdot \Delta t} (2)

Where:

\dot Q_{L} - Heat rate received from the water, in watts.

\dot W - Power, in watts.

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

V - Volume of water, in cubic meters.

c_{w} - Specific heat of water, in joules per kilogram-degree Celsius.

\Delta T - Temperature change, in degrees Celsius.

\Delta t - Cooling time, in seconds.

If we know that \rho = 1000\,\frac{kg}{m^{3}}, V = 1.45\times 10^{-3}\,m^{3}, c_{w} = 4187\,\frac{J}{kg\cdot ^{ \circ}C}, \Delta T = 10\,^{\circ}C, \dot W = 10.7\,W and \Delta t = 2520\,s, then the coefficient of refrigeration of the refrigerator is:

COP = \frac{\rho\cdot V\cdot c_{w}\cdot \Delta T}{\dot W \cdot \Delta t}

COP = 2.251

The coefficient of performance of the refrigerator is 2.251.

6 0
3 years ago
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