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Ilia_Sergeevich [38]
4 years ago
9

Object A has a length of 3 cm, a width of 2 cm, and a height of 4 cm. Object B is dropped into a graduated cylinder. It displace

s 19 mL of water. The volume of object A is:
A: greater than the volume of object B
B: less than the volume of object B
C: equal to the volume of object B
Physics
1 answer:
MA_775_DIABLO [31]4 years ago
4 0

Answer:

  A:  greater than the volume of object B

Explanation:

Based on the dimensions, the volume of object A is ...

  (3 cm)(2 cm)(4 cm) = 24 cm³ = 24 mL

If object B displaces 19 mL of water, we presume that is its volume. Since 24 mL is more than 19 mL, we conclude ...

  The volume of object A is greater than the volume of object B.

_____

<em>Comment on density</em>

We have presumed that object B is completely submerged. If it is not, then the relative volumes will depend on the densities. If the density of object B is less than about (19/24) g/mL, its volume may very well be larger than that of object A.

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Answer:

Alpha = ω^2 R    where R is radius of blade

g = w^2 r      where r is distance from center

ω^2 R = 11.5 ω^2 r

R / r = 11.5 / 9.8 = 1.17

Or r = .852 R

Since the angular acceleration depends on both R and ω it seems that one can only get r as it depends on R

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How many focal points does a lens have?
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Answer and Explanation:

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4 years ago
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We are given the description of an enclosure that has fencing on 3 sides and can use 3,056 yards of fencing. We are to find the
shusha [124]

Answer:

The sum of the lengths of the sides is 2292 yards and the sum of the lengths of the triangle is 3056 yards

Explanation:

Since y represents the length of fence that is opposite (parallel) to the river and x represent the length of fence perpendicular to the river.

Therefore since we can use 3,056 yards of fencing

Side perpendicular to the river = x and,

Side opposite to the river = y = 3056 - 2x

The area of the rectangle formed (A) = Perpendicular side × Parallel side

∴ A = x(3056 - 2x) = 3056x - 2x²

A = 3056x - 2x²

To maximize the area, A' (dA/dx) = 0

∴ A' = 3056 - 4x = 0

3056 - 4x = 0

4x = 3056

x = 764 yards

y =  3056 - 2x = 3056 - 2(764) = 1528 yards.

Side perpendicular to the river = 764 yards and,

Side opposite to the river = 1528 yards

The sum of the lengths of the sides = 764 + 1528 = 2292 yard and the sum of the lengths of the triangle = 764 + 764 + 1528 = 3056 yards

5 0
3 years ago
Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat
QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

                               j_{H_2O}=\frac{D}{l} \bigtriangleup c

where

  • j_{H_2O} is the mass flux of the water which is to be calculated.
  • D is diffusion coefficient which is given as 0.25 cm^2/s
  • l is the thickness of the film which is 0.15 cm thick.
  • \bigtriangleup c is given as

                                \bigtriangleup c= \frac{P_{sat}-P_a}{RT}

In this

  • P_{sat} is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa
  • P_a is the air pressure which is given as 0.5 times of P_{sat}
  • R is the universal gas constant as 8.314 kJ/kmol-K
  • T is the temperature in Kelvin scale which is 20+273= 293K

By substituting values in the equation

                                    \bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3

Converting \bigtriangleup c into cm^3/cm^3

As 1 mole of water 18 cm^3 so

                               \bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6}  cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6}  cm^3/cm^3

Putting this in the equation of mass flux equation gives

                            j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6}  cm/s

For calculation of water level drop in a day, converting mass flux as

                     j_{H_2O}=14.4 \times 10^{-6}  \times 24 \times 3600  cm/day\\ j_{H_2O}=1.24  cm/day

So the water level will drop by about 1.24 cm in 1 day.

7 0
3 years ago
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