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-Dominant- [34]
3 years ago
12

Which statement is true regarding flexibility?

Physics
2 answers:
Rom4ik [11]3 years ago
8 0

The answer is B. A joint's range of motion will be lost if the joint is not used regularly.


dmitriy555 [2]3 years ago
4 0
<span>The answer is B. A joint's range of motion will be lost if the joint is not used regularly.</span>
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Which of these statements is true?
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C.) Meiosis involves two cycles of cell division

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4 years ago
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What are the three points of the fire triangle
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1-fuel
2-heat
3-oxygen
5 0
3 years ago
What difficulty will you encounter if you only have data from two recording station?
denpristay [2]

<span>If you have only two data from two recording stations then you will be having a hard time finding the correct location of the epicenter. This is because triangulation method requires 3 recording station. If you have 2 recording station, the 2 circles will intersect at 2 points giving you 2 locations that could possibly be the epicenter.</span>

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3 years ago
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If you stood on a planet with four times the mass of Earth, and twice Earth's radius, how much would you weigh?
nikdorinn [45]

Answer:

1/4 times your earth's weight

Explanation:

assuming the Mass of earth = M

Radius of earth = R

∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = \frac{GM}{R^{2} }

where G is the gravitational constant

Gravitational force of the planet = \frac{G4M}{(4R)^{2} }

                                                       =\frac{G4M}{16R^{2} }

                                                       =\frac{GM}{4R^{2} }

recall, gravitational force of earth is given as = \frac{GM}{R^{2} }

∴Gravitational force of planet = 1/4 times the gravitational force of the earth

you would weigh 1/4 times your earth's weight

3 0
3 years ago
A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material
denis-greek [22]

The radius of the cylinder is equal to half the diameter:

r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm

The volume of the cylinder is given by:

V=\pi r^2 h=\pi (27.5 mm)^2 (54.0 mm)=1.28 \cdot 10^5 mm^3

where h is the heigth of the cylinder. Converting into meters,

V=1.28 \cdot 10^{-4} m^3

And the density of the material will be given by the ratio between the mass and the volume:

d=\frac{m}{V}=\frac{1 kg}{1.28 \cdot 10^{-4} m^3}=7812.5 kg/m^3

5 0
3 years ago
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