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3 years ago

Explain the relationship between the distance between the string and your elbow and the effort required to lift the mass.

Physics

2 answers:

MissTica3 years ago

5 0

Answer:

Sorry I I’m a little lost

Explanation:

aleksandr82 [10.1K]3 years ago

4 0

Answer:

hmm the relation ship would be bad 1 because its not cooperating because I would have to do allot of force to pick up the mass with a tiny string, what an idiot, 2nd the distance is far now that its not a good relationship

Explanation:

hope this helps old pal ( :

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The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

5 0

3 years ago

One of the characteristic of sound enogy Similar<br>to light energy is?

Nastasia [14]

Answer:

They both produce heat energy.

6 0

3 years ago

Please help 40 points!!!!

Nutka1998 [239]

The answers to all the research related questions are written below.

Atom is the smallest unit of the element. Different elements have different size atoms and same element have same size atoms.

James Chadwick tracked down the neutron in 1932 and was conceded the Nobel Prize for science in 1935 held in a German prison camp for all of World War 1, he driven the British gathering inside the Manhattan Extend, in which the UK and Canada maintained the USA's World War 2 effort to build the world with nuclear bomb.

Chadwick and Rutherford were the first scientists who measured the nucleus radius using the alpha particles. He did Gold foil experiment to find the size of nucleus.

The contributions are present in the modern atomic model. They are: All matter consists of atoms. Atoms of the same element are the same in size and atoms of different elements are different. Atoms combine in whole-number ratios to form compounds.

Learn more about atoms.

brainly.com/question/1566330

#SPJ1

8 0

2 years ago

The density of a certain type of plastic is 0.75 g/cm3 . If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick,

Oksanka [162]

Answer:

7.5 g

Explanation:

8 0

2 years ago

A student measures the speed of yellow light in water to be 2.00x10^8

max2010maxim [7]

NOTE: The given question is incomplete.

<u>The complete question is given below.</u>

A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.

Solution:

Speed of yellow light in water (v) = 2.00 x 10⁸ m/s

Refractive Index of water with respect to air (μ) = 4/3

Refractive Index = Speed of yellow light in air / Speed of yellow light in water

Or, The speed of yellow light in air = Refractive Index × Speed of yellow light in water

or, = (4/3) × 2.00 x 10⁸ m/s

or, = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s

Hence, the required speed of yellow light in the air will be 3.0 × 10⁸ m/s.

7 0

3 years ago