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3 years ago

10

2 kilogram bowling ball sits on top of a building that is 40 m the object has k e or g p e or both be sure to include the correc

t units

1 answer:

Licemer1 [7]3 years ago

7 0

GPE=mass*GFS*height
2kg*9.8N/Kg*40m -I've used the Grabitational field strength of the earth's
surface

=784J

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Plz Help me with this

AlexFokin [52]

Answer:

Cools ; size

Explanation:

The rate at which magma cools determines the size of the crystals in the new rock. Igneous rocks are formed from the cooling and solidification of molten magma which finds its way to the surface or depth of very low pressure beneath the surface. This place or depth of cooling of magma affects the cooling rate and hence the size of the crystals formed. Igneous rocks formed at depths below the surface have more time to cool and allows more time for Crystal growth and hence produce coarse grained crystal grains called Intrusive igneous rocks which have significantly larger crystals than those formed on the surface which cools rapidly and allowing very little time for crystal growth giving rise to the formation of fine grained crystals and are called extrusive igneous rocks.

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3 years ago

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

Mariulka [41]

Answer:

a) 0.0288 grams

b) $2.6*10^{-10} J/kg$

Explanation:

Given that:

A typical human body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

a)

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

b)

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = $\frac{energy \ absorbed }{mass \ of \ the \ body}$

= $\frac{1.10*10^6*1.6*10^{-14}}{80}$

= $2.2*10^{-10} \ J/kg$

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) = $1.2 *2.2*10^{-10} \ J/kg$

Effective dose (Sv) = $2.6*10^{-10} J/kg$

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3 years ago

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is

Maru [420]

Answer:

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?

Explanation:

Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.

We will convert all parameters in SI units.

Charge = Q = -5.02nC

Q = -5.02× $10^{-9}$ C

As it is clear from question that Sheet is a square (All sides will be of equal length)

Area = A = (21.8× $10^{-2}$ m) (21.8× $10^{-2}$ m) = 4.75× $10^{-4}$ m²

A = 4.75× $10^{-4}$ m²

Surface charge density = Q/A

Surface charge density = (-5.02× $10^{-9}$ C)/(4.75× $10^{-4}$ m²)

Surface charge density = -1.057× $10^{-5}$ C $m^{-2}$

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4 years ago

In this diagram, medium A represents water and medium B represents air. The phenomenon in this diagram, moving from A to B, is c

Readme [11.4K]

Its C) refraction
hope its may help you

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3 years ago

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Peak wavelength of light coming from a star with a temp of 4300k

PSYCHO15rus [73]

from Wein's law

$\frac{k}{ \lambda_{max} } = T \\ \lambda_{max} = \frac{k}{T} \\ \lambda_{max} = \frac{3.898 \times {10}^{ - 3} }{4300} \\ = 9.065 \times {10}^{ - 7}$

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3 years ago