How much do you have to increase the vertical distance to double the impact speed of an object?

An atom of a certain element has 36 protons, 36 electrons, and a mass number of 84. At room temperature, this element is a very

Grace [21]

Answer:

12 neutrons

Explanation:

The number of protons also shows the atomic number. Therefore the element in question is Krypton (Kr), which also is a noble gas.

Neutrons = Mass number - protons - electrons

Here neutrons = 84 - 36 - 36 = 12

5 0

3 years ago

You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 NN . If you t

shtirl [24]

Answer:

2491.23 kg/m³

Explanation:

From Archimedes principle,

R.d = weight of object in air/ upthrust in water = density of the object/density of water

⇒ W/U = D/D' ....................... Equation 1

Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.

Making D the subject of the equation,

D = D'(W/U).................... Equation 2

Given: W = 28.4 N, U = lost in weight = weight in air- weight in water

U = 28.4 - 17.0 = 11.4 N,

Constant: D' = 1000 kg/m³.

Substitute into equation 2,

D = 100(28.4/11.4)

D = 2491.23 kg/m³

Hence the density of the ceramic statue = 2491.23 kg/m³

7 0

3 years ago

Energy resources that exist in limited amounts and, fonts used, cannot be replaced except over the course of millions of years a

kobusy [5.1K]

Energy resources that cannot be replaced are called "non-renewable" energy resources.

5 0

3 years ago

A bag containing 0ºC ice is much more effective in absorbing energy than one containing the same amount of 0ºC water.

rosijanka [135]

Answer:

No temperature change occurs from heat transfer if ice melts and becomes liquid water (i.e., during a phase change). For example, consider water dripping from icicles melting on a roof warmed by the Sun. Conversely, water freezes in an ice tray cooled by lower-temperature surroundings.

Explanation:

Energy is required to melt a solid because the cohesive bonds between the molecules in the solid must be broken apart such that, in the liquid, the molecules can move around at comparable kinetic energies; thus, there is no rise in temperature. Similarly, energy is needed to vaporize a liquid, because molecules in a liquid interact with each other via attractive forces. There is no temperature change until a phase change is complete. The temperature of a cup of soda initially at 0ºC stays at 0ºC until all the ice has melted. Conversely, energy is released during freezing and condensation, usually in the form of thermal energy. Work is done by cohesive forces when molecules are brought together. The corresponding energy must be given off (dissipated) to allow them to stay together Figure 2.

The energy involved in a phase change depends on two major factors: the number and strength of bonds or force pairs. The number of bonds is proportional to the number of molecules and thus to the mass of the sample. The strength of forces depends on the type of molecules. The heat Q required to change the phase of a sample of mass m is given by

Q = mLf (melting/freezing,

Q = mLv (vaporization/condensation),

where the latent heat of fusion, Lf, and latent heat of vaporization, Lv, are material constants that are determined experimentally.

6 0

2 years ago

The diameter of a copper atom is approximately 2.28e-10 m. The massof one mole of copper is 64 grams. Assume that the atoms area

KIM [24]

1) Mass of one copper atom: $1.063\cdot 10^{-22} kg$

2) There are $9.33\cdot 10^{24}$ atoms in the cube

3) Mass of the cubical block: 992 kg

Explanation:

1)

We are told here that the mass of one mole of copper is

$M=64 g$

for

$n=1 mol$ (number of moles)

We also know that the number of atoms inside 1 mole of substance is equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

This means that $N_A$ atoms of copper have a mass of M = 64 g. Therefore, we can find the mass of one copper atom by dividing the total mass by the avogadro number:

$m=\frac{M}{N_A}=\frac{64}{6.022\cdot 10^{23}}=1.063\cdot 10^{-22} kg$

2)

We are told that the diameter of a copper atom is

$d=2.28\cdot 10^{-10} m$

We can assume that the atoms are arranged in a cube, and that they are all attached to each other; so the side of the cube can be written as size of one atom multiplied by the number of atom per side:

$L=Nd$