The maximum distance particles of the medium move when a wave passes through them is wave.

What is the mass, in grams, of a pure iron cube that has a volume of 3.70cm cubed?

Arlecino [84]

To determine the amount in grams of the iron, we need data on the density of iron. From literature, it has a value of <span>p=7.9 g/cm3. We simply multiply the volume to the density. We do as follows:

mass = 3.70 (7.9) = 29.23 g Fe

Hope this answers the question. Have a nice day.</span>

7 0

3 years ago

In freefall, heavier objects fall with a greater acceleration than lighter objects.

Rom4ik [11]

Answer:

That is not true all objects fall at the same speed excepts things like feathers or paper.

6 0

3 years ago

Read 2 more answers

A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th

Mila [183]

Answer:

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

$\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0$

$\frac{3}{(r + a)^2} = \frac{2}{r^2}$

$\frac{r+a}{r} = \sqrt{\frac{3}{2}}$

$1 + \frac{a}{r} =\sqrt{\frac{3}{2}}$

$\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}$

so we will have

$r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

so the x coordinate of this position is given as

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

6 0

3 years ago

Help me with number 3 and 4 pleaseeeeeeeee

Andrej [43]

i think its b cause if you were talking about ice it freezes the melts the evaporates to gas

6 0

3 years ago

Read 2 more answers

In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init

liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,

n = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,

U = − 0.061 J

The charge on an electron,

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:

q ₁ = n q

= 1 ×10 ¹³ * − 1.6 × 10 ⁻¹⁹

= -1.6 × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6 × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and q ₂ are the two charges.

r is the distance between the charge and the point.

K = 8.98755 × 10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 = (8.98755 × 10 ⁹ * (-1.6 × 10⁻⁶)²) / r

r = (18.41 × 10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

4 0

3 years ago