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3 years ago

A chunk of paraffin (wax) has a mass of 50.4 grams and a volume of 57.9 cm3. What is the density of

1 answer:

6 0

$Formula\ for\ density:\\\\
p=\frac{m}{V}\\p-density,\\m-mass,\\V-volume\\\\
Data:\\
m=50,4grams\\
V=57,9cm^3\\\\
p=\frac{50,4g}{57,9cm^3}=0,87\frac{g}{cm^3}\\\\Density\ of\ paraffin\ is\ equal\ to\ 0,87\frac{g}{cm^3}.
$

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How do i find the number of leptons in an atom??

Serjik [45]

Answer:

In particle physics, a lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions.[1] Two main classes of leptons exist: charged leptons (also known as the electron-like leptons or muons), and neutral leptons (better known as neutrinos). Charged leptons can combine with other particles to form various composite particles such as atoms and positronium, while neutrinos rarely interact with anything, and are consequently rarely observed. The best known of all leptons is the electron.

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2 years ago

A wave amplitude 0.36m interferes with a second wave of amplitude 0.22m traveling in the same direction. What is the largest res

DerKrebs [107]

The largest resultant amplitude would be that created by constructive interference, basically when the two waves are of the same phase, so it would be 0.36m+0.22m= 0.58 m.

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3 years ago

Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he

11111nata11111 [884]

Answer:

E = 2k $\frac{\lambda}{ r}$

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

λ = q_ {int} /l

q_ {int} = l λ

we substitute

E A = l λ /ε₀

is area of cylinder is

A = 2π r l

we substitute

E = $\frac{ l \ \lambda}{ \epsilon_o \ 2\pi \ r \ l }$

E = $\frac{\lambda}{ 2\pi \epsilon_o \ r}$

the amount

k = 1 / 4πε₀

E = 2k $\frac{\lambda}{ r}$

5 0

2 years ago

A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling

hodyreva [135]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

4 0

3 years ago

A child (approximately 4 years old) takes her metal “slinky Toy” (a flexible coiled metal spring) and does various tests to dete

anzhelika [568]

Answer:

248

Explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H

$l$ = length of the slinky = 3 m

N = number of turns in the slinky

r = radius of slinky = 4 cm = 0.04 m

Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

Inductance is given as

$L = \frac{\mu _{o}N^{2}A}{l}$

$130\times 10^{-6} = \frac{(12.56\times 10^{-7})N^{2}(0.005024)}{3}$

N = 248

8 0

2 years ago