3 years ago

Approximately how many atoms are there along a 9.5 cm line

1 answer:

4 0

First convert 5.5 cm to meters.

(5.5 cm / 1) x (m / 100 cm) = 0.055 m

A typical atom is about 1.0E-10 m in diameter; thus:

0.055 m / 1.0E-10 m = 5.5E8 atoms or 550,000,000 end-to-end atoms in 5.5 cm

You might be interested in

X<br> x<br> 2.5 N<br> 1.7 N<br> 2.5 N

Olin [163]

Answer:

2.5*1.7/2.5

Explanation:

6 0

3 years ago

Electromagnets are created by?

AysviL [449]

A coil of insulated wire around an iron core

7 0

4 years ago

Read 2 more answers

A train is accelerating at a rate of 2m/s2. if it's initial velocity is 20 m/s , what is its velocity after 30 seconds

netineya [11]

(v₂-v₁)/t = a

(v₂-20)/30 = 2

v₂-20 = 60

v₂ = 80 m/s

6 0

3 years ago

A 0,9 -kg object attached to the end of a string swings in a vertical circle (radius = 75 cm). At the top of the circle the spee

omeli [17]

0.6 cm is the answer add it up and find the m/s hope this helps

6 0

3 years ago

A velocity selector, in which charged particles of a specific speed pass through undeflected while those of greater or lesser sp

uranmaximum [27]

Answer:

a) 351351.35m/s

b) 1.044*10^{-8}kg/C

Explanation:

a) Electric force and magnetic force over the charge must have the same magnitude. From there we can compute the seep of the charge.

$F_E=F_B\\\\qE=qvB\\\\v=\frac{E}{B}=\frac{1.95*10^{5}V/m}{0.555T}=351351.35\frac{m}{s}$

b) the mass-charge ratio is given by:

$\frac{m}{q}=\frac{rB}{v}=\frac{(6.61*10^{-3}m)(0.555T)}{351351.35m/s}=1.044*10^{-8}\frac{kg}{C}$

hope this helps!!

5 0

3 years ago