3 years ago

How many protons are in the radioactive isotope 40/19K?

1 answer:

8 0

First we have to establish that the number of protons is equivalent to the atomic number of element. Here I am assuming that you are referring to Potassium (K) - 40. Potassium, stable or unstable has 19 protons.

You might be interested in

0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distanc

weqwewe [10]

Answer:

1.6 J

Explanation:

Work = change in energy

W = ΔKE

Fd = KE

(0.8 N) (2 m) = KE

KE = 1.6 J

4 0

3 years ago

When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in th

Y_Kistochka [10]

Your thumb points in the direction of the charge.

6 0

3 years ago

Read 2 more answers

Stronauts in space need to communicate with each other by radio links because

dybincka [34]

Answer:

Because there is nothing out in space , the sound waves from one astronaut's whistling can't travel over to the other astronaut's ears.

7 0

3 years ago

Which is the correct order to use in writing the name of an ionic compound?

muminat

Answer:anion followed by cation

Explanation:

8 0

3 years ago

Read 2 more answers

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

$U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m$

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J$

So, the required work done is 16 J.

7 0

3 years ago