Represent the following sentence as an algebraic expression, where "a number" is the

If an unknown element has a mass of 17 and contains 6 neutrons, how many protons does it have ?

lbvjy [14]

The mass number is the total number of protons and neutrons within an atom and since we know that the unknown element has 6 neutrons, we can simply subtract the number of neutrons from the mass number to get the number of protons.

17 - 6 = 11

There are 11 protons in this unknown element.

Extra:

The number of protons (+) and electrons (-) are equal in a neutral atom so since you know that there are 11 protons you also know that there are 11 electrons. On the periodic table, the element with 11 electrons is Na or Sodium.

Hope this helps! :)

6 0

3 years ago

If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *

Tanzania [10]

Answer:

Less than 400 N

Explanation:

The force of friction is a resistive force that opposes the motion of an object. It is caused by the presence of microbumps on the surfaces of the object, so that when for instance we try to make an object sliding along the surface, there is a force opposite to the push we are applying, and this force is the force of friction.

There are two types of force of friction:

- Force of static friction: this occurs when the object is still at rest. Its value changes according to the push applied: therefore, its value ranges from zero up to a maximum value given by

$F_s = \mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the object

g is the acceleration due to gravity

- Force of kinetic friction: this occurs when the object is already in motion. Its magnitude is given by

$F_k=\mu_k mg$

where

$\mu_k$ is the coefficient of kinetic friction

Generally for most surfaces, $\mu_k < \mu_s$ , so the force of static friction is usually larger than the force of kinetic friction (that's why it generally requires more force to put the object in motion, rather than keeping it in motion).