All categories

3 years ago

How are reflectivity and solubility related?

Physics

2 answers:

ser-zykov [4K]3 years ago

5 0

He or she said bc i searched it up and they right

Lubov Fominskaja [6]3 years ago

4 0

Answer:

they are related because..

Explanation:

Physical properties can be observed or measured without changing the composition of matter. Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others

You might be interested in

A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What

icang [17]

Answer:

Friction between the box and the floor is 25N to the left

Explanation:

There are two forces acting on the box along the horizontal direction:

- The force of push applied by the worker, in the forward direction, F

- The force of friction, $F_f$ , acting in the opposite direction (backward)

So the net force acting on the box is

$F_{net}=F-F_f$

According to Newton's second law of motion, the net force on an object is equal to the product between its mass (m) and its acceleration (a), so we can write:

$F_{net}=ma$

And so

$F-F_f = ma$

However, in this case the box is moving at constant speed; this means that its acceleration is zero:

$a=0$

Therefore we have:

$F-F_f=0$

Which means

$F_f=F$

And since we are told that

$F=25 N$

This means that the force of friction is also 25 N:

$F_f=F=25 N$

6 0

3 years ago

Read 2 more answers

A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.

Maksim231197 [3]

Answer:

$v=33.764m/s$

Explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

Friction Force F=3.6 N

To find

Velocity v

Solution

From the Centripetal force we know that

$F_{c}=\frac{mv^{2} }{r}$

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

$F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s$