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3 years ago

What is the definition of physical change

1 answer:

jeka943 years ago

7 0

They are used to separate mixtures into their component compounds but can no usually be used to deprecate compounds into chemical elements or simpler compounds

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If a 15g irregular shaped object is submerged in a graduated cylinder of water, the level rises from 10 ml to 25 ml, what is the

SashulF [63]

Answer:

one im so sry i have no idea. I have been researshing for about 30min and i cant find anything im so sry:/

Explanation:

6 0

4 years ago

If the velocity is 50 and the time is 5 seconds what is the acceleration?

disa [49]

Just divide the both, you will get the answer!

does it sound rude?

im sorry for that!

3 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

3 years ago

The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potentia

MAVERICK [17]

Answer:

Energy Lost for group A's car = 0.687 J

Energy Lost for group B's car = 0.55 J

Explanation:

The exact question is as follows :

Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.

To find - How much energy is lost due to heat for group A's car ?

How much for Group B's car ?

Solution -

We know that,

GPE = 1 Joule (Potential Energy)

Now,

For Group A -

Energy Lost = GPE - KE

= 1 J - 0.313 J

= 0.687 J

So,

Energy Lost for group A's car = 0.687 J

Now,

For Group B -

Energy Lost = GPE - KE

= 1 J - 0.45 J

= 0.55 J

So,

Energy Lost for group B's car = 0.55 J

8 0

3 years ago

Se aplica una fuerza neta de 25 N a una masa de 10 kg. ¿Cuál es la aceleración dada a la masa?

lys-0071 [83]

Answer:

Se queda

Explanation:

Porque 10 kg es más que 25N

4 0

3 years ago