All categories

4 years ago

If there is 8 g of a substance before a physical change , how much will there be afterwards?

2 answers:

8 0

A substance undergoing a physical change will still weigh the same even after the change. This is in accordance to the law of conservation of mass which states that mass is neither created nor destroyed. so an 8 g substance remains of the same weight even after undergoing a physical change.

stiks02 [169]4 years ago

6 0

Answer: It will be 8 g afterwards

Explanation:

A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.

Example: Melting of ice

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Thus if there is 8 g of a substance before a physical change, 8 g of it will there be afterwards.

You might be interested in

When a bond Norma between two or more elements, what part of the atom is forming the bond?

Mumz [18]

The electrons are the only part of the atom that forms the bond

7 0

4 years ago

What causes all planets to orbit it circles?

Leya [2.2K]

Answer:

because of Gravity

Explanation:

The Sun's gravity pulls on the planets, just as Earth's gravity pulls down anything that is not held up by some other force and keeps you and me on the ground.

4 0

3 years ago

A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Vadim26 [7]

Answer:

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

$F_{c} = F_{G}$

$\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}$

$v = \sqrt{\frac{gr^{2}}{r+h}$

Where:

g is the gravity = 9.81 m/s²

r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km

$v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s$

b) The period of its revolution is:

$T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h$

c) The gravitational force acting on it is given by:

$F = \frac{GMm}{(r + h)^{2}}$

Where:

M is the Earth's mass = 5.97x10²⁴ kg

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

$F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N$

I hope it helps you!

3 0

3 years ago

A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a speed of 8.0 m/swhen it pass

arlik [135]

Answer:

a) $v_{60^{o}} =4.98 m/s$

b) $\theta_{max}=79.34^{o}$

Explanation:

This problem can be solved by doing an energy analysis on the given situation. So the very first thing we can do in order to solve this is to draw a diagram of the situation. (see attached picture)

So, in an energy analysis, basically you will always have the same amount of energy in any position of the pendulum. (This is in ideal conditions) So in this case:

$K_{lowest}+U_{lowest}=K_{60^{o}}+U_{60^{0}}$

where K is the kinetic energy and U is the potential energy.

We know the potential energy at the lowest of its trajectory will be zero because it will have a relative height of zero. So the equation simplifies to:

$K_{lowest}=K_{60^{o}}+U_{60^{0}}$

So now, we can substitute the respective equations for kinetic and potential energy so we get:

$\frac{1}{2}mv_{lowest}^{2}=\frac{1}{2}mv_{60^{o}}^{2}+mgh_{60^{o}}$

we can divide both sides of the equation into the mass of the pendulum so we get:

$\frac{1}{2}v_{lowest}^{2}=\frac{1}{2}v_{60^{o}}^{2}+gh_{60^{o}}$

and we can multiply both sides of the equation by 2 to get:

$v_{lowest}^{2}=v_{60^{o}}^{2}+2gh_{60^{o}}$

so we can solve this for $v_{60^{o}}$ . So we get:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

so we just need to find the height of the stone when the pendulum is at a 60 degree angle from the vertical. We can do this with the cos function. First, we find the vertical distance from the axis of the pendulum to the height of the stone when the angle is 60°. We will call this distance y. So:

$cos \theta = \frac{y}{4m}$