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g100num [7]
4 years ago
14

What is the difference between matter and energy

Physics
1 answer:
Mariulka [41]4 years ago
3 0

Answer:

Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.

Explanation:

You might be interested in
Which statement is true for objects in dynamic equilibrium?
lakkis [162]

Answer:

The correct answer to the question is objects have zero acceleration.

Explanation:

Before answering the question, first we have to understand dynamic equilibrium .

A body moving with uniform velocity is said to be in dynamic equilibrium if the net external forces acting on the body is zero. Hence, the body is under balanced forces.

If the external forces acting on a body is not balanced, then the body will accelerate which will destroy its equilibrium condition. Hence, the necessary and sufficient condition for a body to be in dynamic equilibrium is that the forces are balanced.

When a body is in dynamic equilibrium, the body moves with uniform velocity along a straight line unless and until it is compelled by some external unbalanced forces.

Hence, the rate of change of velocity or acceleration of the body will be zero.

5 0
4 years ago
Read 2 more answers
Is direction the length of the route between two points ?
zepelin [54]

Answer:

No distance is the length between two routes

Explanation:

Distance is the length of the route between two points. ... Direction is just as important as distance in describing motion. A vector is a quantity that has both size and direction. It can be used to represent the distance and direction of motion.

6 0
3 years ago
How do we write 0.00846 as a scientific notation?<br>​
Nadusha1986 [10]

Answer:

Explanation:

m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.

3 0
3 years ago
A mass of 4.10 kg is suspended from a 1.69 m long string. It revolves in a horizontal circle as shown in the figure.
nikklg [1K]

The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R

where m=4.10\,\rm kg, v=2.85\frac{\rm m}{\rm s}, and R is the radius of the circular path.

As shown in the diagram, we can see that

\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)

where r=1.69\,\rm m, so that

F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}

Solve for \theta :

\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0

Complete the square:

\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}

Plugging in the known quantities, we end up with

\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27

The second case has no real solution, since -1\le\cos(\theta)\le1 for all \theta. This leaves us with

\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}

7 0
2 years ago
To determine<br> the force of an object what two measurements<br> are multiplied?
IrinaVladis [17]

Answer:

Mass and acceleration

7 0
3 years ago
Read 2 more answers
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