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bagirrra123 [75]
3 years ago
15

Does decomposing water require high activation energy

Physics
1 answer:
skelet666 [1.2K]3 years ago
6 0
decomposing water does not require High activation energy
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PHYSICS 50 POINTS PLEASE HELP
tangare [24]

Answer:

One way to look at Newton’s three laws of motion is this:

The third law states what forces are. That is, all forces are interactions between two different objects. If one object is interacting with another, then equal and opposite forces act on each object. So no force acts alone. When you exert a force on something, it is exerting the identical force back on you.

The first and second laws deal with the consequences of the forces that act on an object. The first law says that in the absence of a net force on an object, it simply continues doing whatever it was already doing. If it is at rest, it will remain at rest. If it is in motion, it will continue with that same motion - at constant speed and in the direction it was already traveling.

The second law says what happens if there is a net force on the object. In that case, the object accelerates - either by changing its speed, its direction, or both - in proportion and in the direction of the net force that acts on it. The amount of acceleration depends the object’s mass. That is, the larger the mass the smaller the acceleration for a given net force. The first and second laws can be summarized in the mathematical expression

F = ma

where F is the vector sum of all the forces that act on the object at any given moment (i.e., the net force), m is the mass of the object, and a is the acceleration of the object due to the net force at that moment - and is always in the same direction of the net force.

And notice that in a way, the first law is then “contained” within the second. That is, if the net force is zero on an object, then so is the acceleration. That is, either the object is (still) at rest or, if already in motion, the velocity didn’t change, in either case, the acceleration was zero.

Explanation:

4 0
2 years ago
Read 2 more answers
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
A 150 kg safe on frictionless casters is being raised at 1.20m to the bed of a truck using planks 4m long. The force needed to p
Vesnalui [34]
The only thing you need to know in order to solve this task is that <span>plank length (which is force x), should equal the increase in potential energy, so what we have now : (mass)* g * (height).
It has to look like that:  </span>
<span>F * 3.0 = 150 x 9.81 x 1.20 
Then solve for F, the result should be in newtones = 588N

Do hope it makes sense.</span>
3 0
2 years ago
How far will a train with a speed of 20 m/s travel in: 10s
Citrus2011 [14]

Answer:

The correct answer would be 200m

Explanation:

20x10=200

the previous answer is incorect from the other user.

8 0
3 years ago
If a 1.00 kg body has an acceleration of 2.44 m/s2 at 53° to the positive direction of the x axis, then what are (a) the x comp
Ilia_Sergeevich [38]

(a) Fx = 1.464 N

(b) Fy = 1.952 N

(c) F(x, y) = 1.464 i + 1.952 j

Given

Mass = 1kg

Acceleration = 2.44 m/s2

Angle with positive X axis = 53°

As we know

F = ma

By substituting value

F= 1×2.44 N

F= 2.44 N

(a)   Component of force in X direction

Fx = F Cosθ

Fx = 2.44 Cos(53°)

Fx = 2.44 × 0.60 = 1.464 N

(b) Component of force in Y direction

Fy = F Sinθ

Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N

(c) Net force in vector notation

F(x, y) = 1.464 i + 1.952 j

Thus we got net force.

#SPJ4

For details visit www.brainly.com

6 0
2 years ago
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