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Vitek1552 [10]
3 years ago
6

During combustion reactions, explain why the energy of the reactants must exceed the total energy of the products​

Physics
1 answer:
maxonik [38]3 years ago
8 0

Answer:

In these reactions the products are higher in energy than the reactants. ... This barrier is due to the fact that to make CO2 and H2O we have to break 4 carbon-hydrogen bonds and some ...

Explanation:

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A student carried out an experiment adding different weights to a spring and recording the results. Look at the table of results
MAXImum [283]

Answer:

0.25 m.

Explanation:

We'll begin by calculating the spring constant of the spring.

From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:

Force (F) = 0.1 N

Extention (e) = 0.125 m

Spring constant (K) =?

F = Ke

0.1 = K x 0.125

Divide both side by 0.125

K = 0.1/0.125

K = 0.8 N/m

Therefore, the force constant, K of spring is 0.8 N/m

Now, we can obtain the number in gap 1 in the diagram above as follow:

Force (F) = 0.2 N

Spring constant (K) = 0.8 N/m

Extention (e) =..?

F = Ke

0.2 = 0.8 x e

Divide both side by 0.8

e = 0.2/0.8

e = 0.25 m

Therefore, the number that will complete gap 1is 0.25 m.

5 0
3 years ago
An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

3 0
3 years ago
you throw an object straight up at 15m/s. how fast is it going when it gets back to the height from which you threw it?
laiz [17]
At the same speed because it will slow down as it approaches the peak then speed up as it goes down again


it will be going 15m/s when it gets to the same height if we neglect air resistance and the object doesn't hit something
8 0
3 years ago
I believe Newton's 1st law is
Bogdan [553]

Answer:

The first law states that if the net force is zero, then the velocity of the object is constant.

5 0
3 years ago
A jetliner has a velocity of 95 m/s. What is the displacement of the jetliner at t=3.0 seconds?
charle [14.2K]
Distance = speed / time

speed = 95 m/s
time = 3 s

distance = 95 / 3 m

displacement = 95/3 m or 32 m (2 s.f.)
5 0
3 years ago
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