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3 years ago

How would you expect energy to change her in a chemical reaction?

1 answer:

6 0

The correct answer is option (c) i.e. Energy will be converted into another form.
<span>
During a chemical reaction, energy remains unchanged, It just gets converted into another form and doesn’t increases or decreases. Excess energy is released in the form of heat, light during a chemical reaction and thus, proving the law of conservation of energy which says, energy can neither be created nor destroyed. It just gets converted from one form to another form. </span>

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What is thermodynamics

icang [17]

Answer:

The science of the conversions between heat and other forms of energy.

Explanation:

5 0

3 years ago

Read 2 more answers

A car starts from rest and after 10 seconds is traveling at 20 m/s. Assuming that it continues to accelerate at the same rate it

Alenkinab [10]

Assuming that it continues to accelerate at the same rate it will take another 10 seconds to reach 40 m/s.

Answer:

Explanation:

Since the first question states that there is a change in the velocity from rest to 20 m/s in 10 seconds time interval. So the acceleration experienced by the car during this 10 seconds should be determined first as follows:

Acceleration = (final velocity-initial velocity)/Time

Acceleration = (20-0)/10 = 2 m/s².

So this means the car is traveling with an acceleration of 2 m/s².

As it is stated that the car continues to move with same acceleration, then in the second case, the acceleration is fixed as 2 m/s², initial velocity as 20 m/s and final velocity as 40 m/s. So the time taken for the car to reach this velocity with the constant acceleration value will be as follows:

Time = Change in velocity/Acceleration

Time = (40-20)/2 = 20/2=10 s

So again in another 10 seconds by the car to reach 40 m/s from 20 m/s. Similarly the car will take a total of 20 seconds to reach from rest to 40 m/s value for velocity.

5 0

4 years ago

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

A DC-9 aircraft leaves an airport from a runway whose bearing is N4343degrees°E. After flying for one half 1 2 mile, the pilot

bearhunter [10]

Answer:

103.4° or S76.6°E

Explanation:

The direction N43°E is perpendicular to the direction south-east when the plane turn 90° and heads in the south-east direction.

Since the distance 1/2 mile N43°E is perpendicular to the distance 1 mile south-east, we have a right angled triangle.

So, the angle θ between the aircraft's new position and old position is gotten from tanθ = 1 ÷ 1/2 = 2

θ = tan⁻¹(2) = 63.43°

So, the total angle from North to its new position is 40° + 63.43° = 103.43°

Since we need the south-east bearing, the angle from south is 180° - 103.43° = 76.57° ≅ 76.6°

So, our bearing is 103.4° or S76.6°E

3 0

4 years ago

How are gravity and air resistance alike? How are they different?

Iteru [2.4K]

Air resistance is a force that slows down objects moving through the air. ... Objects are affected by gravity the same, but air resistance can affect the speed of an object's descent. An object with more surface area will have more air resistance.

8 0

3 years ago