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2 years ago

Explain the conservation of energy in terms of the energy transformation(s) that occur when you strike (light) a match.

Physics

1 answer:

Amiraneli [1.4K]2 years ago

8 0

Answer:

The conservation of energy of energy states that energy can neither be created nor destroyed but can be transferred from one form to other. It also states that energy of bodies remain constant in a closed system.

While striking (light) a match, the mechanical energy is utilized is transformed to thermal energy, this thermal energy then releases the chemical energy stored in match. The chemical energy at the ends transform to thermal energy which is visible as as light.

So, in this the energy transformation that occurs remain constant.

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I just don't know how to do the question (a) and (b)

ozzi

Using the equation

$F=ma$

we can observe that you have to apply a non-zero net force to an object in order to make it accelerate. In fact, if the net force is zero you have

$0=ma \iff a=0$

Since we're assuming $n\neq 0$

Now, if the 12N force is applied, the object moves with a constant speed. A constant speed means no acceleration, since by definition the acceleration is a change in speed.

If this sounds counterintuitive to you (why I'm applying a force but I have to acceleration?) think of when we drive a car: even if you want to keep your speed constant, you still have to use the gas pedal, just enough so that the push of the motor balances exactly the road/wheels friction. If you give less gas, the friction becomes stronger, and the car slows down. If you give more gas, the motor push becomes stronger, and the car accelerates.

Back to your exercise: constant speed means to acceleration, so the net force must be zero. This implies that the friction force is exactly 12N.

If the force is increased to 18N, there will be a net force of 6N pushing the object, causing it to accelerate. Using again the same equation of before, and plugging the 3kg mass in the equation, we have

$F=ma \iff 6=3a \iff a=2$

So, the object moves with constant acceleration and initial speed of 10m/s for 0.2 seconds. It's final speed will be

$v = v_0+at = 10+2\cdot 0.2 = 10.4$

5 0

3 years ago

Which of the following statements is FALSE?

kobusy [5.1K]

The false statement would be :
C. constructive interference occurs when the crest of one wave overlaps the trough of another and their individual effects are reduced or cancelled out.

Constructive interference should be when the crest of one wave overlaps another and their individual effects are add up, becoming a single wave of increased amplitude<span />

6 0

3 years ago

when the sun and moon are in a line with the earth, the: group of answer choices arrival of high tide will be delayed. gravitati

Alecsey [184]

When the sun, moon, and Earth are lined up (during a new or full moon), the solar tide adds to the lunar tide to produce extremely high tides and very low tides, both of which are known as spring tides.

Basically describes a situation in astronomy where three celestial bodies align in a straight line as part of a gravitational system. The phrase is frequently used to describe how the Sun, Moon, and Earth are in a straight line.

The moon is responsible for causing high and low tides. The tidal force is produced by the moon's gravitational pull. Earth and its water protrude outward on both the side that is closest to and farthest from the moon as a result of the tidal force. These watery peaks are high tide

To know more about high tides

brainly.com/question/11243732

#SPJ4

8 0

1 year ago

What is a plane mirror ?

g100num [7]

Answer:

A plain mirrior is a mirrior with flat reflective surface.

hope it is helpful for you.

6 0

3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

3 years ago