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4 years ago

12

Which of the following sequences correctly displays the energy transformation from the inside of a battery to the igniter inside

a car's engine?
A.
Chemical Energy Mechanical Energy Heat Energy
B.
Chemical Energy Electrical Energy Heat Energy
C.
Heat Energy Electrical Energy Chemical Energy
D.
Electrical Energy Chemical Energy Heat Energy

1 answer:

Nadusha1986 [10]4 years ago

4 0

The CORRECT answer is actually

B.

Chemical Energy Electrical Energy Heat Energy

Explanation:

There are chemicals inside a battery. They produce electrical energy (you can tell this because batteries have positively-charged and negatively-charged ends). Electrical energy is then transformed into heat energy so the igniter will work.

You might be interested in

When two substance mixed together what does not indicate that a chemical reaction has

nevsk [136]

Answer:

When two substances are mixed together, a color change is observed. This observation is not enough evidence that there is a chemical reaction happening because color changes are not indicators of chemical reaction:

4 0

3 years ago

Jason has 13720 J of gravitational potential energy standing at the top of a cliff over the lake. If he jumps off the cliff and

Anna [14]

The conservation of energy and Newton's second law allows us to find the results about Jason's falling motion are;

The energy when reaching the water is K = 13720 J

The average force of the water to stop it is: F = 2744 N

<h3>Energy conservation.</h3><h3> </h3>

The conservation of energy is one of the most important principles of physics, stable that if there is no friction force, mechanical energy is conserved at all points.

Mechanical energy is the sum of kinetic energy plus potential energy.

Let's look for the energy at two points

Starting point. Get higher.

Em₀ = U = 13720 J

Final point. Lower down.

$Em_f$ = K

Friction in the air is negligible, so energy is conserved.

$Em_o= Em_f$

K = 13720J

<h3>Kinematics and Newton's law.</h3><h3> </h3>

They indicate that it stops 5m under the water, if we assume that the water acts with a constant force, we can use kinematics and Newton's second law to find this force.

The kinematics expression to find the acceleration is

v² =v₀² – 2ay

When it stops the speed is zero.

$a = \frac{v_o^2}{2y}$

Newton's second law is:

F = ma

F = m ( $\frac{v_o^2}{2y}$ )

The expression for the kinetic energy is:

K = ½ m v₀²

$v_o^2 = \frac{2K}{m}$

Let's substitute.

F = $m (\frac{2K}{m}) \frac{1}{2y}$

$F= \frac{K}{y}$

Let's calculate.

F= $\frac{13720}{5}$

F = 2744N

In conclusion using conservation of energy and Newton's second law we can find the results about Jason's falling motion are;

The energy when reaching the water is K = 13720 J

The average force of the water to stop it is: F = 2744 N

Learn more about energy here: brainly.com/question/14274074

6 0

2 years ago

A cannonball is fired across a flat field at an angle of 43 degrees with an initial speed 32 m/s and height of 12 m.

eimsori [14]

1) $x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2$

The initial data of the projectile are:

$y_0 = 12 m$ is the initial height

$v_0 = 32 m/s$ is the initial speed of the projectile, so its components along the x- and y- directions are

$v_{0x} = v_0 cos \theta = (32 m/s)(cos 43^{\circ})=23.4 m/s\\v_{0y} = v_0 sin \theta = (32 m/s)(sin 43^{\circ})=21.8 m/s$

The motion of the cannonball along the x-direction is a uniform motion with constant speed, while on the y-direction it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 downward. So, the two equations of motion of the projectile along the two directions are:

$x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2$

2) 4.94 s

To determine how long the cannon ball was in the air, we need to find the time t at which the cannonball hits the ground, so the time t at which y(t)=0:

$0=12+21.8t-4.9 t^2$

Solving the equation with the formula, we have:

$t_{1,2}=\frac{-21.8\pm \sqrt{(21.8)^2-4(-4.9)(12)}}{2(-4.9)}$

which has two solutions:

t = -0.50 s

t = 4.94 s

Discarding the first solution which is a negative time so it has no physical meaning, the correct solution is

t = 4.94 s

3) 115.6 m

To determine how far the cannonball travelled, we need to find the value of the horizontal position x(t) when the ball hits the ground, at t=4.94 s. Substituting this value into the equation of motion along x, we find:

$x=v_{0x}t=(23.4 m/s)(4.94 s)=115.6 m$

4) 2.22 s

The cannonball reaches its maximum height when the vertical velocity becaomes zero.

The vertical velocity at time t is given by

$v_y(t)= v_{0y} -gt$

where

g = 9.8 m/s^2 is the acceleration due to gravity

Substutiting $v_y(t)=0$ and solving for t, we find

$t=\frac{v_{0y}}{g}=\frac{21.8 m/s}{9.8 m/s^2}=2.22 s$

5) 36.2 m

The maximum height reached by the cannon is equal to the vertical postion y(t) when the vertical velocity is zero, so when t=2.22 s. Substituting this value into the equation of the vertical motion, we find:

$y(t)=y_0 + v_{0y}t-\frac{1}{2}gt^2=12+(21.8)(2.22)-(4.9)(2.22)^2=36.2 m$

3 0

4 years ago

450 nm of light falls on a single slit of width 0.30 mm. What is the angular width of the central diffraction peak

Andre45 [30]

Angular width is 3 x 10^-3

Let D be the distance between source and screed d the distance between coherent source then for central diffraction maxima,

where λ is wavelength

Given:

λ = 450 nm = 450×10^−9m

d = 0.3x10^−3m, D = 1m

W = 2 x 450×10^−9/0.3x10^−3*1

To Find:

Angular width

Solution: The width of the central maxima is nothing but the difference between the positions of the first two minima. Hence we will use the expression for the position of minima and accordingly obtain the expression of the width of central maxima and secondary maxima

θ = W/D

θ = 2 x 450×10^−9/0.3x10^−3*1/1 = 3 x 10^-3

Hence, angular width is 3 x 10^-3

Learn more about Angular width here:

brainly.com/question/25292087

#SPJ4

4 0

2 years ago

a shot putter accelerates a 7.3kg shot from rest to 14m/s in 1.5r seconds. what average power was developed?

Sedaia [141]

Explanation:

$power = \frac{energy \: expended}{time}$

m = 7.3kg

u = 0

v = 14m/s

t = 1.5sec

P = (0.5×7.3×14²) ÷ 1.5

P = 476.93

P = 477 watt

3 0

4 years ago