All categories

3 years ago

Your car burns gasoline as you drive up a mountain road. What energy transformation is taking place?

Physics

2 answers:

Alika [10]3 years ago

7 0

Answer:

the chemical energy of the gasoline is converted into kinetic energy of the car.

Explanation:

xeze [42]3 years ago

4 0

Chemical potential energy is transformed in to mechanical kinetic energy

You might be interested in

Water that flows from behind a large dam can cause machines to produce electricity. What change takes place?

lyudmila [28]

The mechanical energy in the falling water is used to spin the generator, and gets transformed into electrical energy. That's the first choice on the list.

8 0

3 years ago

Potential Energy Equation (GPE = mgh)

Shkiper50 [21]

Answer:

Explanation:

weight=25x9.2=230

25x9.2x10=2300

2300/230=10

6 0

2 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago

Please help with the working of this question

Vilka [71]

Answer:

Principle of conservation of linear momentum is state as provided no external force acts on a system of colliding bodies the total linear momentum of the bodies (in a given direction) remains constant

8 0

2 years ago

A girl rolls a basketball across a basketball court. The ball slowly decelerates at a rate of -0.20m/s^2. If the initial velocit

Dafna1 [17]

By equation of motion :

$2as=v^2-u^2\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{0^2-2^2}{2\times( -0.2)}\\\\s=10\ m$

Now, it is given that it stops after 5 seconds of motion and time at that point is 12:00.

So, time when it started is 11 : 59 : 55.

Therefore, distance travelled is 10 m.

Hence, this is the required solution.

7 0

2 years ago