All categories

3 years ago

Which of these has the least chemical energy?

Physics

1 answer:

madreJ [45]3 years ago

4 0

The answer is option D a gallon of gasoline.

Explanation:

A gallon of gasoline has the least chemical energy. The energy content of gallon of gasoline is about 132,000 Btu. The gasoline produced is equivalent in energy terms to 4 kilowatt hours.

Gasoline has chemical potential energy stored in chemical bonds. Gasoline is called as gas or petrol, mixture of volatile, flammable liquid hydrocarbons are used as fuel for internal- combustion engines. It is used as solvent for oils and fats.

Gasoline gallon equivalent is the alternative fuel taken to equal the energy content of one liquid gallon of gasoline.

You might be interested in

what is the magnitude of the electric force between charges of 0.25 C and 0.11 C at a separation of 0.88 m? if the separation be

leonid [27]

$F = k \cdot \frac{q_1 q_2}{d} = 9 \cdot 10^{9} \cdot \frac{0.25 \cdot 0.11}{0.88} = 144 \cdot 5^{9} \ N.$

If the separation between the charges is increased then the magnitude of the force will increase in fact how the distance is being used in that formula.

6 0

2 years ago

Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

igomit [66]

Let's break the question into two parts:

1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.

1. Ramp Scenario:
In an incline, the only component of cart's weight(mg) that is in the direction of motion is $mgsin \alpha$ . Therefore the effort force in this case must be equal or greater than $mgsin \alpha$ .

Now we need to find $\alpha$ . $\alpha$ is the angle between the incline of the ramp and the ground.

Since the height is 5m and the length of the ramp is 8m, $sin \alpha$ would be 5/8 or 0.625. Now that you have $sin \alpha$ , mutiple it with mg.

=> m*g* $sin \alpha$ = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.

2. Lever Scenario:
Just apply "moment action" in this case, which is:
$F_{e} d_{e} = F_{r} d_{r}$

$F_{e}$ = ?

$F_{r}$ = mg = 20 * 10 = 200N
$d_{e}$ = 10m
$d_{r}$ = 1m

Plug-in the values in the above equation:
$F_{e}$ = 200/10= 20N

As 20N << 125N, the best choice is to use lever.

4 0

3 years ago

The greatest wind speed recorded on Earth’s surface is 231 miles per hour. What is the speed in kilometers per hour?

Contact [7]

It would be 371.758hole this helps

4 0

3 years ago

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

3 years ago

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn

elena-s [515]

Answer:

The change in potential energy is $\Delta PE = - 3.8*10^{-16} \ J$

Explanation:

From the question we are told that

The magnitude of the uniform electric field is $E = 950 \ N/C$

The distance traveled by the electron is $x = 2.50 \ m$

Generally the force on this electron is mathematically represented as

$F = qE$

Where F is the force and q is the charge on the electron which is a constant value of $q = 1.60*10^{-19} \ C$

Thus

$F = 950 * 1.60 **10^{-19}$

$F = 1.52 *10^{-16} \ N$

Generally the work energy theorem can be mathematically represented as

$W = \Delta KE$

Where W is the workdone on the electron by the Electric field and $\Delta KE$ is the change in kinetic energy

Also workdone on the electron can also be represented as

$W = F* x *cos( \theta )$

Where $\theta = 0 ^o$ considering that the movement of the electron is along the x-axis

$\Delta KE = F * x cos (0)$

substituting values

$\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)$

$\Delta KE = 3.8*10^{-16} J$

Now From the law of energy conservation

$\Delta PE = - \Delta KE$

Where $\Delta PE$ is the change in potential energy

Thus

$\Delta PE = - 3.8*10^{-16} \ J$

7 0

3 years ago