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Karolina [17]
2 years ago
5

16. A 7500 kg 18-wheeler traveling at 20 m/s exits onto the runaway truck ramp on the freeway.

Physics
1 answer:
miskamm [114]2 years ago
7 0

Answer:

<em>765,000 Joule</em>

Explanation:

<u>Principle of Conservation of Energy </u>

The total energy in an isolated system cannot be created or destroyed, but transformed. Moving objects have kinetic energy, objects placed in some height above a reference level have gravitational potential energy. When they change their motion variables, one energy converts into the other, but if the numbers don't fit, we know there was some other type of energy acting into the system. The most common reason for energy 'losses' is the thermal energy, produced when objects move in rough surfaces or take friction from the air.

The 7,500 kg truck is originally traveling at 20 m/s to a certain height we'll set to 0. Thus, its total energy is  

\displaystyle E_1=\frac{mv^2}{2}

\displaystyle E_1=\frac{7,500\ 20^2}{2}

E_1=1,500,000\ Joule

When it comes to a stop, its speed is 0 and its height is 10 m higher than before. It means all the kinetic energy was transformed into other types of energy. The gravitational potential energy is

U=mgh=(7,500)(9.8)(10)=735,000\ Joule

Since this number is not equal to the previous value of the energy, the difference is due to thermal energy dissipated by friction

E_t=1,500,000\ Joule-735,000\ Joule=765,000\ Joule

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To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V
aleksley [76]

Answer:

The maximum energy stored in the combination is 0.0466Joules

Explanation:

The question is incomplete. Here is the complete question.

Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Energy stored in a capacitor is expressed as E = 1/2CtV² where

Ct is the total effective capacitance

V is the supply voltage

Since the capacitors are connected in series.

1/Ct = 1/C1+1/C2+1/C3

Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF

1/Ct = 1/11.7 + 1/21.0 + 1/28.8

1/Ct = 0.0855+0.0476+0.0347

1/Ct = 0.1678

Ct = 1/0.1678

Ct = 5.96μF

Ct = 5.96×10^-6F

Since V = 125V

E = 1/2(5.96×10^-6)(125)²

E = 0.0466Joules

8 0
2 years ago
If a force always acts perpendicular to an object's direction of motion, that force cannot change the object's kinetic energy.
laiz [17]
This is very good conceptual question and can clear your doubts regarding work-energy theorem.
Whenever force is perpendicular to the direction of the motion, work done by that force is zero.
According to work-energy theorem,
Work done by all the force = change in kinetic energy.

here, work done = 0.
Therefore, 
0=change in kinetic energy
This means kinetic energy remains constant.
Hope this helps
5 0
2 years ago
Describe the energy transformations that take place when a skier starts skiing down a hill but after a time is brought to rest b
Andrews [41]
<span>The skier will transform their gravitational energy into mostly kinetic energy (with a minor amount transformed into heat from the friction of the skis across the snow and air friction). Once the skier hits the snowdrift, their kinetic energy is transferred into the snow which moves when they strike it due to the kinetic energy that is now in the snow. Along with again a minor amount of heat energy transferred as they move through the snowdrift.</span>
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2 years ago
A man climbs a ladder. Which two quantities can be used to calculate the energy stored of the man at the top of the ladder.
Dvinal [7]

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Explanation: you calculate it by the force you applied times the distance you moved

8 0
2 years ago
Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als
lesantik [10]

Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would  be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

3 0
2 years ago
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