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3 years ago

Why doesn't the principle of mechanical energy conservation hold in situations when frictional forces are present?

1 answer:

7 0

When you are talking about the Principle of mechanical Energy Conservation, it is really only including the kinetic and potential energy in a total system. When frictional forces are present, although the conservation of energy law is still present, it does not work when it comes to the conservation of mechanical energy as there is another type of energy that is factored in. As friction acts on the object, that transition from potential to kinetic as it slide/falls will be completely different as some of that energy is being transformed into thermal energy. Which breaks the conservation of mechanical energy.

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You are on the roof of the physics building, 46.0 above the ground . Your physics professor, who is 1.80 tall, is walking alongs

Fudgin [204]

Answer:

3.6m

Explanation:

if you are at a building that is 46m above the ground, and the professor is 1.80m, the egg must fall:

46m - 1.80m = 44.2m

the egg must fall for 44.2m to land on the head of the professor.

Now, how many time this takes?

we have to use the following free fall equation:

$h=v_{0}t+\frac{1}{2}gt^2$

where $h$ is the height, $v_{0}$ is the initial velocity, in this case $v_{0}=0$ . $g$ is the acceleration of gravity: $g=9.81m/s$ and $t$ is time, thus:

$h=\frac{1}{2}gt^2$

clearing for time:

$2h=gt^2\\\frac{2h}{g}=t^2\\\sqrt{\frac{2h}{g}} =t$

we know that the egg has to fall for 44.2m, so $h=44.2$ , and $g=9.81m/s$ , so we the time is:

$t=\sqrt{\frac{2(44.2m)}{9.8m/s^2} }=\sqrt{\frac{88.4m}{9.81m/s^2} } =\sqrt{9.011s^2}= 3.002s$

Finally, if the professor has a speed of $v=1.2m/s$ , it has to be at a distance:

$d=vt$

and t=3.002s:

$d=(1.2m/s)(3.002s)=3.6m$

so the answer is the professor has to be 3.6m far from the building when you release the egg

7 0

3 years ago

You are pulling a 80 kg box with a rope. The force you exert on the box is 90 N at 30 degrees from the ground. What would be the

Feliz [49]

Answer:

Coefficient of dynamic friction= md= 0.09931

Explanation:

To determine the coefficient of dynamic friction we must first match the friction force that is permendicular to the normal force of the block and opposite to the drag force, to the component of the drag force in this same direction. This component on the X axis of the drag force will be:

F= 90N × cos(30°) = 77.9423N

This component on the X axis of the drag force must be equal to the dynamic friction force that is equal to the coefficient of dynamic friction by the normal force of the block weight:

F= md × m × g= 77.9423N

m= mass of the block

md= coefficient of dynamic friction

g= gravity acceleration

F= md × 80kg× 9.81 (m/s²)= 77.9423(kg×m/s²)

md= (77.9423(kg×m/s²) / 784.8 (kg×m/s²)) = 0.09931

7 0

3 years ago

Sarah, who has a mass of 55 kg, is riding in a car at 20 m/s. She sees a cat crossing the street and slams on the brakes! Her se

lions [1.4K]

Answer:

-2200 N

Explanation:

Here we can use the impulse theorem, which states that the impulse exerted on Sarah (product of force and duration of collision) is equal to Sarah's change in momentum:

$I=\Delta p\\F \Delta t = m \Delta v$

where

F is the average force

$\Delta t$ is the duration of the collision

m is the mass

$\Delta v$ is the change in velocity

In this problem:

m = 55 kg

$\Delta v = 0-20 = -20 m/s$

$\Delta t = 0.5 s$

Solving the formula, we find the force exerted by the seatbelt on Sarah:

$F=\frac{m\Delta v}{\Delta t}=\frac{(55)(-20)}{0.5}=-2200 N$

And the negative sign means the direction is opposite to that of Sarah's initial motion.

6 0

3 years ago

In metals, some electrons from the outer shells of atoms are shared with other atoms, with the result that electrons can freely

Kazeer [188]

<h2>Answer:Option F</h2>

Explanation:

Metals have few number of electrons in their valence shell.

The force with which the nucleus of the metal atom is pulling the electron towards itself is lower on the valence shell because the distance between the nucleus an valence shell is large.

So,there are a lot of free electrons in metals.

When the ends of metals are at different potential,electrons flow from one end to another to balance the potential.This feature of metals allows them to conduct electricity.

Similarly,when the ends of metals are at different temperatures,conduction will take place and electrons are the carriers of conduction.This feature of metals allows them to conduct heat.

So,metals are good conductors of heat and electricity.

4 0

4 years ago

Helppppp pleaseeee 25 points please?????

Sedbober [7]

Answer:

Where is question 12, we need it to answer this question

Explanation:

6 0

3 years ago