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2 years ago

What energy transformation takes place when an object is pushed up an inclined plane?

Physics

2 answers:

sweet [91]2 years ago

8 0

Answer:

There are two main phases or positions for any object to have in order to posses some energy, and that is either due to its height,H or due to the motion in which the body performs displacement from point A to B. So, lets say there is a object having a specific mass, m inside any region.
Then if we displace that very object from its initial point to the final point of rest at a certain, during these two points there is kinetic energy(K.E) for the body to have, after which the whole scenario changes when it comes to a rest position. Now, the parameters are much more changed for the object and its because of the height,h and the gravity that is applying on the object placed at a certain height,h. So, we got the very equation for calculating the potential energy(P.E) of that body or object, which is P.E= (m)×(g)×(h).
And at any certain height,h the factor of gravity comes into play which brings more energy,E into the whole system of the body and to acquire a certain possession of energy inside the space.

Fofino [41]2 years ago

5 0

When you are pushing an object up an inclined plane, the object is gaining gravitational potential energy as it is gaining height. The kinetic energy of the object decreases and converts into that potential energy as you go up. When you have stopped, all of the kinetic energy of the object has fully been converted to gravitational potential energy.

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3 years ago

Which of the following statements BEST explains why ball-and-socket joints have the greatest range of motion?

galina1969 [7]

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2 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

2 years ago

23 grams of sodium reacted with 35.5 grams of chlorine. Calculate the mass of the sodium chloride compound formed

xxTIMURxx [149]

According to law of conservation of mass within a reaction,
The mass of the compound formed is (23+35.5) grams means 58.5 grams of sodium chloride[NaCl] will be formed.

3 0

3 years ago

Read 2 more answers

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0

3 years ago