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2 years ago

Walking converts what type of energy into mechanical energy

2 answers:

8 0

Walking converts "Chemical energy" into "Mechanical energy"

That chemical energy stored in our body as "ATP"

Hope this helps!

densk [106]2 years ago

4 0

The answer is chemical energy.

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A ball is thrown at an angle of 38 degrees to the horizontal. What happens to the magnitude of the ball's vertical acceleration

Brut [27]

Answer:

I belive it is 38 degrees

Explanation:

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2 years ago

What motion makes the acceleration unchanged?

Rashid [163]

Answer:

The acceleration is equal to the net force divided by the mass. If the net force acting on an object doubles, its acceleration is doubled. If the mass is doubled, then acceleration will be halved. If both the net force and the mass are doubled, the acceleration will be unchanged.

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2 years ago

Read 2 more answers

3. Provide two examples of static electric charge.

Rzqust [24]

Answer: 1. walking across a carpet and touching a metal door handle 2. pulling your hat off and having your hair stand on end.

Explanation

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3 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.

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

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2 years ago

What is Secular Music's instrument's?

nasty-shy [4]

Answer:

Drums, harps, recorders, and bagpipes.

Explanation:

8 0

2 years ago