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trapecia [35]
2 years ago
10

How would I turn the formula (1/2)mv^2=mgh into a formula to find final velocity?

Physics
2 answers:
kirill115 [55]2 years ago
4 0

Answer:

\sqrt{2gh}

Explanation:

(1/2)mv^2=mgh

mv^2=2mgh

v^2=(2mgh)/m

v^2=2gh

v=\sqrt{2gh}

IrinaK [193]2 years ago
3 0

Answer:

Explanation:

\frac{1}{2}mv^{2} =mgh\\v^{2} =2gh\\v=\sqrt{2gh}

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The conductor is initially electrically neutral, and then a charge q is placed at the center of the hollow space. Suppose the co
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The total charge on the interior of the conductor is zero.

The total charge on the exterior of the conductor is 8q.

<h3>Total charge on the interior</h3>

Due to large number of electrons available for conduction in a conductor, most of the electrons moves to surface leaving zero net charge inside the conductor.

Thus, the total charge on the interior of the conductor is zero.

<h3>Total charge on the exterior</h3>

The total charge on the exterior of the conductor is calculated as follows;

Q = q + 7q = 8q

Thus, the total charge on the exterior of the conductor is 8q.

Learn more about net charge on interior and exterior of conductors here: brainly.com/question/14653264

7 0
2 years ago
A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag
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Answer:

(a)0.0675  J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675  J

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

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3 years ago
Where do you see triangulation used on this structure? Explain how triangulation​
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3 0
3 years ago
What is potential energy?
GenaCL600 [577]

Answer:

energy that is stagnant and cannot be changed

4 0
3 years ago
Read 2 more answers
The kinetic energy of a body of mass 15 kg is 30 joule. What is its momentum?
lys-0071 [83]

This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum.  So here they are:

Kinetic energy = (1/2) · (mass) · (speed²)

Momentum = (mass) · (speed)

So, now ... We know that

==> mass = 15 kg,  and

==> kinetic energy = 30 Joules

Take those pieces of info and pluggum into the formula for kinetic energy:

Kinetic energy = (1/2) · (mass) · (speed²)

30 Joules = (1/2) · (15 kg) · (speed²)

60 Joules = (15 kg) · (speed²)

4 m²/s² = speed²

Speed = 2 m/s

THAT's all you need !  Now you can find momentum:

Momentum = (mass) · (speed)

Momentum = (15 kg) · (2 m/s)

<em>Momentum = 30 kg·m/s</em>

<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum.  When I saw this, I wondered whether that's always true.  So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>

8 0
2 years ago
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