All categories

2 years ago

How does water flowing over a waterfall involve both kinetic energy and potential energy?

Physics

2 answers:

Vinvika [58]2 years ago

7 0

Answer:

While the water falls v increases and h decreases, so the kinetic energy increases and the gravitational potential energy decreases, and this happens in a way that the total energy is always the same. (If there is no friction)

Explanation:

goldenfox [79]2 years ago

5 0

Answer:yee what he said!

Explanation:

You might be interested in

If I were a geologist working in South America along one of those plate boundaries, what might I see if I were looking at sedime

faust18 [17]

Answer:

you'll see rocks that have been there for years

4 0

2 years ago

Which statement describes the distribution of charge in an atom? (1) A positively charged nucleus is surrounded by one or more n

pantera1 [17]

Answer: option B

Explanation: when a neutral atom loses an electron or gains a positive charge electron, it becomes a positive ion (positively charged) and when an neutral atom gains an electronic charge or losses a positive charge electron, it becomes a negative ion (negatively charged).

4 0

3 years ago

Read 2 more answers

Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i

Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

$L = L_i_n_i + L_3_s$

$L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt$

$L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s$

Moment if inertia is

$I = 2ml^2$

$I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2$

Angular speed

ω = L/I

$= 38 / 5.75\\\\=6.61$

The speed of each ball is

V = ωL

$= 6.61\times0.5\\\\= 3.31m/s$

7 0

3 years ago

A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m

ArbitrLikvidat [17]

Answer:

Explanation:

The sum of force along both components x and y is expressed as;

$\sum Fx = ma_x \ and \ \sum Fy = ma_y$

The magnitude of the net force which is also known as the resultant will be expressed as $R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}$

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

$a_x = \frac{d^2 x }{dt^2}$

$a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}$

Similarly,

$a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2$

$\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N$

$R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N$

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0

3 years ago

The unusually bright centers found in some galaxies are called ______.

Nimfa-mama [501]

<span>Active Galactic Nuclei.</span>

4 0

2 years ago