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3 years ago

A stone is thrown straight up in the air. If g represents the acceleration due to

Physics

1 answer:

maw [93]3 years ago

7 0

Answer:

The correct false statement is;

As the stone rises, its potential energy decreases

Explanation:

When the stone is thrown straight up into the air, the initial velocity, u, progressively decreases according to the following equation;

v = u - g·t

Where;

v = The final velocity

g = The acceleration due to gravity = 9.81 m/s²

t = The time taken

Therefore whereby, the kinetic energy. K.E. is given by the equation , K.E. = 1/2 × m × u², as u reduces, the kinetic energy reduces

However, the potential energy, P.E. is given as P.E. = m × g × h

Where;

h = Height

m = The mass of the stone

Therefore, as the stone rises, the potential energy increases.

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An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

galina1969 [7]

Answer:

it moves 25 inches.

Explanation:

the east west bit isn't important, ignore it. if an ant starts at 6 then moves to 19 then we need to subtract 19 from 6, that's 13. then it moves to 7. the difference between 19 and 7 is 12. add that to 13 and you get 25. it's important to remember that there is no such thing as negative distance. if it moved, then it counts.

3 0

3 years ago

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

3 years ago

Read 2 more answers

The dot or scalar product of two (3d) vec- tors ⃗a = ⟨a1,a2,a3⟩ and ⃗b = ⟨b1,b2,b3⟩ is defined as

Neko [114]

Yes, yes, we know all of that. It certainly took you long enough to
get around to asking your question.

If
   a = (14, 10.5, 0)
and
   b = (4.62, 9.45, 0) ,

then, to begin with, neither vector has a z-component, and they
both lie in the x-y plane.

Their dot-product a · b = (14 x 4.62) + (10.5 x 9.45) =

         (64.68)   +   (99.225) = 163.905 (scalar)

I feel I earned your generous 5 points just reading your treatise and
finding your question (in the last line). I shall cherish every one of them.

7 0

3 years ago

What do scientist do to receive suggestions an criticism of their research from other scientists?

barxatty [35]

They PUBLISH !

They provide complete written, detailed, technical descriptions of their thoughts, hypotheses, experiments, data, and conclusions, in publications that are read by other scientists around the world.

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3 years ago

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Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector

garri49 [273]

Answer:

$5.2\ \text{m/s}$