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3 years ago

when in object moves stop moving changes speed or changes direction how do scientist describe that condition?

Physics

2 answers:

earnstyle [38]3 years ago

8 0

i think the right answer is unbalanced

Kruka [31]3 years ago

6 0

im taking test rn nd this question was on there, im saying unbalanced force, if it not correct i will put right answer, but im pretty sure the answer is unblanced

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Pluto has a shape that is nearly round, and it orbits the Sun. It has five known moons. Why is it called a dwarf planet and not

Vikentia [17]

Pluto has a shape that is nearly round, and it orbits the Sun. It has five known moons. It is called a dwarf planet and not a planet because its orbit is not cleared of like-sized or larger objects

Answer: Option B

Explanation:

According to the International Astronomical Union (IAU), Pluto has been mentioned as Dwarf Planet as it doesn't hold the standardisation to be counted as a Planet.

According to the standard criteria, a planet must have been surrounded by the orbit of the Sun, it must have significant mass to have a gravitational pull and hence, circular shape, and should have isolated surrounding.

But, Pluto has various celestial bodies in its vicinity that are larger in size. Hence, as concluded by the Union, any celestial body that does not has clear surrounding, is regarded as Dwarf Planet just like Pluto.

6 0

3 years ago

Note: The rope is 20 m long. Answer like this: (1. 2._____ etc)

qaws [65]

Answer:

1. Potential energy, 2. Potential and kinetic energy, 3. Potential and kinetic energy, 4. Potential and kinetic energy, 5. Potential energy

Explanation:

We note that the total mechanical energy (M.E.) of the body is given as follows;

M.E. = K.E. + P.E. = Constant

Where;

K.E. = The kinetic energy of the body = (1/2)·m·v²

P.E. = The potential energy of the body = m·g·h

m = The mass of the person

v = The velocity with which the person is in motion

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

h = The height of the person above the ground

The length of the rope = 20 m

The initial height at location 1, h₁ = 40.0 m

At location 1, the velocity, v₁ = 0.00 m/s

The mechanical energy, M.E. = K.E.₁ + P.E.₁

∴ K.E.₁ = 0 and P.E.₁ = m ×9.81×40

M.E. = (1/2) ×m ×0² + m ×9.81×40

∴ M.E. = 0 + P.E.₁ the type of energy present at location 1 is only potential energy

At location 2, the velocity, v₂ = 10.0 m/s

The mechanical energy, M.E. = K.E.₂ + P.E.₂ = (1/2) ×m ×10² + m ×9.81×40

∴ K.E.₂ = 50·m and P.E.₂ = m ×9.81×35 = 343.35·m

M.E. = 50·m + 343.35·m the type of energy at location 2 is both kinetic energy, K.E. and potential energy, P.E.

At location 3, the velocity, v₃ = 20.0 m/s

The mechanical energy, M.E. = K.E.₃ + P.E.₃ = (1/2) ×m ×20² + m ×9.81×20

∴ K.E.₃ = 200·m and P.E.₃ = m ×9.81×20 = 196.2·m

M.E. = 200·m + 196.2·m the type of energy at location 3 is both kinetic energy, K.E. and potential energy, P.E.

At location 4, the velocity, v₄² = 350.0 m²/s², h₄ = 15.0 m

The mechanical energy, M.E. = K.E.₄ + P.E.₄ = (1/2) × m ×350 + m ×9.81×15

∴ K.E.₄ = 175·m and P.E.₄ = m×9.81×15 = 147.15·m

M.E. = 175·m + 147.15·m the type of energy at location 4 is both kinetic energy, K.E. and potential energy, P.E.

At location 5, the velocity, v₅ = 0 m/s, h₅ = 10.0 m

The mechanical energy, M.E. = K.E.₅ + P.E.₅ = (1/2) × m × 0 + m ×9.81×10

∴ K.E.₅ = 0·m and P.E.₅ = m×98.1 = 98.1·m

M.E. = 0·m + 98.1·m the type of energy at location 5 is only potential energy, P.E.

Therefore, we have;

$\left|\begin{array}{ccc}Location&&Type(s) \ of \ Energy \ Presents\\1&&Potential \ Energy\\2&&Potential \ and \ Kinetic \ Energy\\3&&Potential \ and \ Kinetic \ Energy\\4&&Potential \ and \ Kinetic \ Energy\\5&&Potential \ Energy\end{array} \right |$