According to the scientific method, how does a physicist formulate and objectively test hypotheses?

Physics

1 answer:

AlladinOne [14]3 years ago

4 0

Answer:

According to the scientific method, physicist formulate and objectively test hypotheses through experiments.

Explanation:

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Newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its su

skad [1K]

Answer:

$g_n=\dfrac{2}{9}g$

Explanation:

M = Mass of Earth

G = Gravitational constant

R = Radius of Earth

The acceleration due to gravity on Earth is

$g=\dfrac{GM}{R^2}$

On new planet

$g_n=\dfrac{G2M}{(3R)^2}\\\Rightarrow g_n=\dfrac{2GM}{9R^2}$

Dividing the two equations we get

$\dfrac{g_n}{g}=\dfrac{\dfrac{2GM}{9R^2}}{\dfrac{GM}{R^2}}\\\Rightarrow \dfrac{g_n}{g}=\dfrac{2}{9}\\\Rightarrow g_n=\dfrac{2}{9}g$

The acceleration due to gravity on the other planet is $g_n=\dfrac{2}{9}g$

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2 years ago

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Help can’t find the answer no where

katen-ka-za [31]

i'm stuck on that question also

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3 years ago

An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$