IF there is no air resistance, then he could drop a feather, a piece
of Kleenex, a school bus, and a battleship. If he dropped them all
at the same time from the same height, they would all hit the ground
at the same time.

8 0

4 years ago

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The atmosphere of Jupiter is essentially made up of hydrogen, H2. For H2, the specific gas constant is 4157 J/(kg K). The accele

Alenkinab [10]

Answer:

h=17357.9m

Explanation:

The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.

To calculate this, you need to use the barometric formula:

$P=P_0e^{-\frac{Mg}{RT}h}$

Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.

Furthermore, the specific gas constant is defined by:

$R_{H_2}=\frac{R}{M}$

Therefore yo can write the barometric formula as:

$P=P_0e^{-\frac{g}{R_{H_2}T}h}$

at the surface of the planet (h =0) the pressure is $P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}$

applying to the previuos equation:

$\frac{P_0}{2} =P_0e^{-\frac{g}{R_{H_2}T}h}$

solving for h:

h=17357.9m

3 0

4 years ago

Light rays in a material with index of refrection 1.29 1.29 can undergo total internal reflection when they strike the interface

deff fn [24]

Answer:

The second material's index of refraction is 1.17.

Explanation:

Given that,

Refractive index of the material, n = 1.29

Critical angle is 65.9 degrees.

We need to find the second material's index of refraction. We know that at critical angle of incidence, angle of refraction is equal to 90 degrees. Using Snell's law as:

$n_1\sin \theta_c=n_2\sin (90)\\\\n_2=n_1\sin \theta_c\\\\n_2=1.29\times \sin (65.9)\\\\n_2=1.17$

So, the second material's index of refraction is 1.17.

7 0

3 years ago

A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat

postnew [5]

Answer:

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

The length of the pole, l = 10.0 m

The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let ' $F_A$ ' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

$F_A$ × 10.0 m - W × 4.0 m = 0

∴ $F_A$ × 10.0 m = W × 4.0 m = 600 N × 4.0 m

$F_A$ × 10.0 m = 600 N × 4.0 m

∴ $F_A$ = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', $F_A$ = 240 N.

8 0

3 years ago