What is 3,278,000 as scientific notation?

En la Tierra un volcán puede expulsar rocas verticalmente hasta una altura máxima H. A) ¿A qué altura (en términos de H) llegarí

Nonamiya [84]

A) 2.64 H

The maximum height that the expelled rock can reach can be found by using the equation:

$v^2-u^2 = 2gd$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g is the acceleration of gravity

d is the maximum height

Solving for d,

$d=\frac{-u^2}{2g}$

We see that the maximum heigth is inversely proportional to g. On the Earth,

$d=H$ and $g=g_e = -9.81 m/s^2$

So we can write:

$\frac{H}{H'}=\frac{g_m}{g_e}$

where H' is the maximum height reached on Mars, and $g_m = -3.71 m/s^2$ is the acceleration of gravity on Mars. Solving for H',

$H' = \frac{g_e}{g_m}H = \frac{-9.81}{3.71}H=2.64 H$

B) 2.64T

The time after which the rock reaches the maximum height can be found by using

$v=u+gt$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

Solving for t,

$t=\frac{v-u}{g}$

The total time of the motion is twice this value, so:

$t=2\frac{v-u}{g}$

So we see that it is inversely proportional to g.

On the Earth, t = T. So we can write:

$\frac{T}{T'}=\frac{g_m}{g_E}$

where T' is the total time of the motion on Mars. Solving for T',

$T' = \frac{g_e}{g_m}T=\frac{-9.81}{-3.71}T=2.64T$

4 0

3 years ago

A light ray moving from oil (n=1.38) into an unknown material. If the light is incident at 35 and refracts at 23, what is the in

Nastasia [14]

The index of refraction of the unknown material in which a ray of light is incident at 35° and refracted at 23° is 2.03

<h3>Snell's law</h3>

index of refraction (n) = Sine i / Sine r

n = Sine i / Sine r

Where

i is the angle of incidence
r is the angle of refraction

<h3>How to determine the refractive index </h3>

From the question given above, the following data were obtained:

Index of refraction of oil (nₒ) = 1.38
Angle of incidence (i) = 35°
Angle of refraction (r) = 23°
Index of refraction of unknown material (nᵣ) =?

nₒSine i = nᵣSine r

1.38 × Sine 35 = nᵣ × Sine 23

Divide both side by Sine 23

nᵣ = (1.38 × Sine 35) / Sine 23

nᵣ = 2.03

Thus, the index of refraction of the unknown material is 2.03

Learn more about Snell's law:

brainly.com/question/25758484

7 0

2 years ago

4Fe + 3O2 + xH2O → 2Fe2O3 • xH2O

Alecsey [184]

The answer is synthesis

6 0

3 years ago

Read 2 more answers

A racing car has a mass of 1720 kg. the acceleration of gravity is 9.8 m/s 2 . what is its kinetic energy if it has a speed of 1

lesantik [10]

Is this to late to answer?

3 0

3 years ago

Saturated water vapor at 200 kPa is condensed into a saturated liquid via a constant-pressure process inside of a piston-cylinde

evablogger [386]

Answer:

The process is not possible

Explanation:

if we want to determine if the process is possible , we can check with the second law of thermodynamics

ΔS≥ ∫dQ/T

for a constant temperature process ( condensation)

ΔS≥ 1/T ∫dQ

and from the first law of thermodynamics

ΔH = Q - ∫VdP , but P=constant → dP=0 → ∫VdP=0

Q=ΔH

then

ΔS≥ΔH/T

from steam tables

at P= constant = 200 Kpa → T= 120°C = 393 K

at P= constant → H vapor = 2201.5 kJ/kg , H liquid = 1.5302 kJ/kg

, S vapor= 7.1269 kJ/kg , S liquid 1.7022 kJ/kg

therefore

ΔH = H vapor - H liquid = 2201.5 kJ/kg - 1.5302 kJ/kg = 2199.9698 kJ/kg

ΔS = S vapor - S liquid = 7.1269 kJ/kg - 1.7022 kJ/kg = 5.4247 kJ/kg

therefore since

ΔS required = ΔH/T = 2199.9698 kJ/kg/(393 K)= 5.597 kJ/kg K

and

ΔS= 5.4247 kJ/kg ≤ ΔS required=5.597 kJ/kg K

the process is not possible

5 0

3 years ago