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Paraphin [41]
3 years ago
14

La velocidad de la luz en el vacío es c= 3000.000 km\s la luz del sol tarda en llegar a la tierra 8 minutos y 14 segundos

Physics
1 answer:
ehidna [41]3 years ago
3 0

La velocidad correcta de la luz en el vacío es  300.000 km/s .

La distancia = (velocidad) x (duración de tiempo)

Duración de tiempo = 494 segundos, porque cada minuto = 60 segundos

La distancia = (300.000 km/s) x (494 s)

<em>La distancia = 148.200.000 km</em>

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Exposure to the Sun's harmful infrared radiation should be kept to a minimum.
zysi [14]

Of course! If it's harmful, then your exposure to it should be kept
to a minimum.  That's a no-brainer.  But the sun's infrared radiation
is generally less harmful than its ultraviolet radiation is.

7 0
3 years ago
Suppose we repeat the experiment from the video, but this time we use a rocket three times as massive as the one in the video, a
shusha [124]

Answer:

2/3

Explanation:

In the case shown above, the result 2/3 is directly related to the fact that the speed of the rocket is proportional to the ratio between the mass of the fluid and the mass of the rocket.

In the case shown in the question above, the momentum will happen due to the influence of the fluid that is in the rocket, which is proportional to the mass and speed of the same rocket. If we consider the constant speed, this will result in an increase in the momentum of the fluid. Based on this and considering that rocket and fluid has momentum in opposite directions we can make the following calculation:

Rocket speed = rocket momentum / rocket mass.

As we saw in the question above, the mass of the rocket is three times greater than that of the rocket in the video. For this reason, we can conclude that the calculation should be done with the rocket in its initial state and another calculation with its final state:

Initial state: Speed ​​= rocket momentum / rocket mass.

Final state: Speed ​​= 2 rocket momentum / 3 rocket mass. -------------> 2/3

8 0
2 years ago
A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for
Monica [59]

Question

A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g

Part (b) what is the angle of banking of the highway? Give your answer in degrees

Answer:

a. Equation of Tangent

tan(θ) = v²/rg

b. Angle of the banking highway

θ = 0.087°

Explanation:

Given

Radius of the curve = r = 330m

Acceleration of gravity = g = 9.8m/s²

Velocity = v = 8km/h = 8 * 1000/3600

v = 2.22 m/s

a . Write an equation for the tangent of the highway's angle of banking

The Angle is calculated by

tan(θ) = v²/rg

θ = tan-1(v²/rg)

b.

Part (b) what is the angle of banking of the highway? Give your answer in degrees

θ = tan-1(v²/rg)

Substituting the values of v,g and r

θ = tan-1(2.22²/(330 * 9.8)

θ = tan-1(0.001523933209647)

θ = 0.087314873580116°

θ = 0.087°

3 0
3 years ago
Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction
Norma-Jean [14]

Answer:

The ball would hit the floor approximately 0.55\; \rm s after leaving the table.

The ball would travel approximately 5.5\; \rm m horizontally after leaving the table.

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Let \Delta h denote the change to the height of the ball. Let t denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let v_0(\text{vertical}) denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible:\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t.

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}.

The height of the table was 1.5\; \rm m. Therefore, after hitting the floor, the ball would be 1.5\; \rm m \! below where it was before leaving the table. Hence, \Delta h = -1.5\;\rm m.

The equation becomes:

\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}.

Solve for t:

\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55.

In other words, it would take approximately 0.55\; \rm s for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at v(\text{horizontal}) =10\; \rm m \cdot s^{-1}) until the ball hits the floor.

The ball was in the air for approximately t = 0.55\; \rm s and would have travelled approximately v(\text{horizontal})\cdot t \approx 5.5\;\rm m horizontally during the flight.

4 0
2 years ago
Two planets have the same surface gravity, but planet b has twice the radius of planet
vampirchik [111]
Planet A;
m = the mass 
Let r =  the radius 

Planet B:
Let M =  the mass
The radius is 2r (twice the radius of planet A)

The surface gravitational acceleration of planets A and B (they have the same surface gravity) are
g= \frac{Gm}{r^{2}} \, and \, g= \frac{GM}{(2r)^{2}} \\\\ m= \frac{M}{4} \\\\ M=4m

Answer: The mass of planet B is 4m.
3 0
3 years ago
Read 2 more answers
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