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4 years ago

When the moon is directly between the Earth and sun, it is called a _____ eclipse.

Physics

2 answers:

sleet_krkn [62]4 years ago

5 0

C. solar I believe because of the sun is touching the Moons back like hey little buddy

user100 [1]4 years ago

4 0

I think It’s a lunar eclipse
Sorry if I’m wrong

Explanation

When the Moon passes between Sun and Earth, the lunar shadow is seen as a solar eclipse on Earth. When Earth passes
directly between Sun and Moon, its shadow creates a lunar eclipse. Lunar eclipses can only happen when the Moon is opposite the Sun in the sky, a monthly occurrence we know as a full Moon.

Hope it helps!

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The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

8 0

3 years ago

¿Qué es la velocidad?

egoroff_w [7]

Velocidad-es una cantidad vectorial física; Tanto la magnitud como la dirección son necesarias para definirlo.

5 0

3 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

3 years ago

When will the motion of an object be

mafiozo [28]

Answer:

C. The forces are balanced.

Explanation:

Every answer other than C means the object moved.

5 0

3 years ago

Please help :( A ray of laser light strikes a glass surface at an angle θa = 22.5° to the normal. What is the angle θb of the re

pshichka [43]

Answer:

Explanation:

We shall apply law of refraction which is as follows

sin i / sinr = μ , where i is angle of incidence , r is angle of refraction and μ is refractive index

here i = θa = 22.5°

r = θb

μ = 1.77

sin22.5 / sinθb = 1.77

.3826 / sinθb = 1.77

sinθb = .216

θb = 12.5 °.

4 0

4 years ago