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3 years ago

How would the gravitational force between the earth and the moon change if the moon had twice the mass?

1 answer:

7 0

Explanation: The moon's gravity, combined with the waltz of Earth and the moon around their center of mass, forces the oceans into an oval shape, with two simultaneous high tides. ... If the moon were half its mass, then the ocean tides would have been correspondingly smaller and imparted less energy to it.

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Whats the texture of copper wire

bixtya [17]

Copper is a reddish orange, soft metal with a bright metallic luster. It is an excellent conductor of heat and electricity. Copper surfaces when exposed to air gradually turns to a dull, brownish color.

8 0

3 years ago

In fatal crashes, more than __________% of passenger car occupants who were totally ejected from the vehicle were killed.

Ierofanga [76]

Answer:

it would be 83% in a fatal crash.

8 0

3 years ago

A sandbag motionless in outer space is hit by a three-times-as massive sandbag moving at 12 m/s. They stick together. This is an

vekshin1

Answer:

This is an example of inelastic collision and it loses zero(0) % of initial kinetic energy

Explanation:

Let the mass of the motionless sandbag = m₁

Let the mass of the moving sandbag = m₂ = 3m₁

initial velocity of the motionless sandbag, u₁ = 0

initial velocity of the moving sandbag, u₂ = 12 m/s

Let their final velocity, = v

Collision between two particles can either be elastic or inelastic.

Since they stick together after the impact, then the collision is inelastic

Apply the principle of conservation of linear momentum;

Initial kinetic energy = final kinetic energy

¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂v²(m₁ + m₂)

¹/₂m₁(0)² + ¹/₂(3m₁)(12)² = ¹/₂v²(m₁+3m₁)

216m₁ = 2m₁ v²

v² = 108

v = √108

v = 10.392 m/s

Change in kinetic energy = Final kinetic energy - initial kinetic energy

Initial Kinetic energy, KE₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

KE₁ = ¹/₂m₁(0)² + ¹/₂(3m₁)(12)²

KE₁ = 216m₁ J

Final kinetic energy, KE₂ = ¹/₂v²(m₁ + m₂)

KE₂ = ¹/₂(108)(m₁ + 3m₁)

KE₂ = 216m₁ J

ΔKE = KE₂ - KE₁ = 216m₁ J - 216m₁ J = 0%

Therefore, this is an example of inelastic collision and it loses zero(0) % of initial kinetic energy

4 0

3 years ago

Un auto se mueve con velocidad constante de 20 m/s y la mantiene durante 20 s. Después se le aplica una aceleración constante y

pentagon [3]

Answer:

26.67 s

Explanation:

v = Velocidad final = 40 m/s

u = Velocidad inicial = 20 m/s

$t_1$ = Tiempo inicial = 20 s

$t_2$ = Tiempo empleado durante la aceleración

a = Aceleración

s = Desplazamiento del automóvil durante la aceleración = 200 m

De las ecuaciones lineales de movimiento tenemos

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{40^2-20^2}{2\times 200}\\\Rightarrow a=3\ \text{m/s}^2$

$v=u+at_2\\\Rightarrow t_2=\dfrac{v-u}{a}\\\Rightarrow t_2=\dfrac{40-20}{3}\\\Rightarrow t_2=6.67\ \text{s}$

El tiempo necesario para acelerar es 6.67 s

El tiempo total necesario para toda la ruta es $t=t_1+t_2=20+6.67=26.67\ \text{s}$ .

4 0

3 years ago

What is the relationship between particles in soil and flow of water in soil

maxonik [38]

<span>In general the larger the pore space (the higher the porosity) the easier it is</span>

7 0

3 years ago