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3 years ago

In the formation of the solar system, nearly all of the mass of the solar nebula became

Physics

2 answers:

nata0808 [166]3 years ago

7 0

C) the sun
all the mass of the solar nebula became the Sun

maks197457 [2]3 years ago

4 0

Answer:

C. The sun

Explanation:

The solar nebula was a stellar nursery from which sun along with the planets and other bodies were born. The molecular cloud of dust and gases condensed under its own weight and collapsed to form the core. The condensed matter starts spinning and gathering more mass. The temperature and pressure rose until the beginning of the nuclear fusion reaction in the core. The sun was born. Majority of the mass of the solar nebula is the Sun. Rest of the matter accreted into planets and other smaller bodies. The Sun holds 99.8% of the matter of the solar system.

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A 2137 kg car moving east at 12.91 m/s collides with a 3264 kg car moving north. The cars stick together and move as a unit afte

IRINA_888 [86]

Answer:

The speed of the 3264 kg car before collision is 6.32 m/s

Explanation:

Let first car, mass m₁, moving east be moving with velocity, v₁ and second car, mass m₂, moving north be moving with velocity, v₂. Let the velocity of the two cars after collision be v₃

In a Collison (whether elastic or inelastic), momentum is always conserved.

Momentum before collision = momentum after collision.

Momentum before collision = m₁v₁ + m₂v₂

v₁ = (12.91î) m/s in vector form, m₁ = 2137 kg

v₂ = ?, v₂ = (v₂j) m/s in vector form, m₂ = 3264 kg

Momentum before collision = (2137)(12.91î) + (3264)(v₂j) = (27590î + 3264v₂j) kgm/s

Momentum after collision = (total mass of the cars after collision) × (velocity of the stuck-together cars after collision)

Total mass of the cars after collision = m₁ + m₂ = 2137 + 3264 = 5401 kg

Velocity after collision, v₃ = 6.38 m/s In the N36.8°E direction, put in vector form,

v₃ = (6.38 cos 36.8°î + 6.38 sin 36.8°j)

v₃ = (5.11î + 3.82j) m/s

Momentum after collision = 5401 (5.11î + 3.82j) = (27590î + 20641.4j) kgm/s

Momentum before collision = momentum after collision.

(27590î + 3264v₂j) = (27590î + 20641.4j)

Comparing components

3264v₂ = 20641.4j

v₂ = 6.32 m/s

Hope this Helps!!!

6 0

3 years ago

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

Here, the distance covered is $d$ , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

$v_y=u_y + at$

where:

$u_y=0$ is the initial vertical velocity

$a=g$ is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for $v_x,v_y$ , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

4 0

3 years ago

A ball is falling after rolling off a tall roof<br> The ball has what type of energy.

CaHeK987 [17]

Answer:

Linear and rotational Kinetic Energy + Gravitational potential energy

Explanation:

The ball rolls off a tall roof and starts falling.

Let us first consider the potential energy or more specifically gravitational potential energy ( $mgh$ ; $m$ = mass of the ball, $g$ = acceleration due to gravity, $h$ = height of the roof). This energy comes because someone or something had to do work to take the ball to the top of the roof against the force of gravity. The potential energy is naturally maximum at the top and minimum when the ball finally reaches the ground.

Now, the ball starts to roll and falls off the roof. It shall continue rotating because of inertia (Newton's first law). This contributes to the rotational kinetic energy ( $\frac{1}{2}I\omega^2$ ; $I$ =moment of inertia of the ball & $\omega$ = angular velocity).