I NEED HELPP!!!!!!!!!!!!!!

Higher-pitched sounds have blank

kenny6666 [7]

Answer:

D

Explanation:

4 0

3 years ago

Now assume that the mass of object 1 is 2m, while the mass of object 2 remains m. If the collision is elastic, what are the fina

Artemon [7]

Answer:

(v₁, v₂) = [(v/3), (4v/3)]

Or

(v₁, v₂) = (v, 0)

Explanation:

In elastic collisions, the momentum and kinetic energy is usually conserved.

The momentum before collision = momentum after collision

And

Kinetic energy before collision = Kinetic energy after collision

Momentum of object 1 before collision = (2m)v = 2mv

Momentum of object 2 before collision = (m)(0) = 0

Momentum of object 1 after collision = (2m)(v₁) = 2mv₁

Momentum of object 2 after collision = (m)(v₂) = mv₂

So, we have

2mv = 2mv₁ + mv₂

2v = 2v₁ + v₂

v₂ = 2v - 2v₁ (eqn 1)

Kinetic energy of object 1 before collision = (1/2)(2m)(v²) = mv²

Kinetic energy of object 2 before collision = (1/2)(m)(0²) = 0

Kinetic energy of object 1 after collision = (1/2)(2m)(v₁²) = mv₁²

Kinetic energy of object 2 after collision = (1/2)(m)(v₁²) = (mv₂²/2)

So, we have,

mv² = mv₁² + (mv₂²/2)

v² = v₁² + (v₂²/2)

2v² = 2v₁² + v₂² (eqn 2)

Substitute (v₂ = 2v - 2v₁) from (eqn 1) into (eqn 2)

2v² = 2v₁² + (2v - 2v₁)²

2v² = 2v₁² + 4v² - 8vv₁ + 4v₁²

6v₁² - 8vv₁ + 2v² = 0

6v₁² - 6vv₁ - 2vv₁ + 2v² = 0

6v₁(v₁ - v) - 2v(v₁ - v) = 0

(6v₁ - 2v)(v₁ - v) = 0

6v₁ = 2v or v₁ = v

v₁ = (v/3) or v₁ = v

If v₁ = (v/3)

From (eqn 1)

v₂ = 2v - 2v₁

v₂ = 2v - 2(v/3)

v₂ = 2v - (2v/3)

v₂ = (4v/3)

If v₁ = v,

From eqn 1,

v₂ = 2v - 2v₁

v₂ = 2v - 2v = 0

(v₁, v₂) = [(v/3), (4v/3)]

Or

(v₁, v₂) = (v, 0)

8 0

3 years ago

a skier speeds along a flat patch of snow, and then flies horizontally off the edge at 16.0 m/s. He eventually lands on a straig

geniusboy [140]

Answer:

1.63 s

Explanation:

The skier lands on the sloped section when the direction of its velocity is exactly identical to that of the slope, so at $45.0^{\circ}$ below the horizontal.

This occurs when the magnitude of the vertical velocity is equal to the horizontal velocity (in fact, $\tan \theta =\frac{|v_y|}{v_x}$ , and since $\theta=45.0^{\circ}, tan \theta = 1$ and so $|v_y| = v_x$ .

We already know the horizontal velocity of the skier:

$v_x = 16.0 m/s$

And this is constant during the entire motion.

The vertical velocity instead is given by

$v_y = u_y + gt$

where

$u_y = 0$ is the initial vertical velocity (zero since the skier flies off horizontally)

g = 9.8 m/s^2

t is the time

Here we have chosen the downward direction as positive direction.

Substituting $v_y = 16.0 m/s$ , we find the time:

$t=\frac{v_y}{g}=\frac{16.0}{9.8}=1.63 s$

5 0

3 years ago

A 1.94-m-diameter lead sphere has a mass of 5681 kg. A dust particle rests on the surface. What is the ratio of the gravitationa

scoundrel [369]

To develop this problem we will apply Newton's laws regarding gravitational forces, both in space and on earth. From finding this relationship, leaving the variable of the dust mass open, we will find the relationship of the forces between the two surfaces. Our values are,

$\text{Diameter of the lead sphere} = D=1.940m$