This U-shaped valley was produced by the action of a

If a chlorine atom gains or loses a valence electron, it becomes a charged

k0ka [10]

Answer: D. Chloride atom

6 0

4 years ago

A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block

kotegsom [21]

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The velocity at the bottom is $v = 11.76 \ m/ s$

Explanation:

From the question we are told that

The total distance traveled is $d = 1.2 \ m$

The mass of the block is $m_b = 0.3 \ kg$

The height of the block from the ground is h = 0.60 m

According the law of energy

$PE = KE$

Where PE is the potential energy which is mathematically represented as

$PE = m * g * h$

substituting values

$PE = 3 * 9.8 * 0.60$

$PE = 17.64 \ J$

So

KE is the kinetic energy at the bottom which is mathematically represented as

$KE = \frac{1}{2} * m v^2$

So

$\frac{1}{2} * m* v ^2 = PE$

substituting values

=> $\frac{1}{2} * 3 * v ^2 = 17.64$

=> $v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }$

=> $v = 11.76 \ m/ s$

4 0

3 years ago

A hot-air balloon is rising straight up with a speed of 4.6 m/s. a ballast bag is released from rest relative to the balloon at

kherson [118]

The time taken by the ballast bag to reach the ground is 2.18 s

The ballast bag at rest with respect to the balloon has the upward velocity (u) of 4.6 m/s , which is the velocity of the balloon. When it is dropped from the balloon, its motion is similar to an object thrown upwards with an initial velocity u and it falls under the acceleration due to gravity g.

Taking the upward direction as positive and the downward direction as negative, the following equation of motion may be used.

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

The bag makes a net displacement s of 13.4 m downwards, hence

$s=-13.4 m$

Its initial velocity is

$u=+4.6 m/s$

The acceleration due to gravity acts downwards and hence it is negative.

$g=-9.8 m/s^2$

Use the values in the equation of motion and write an equation for t.

$s=ut+\frac{1}{2} at^2 \\ -13.4=4.6t-\frac{1}{2}(9.8)t^2\\ 4.9t^2-4.6t-13.4=0$

Solving the equation for t and taking only the positive value for t,

t=2.18 s

4 0

3 years ago

A racecar traveling at a velocity of 18.5 m/s, accelerates at a rate of 2.47 m/s2 and covers a distance of 79.78 m. Determine th

Vesna [10]

Answer:

The final velocity of the race car is 27.14 m/s

Explanation:

Given;

initial velocity of the race car, u = 18.5 m/s

acceleration of the race car, a = 2.47 m/s²

distance covered by the race car, s = 79.78 m

Apply the following kinematic equation to determine the final velocity of the race car.

v² = u² + 2as

v² = (18.5)² + 2(2.47)(79.78)

v² = 736.363

v = √736.363

v = 27.14 m/s

Therefore, the final velocity of the racecar is 27.14 m/s

4 0

4 years ago

Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. what is the final velocity of each ba

Allushta [10]

The collision is elastic. This means that both momentum and kinetic energy are conserved after the collision.

- Let's start with conservation of momentum. The initial momentum of the total system is the sum of the momenta of the two balls, but we should put a negative sign in front of the velocity of the second ball, because it travels in the opposite direction of ball 1. So ball 1 has mass m and speed v, while ball 2 has mass m and speed -v:
$p_i = p_1-p_2 = mv-mv =0$
So, the final momentum must be zero as well:
$p_f = 0$
Calling v1 and v2 the velocities of the two balls after the collision, the final momentum can be written as
$p_f = mv_1 + mv_2 = 0$
From which
$v_1 = -v_2$

- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is $\frac{1}{2} mv^2$ . Therefore, the total kinetic energy before the collision is
$K_i = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2$
the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
$K_f = K_i = mv^2$ (1)
But the final kinetic energy, Kf, is also
$K_f = \frac{1}{2} mv_1^2 + \frac{1}{2}mv_2^2$
Substituting $v_1 = -v_2$ as we found in the conservation of momentum, this becomes
$K_f = mv_2 ^2$
we also said that Kf must be equal to the initial kinetic energy (1), therefore we can write
$mv_2^2 = mv^2$

Therefore, the two final speeds of the balls are
$v_2 = v$
$v_1 = -v_2 = -v$

This means that after the collision, the two balls have same velocity v, but they go in the opposite direction with respect to their original direction.

8 0

3 years ago