Which phenomenon occurs when light falls on a smooth mirror?

What holds the metal ions together in a lattice?

Anon25 [30]

Metallic bonds! Hope this helps!! :))

3 0

2 years ago

The space between galaxies is continually ________________, thus causing the light from galaxies to become ____________. A. expa

Vinil7 [7]

A. expanding, redder

8 0

3 years ago

Read 2 more answers

HELP ! ILL MARK BRAINLIEST HELP ASAP

irina [24]

Answer:

A

Explanation:

In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.

Hope this helps!

8 0

3 years ago

1. When a particle moves in a circle with constant speed, its acceleration is

omeli [17]

Answer:

Its A.

Explanation:

8 0

3 years ago

A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are rele

ZanzabumX [31]

Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

Given:

Mass of the tennis ball, m = 44.0 g

Mass of the basket ball, M = 594 g

Height of fall, h = 1.08m

Now,

we have

$u^2-u'^2 = 2as$

where, s = distance = h

a = acceleration

u = final speed before the collision

u' = initial speed

since it is free fall case

thus,

a = g = acceleration due to gravity

u' = 0

thus we have

$u^2-0^2 = 2\times9.8\tiimes1.08$

or

$u = \sqrt{21.168}$

or

u = 4.6 m/s

b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed ( $v_f$ ) = v = 4.6 m/s

Then conservation of energy says

$\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2$

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s

u₂ = velocity of the basketball before collision= 4.6 m/s

v₁ = velocity of the tennis ball after collision

v₂ = velocity of the basketball after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

$v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2$

substituting the values in the above equation we get

$v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6$

or

$v = -8.565-3.965$

or

$v = -12.53m/s$

here negative sign depicts the motion of the ball in the upward direction

now the kinetic energy of the tennis ball

$K.E = \frac{1}{2}mv^2$

or

$K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2$

or

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

or

3.45 = 44 × 10⁻³ × 9.8 × h

h = 8.01 m

5 0

3 years ago