Write a reaction equation to show HCO3 acting as an acid

The amplitude of a lightly damped oscillator decreases by 4.2% during each cycle. What percentage of the mechanical energy of th

Nimfa-mama [501]

Answer:

Im actually depressed

Explanation:

jk thx for the points thoo

6 0

3 years ago

A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco

oee [108]

S=(0x4)+(0.5x4.81x4x4)
S=0.78.48

The depth is approximately 78 meters.
(My brain hurts now) :P Good Luck!

4 0

3 years ago

Which of the following statements is false? Group of answer choices Light that is high in energy has a high frequency. The atomi

MArishka [77]

Answer:

The last statement is false.

Explanation:

Photons (Electromagnetic radiation) are released when electrons drop from a higher energy lever to a lower energy level. Therefore the opposite insinuated by the last statement is wrong.

4 0

3 years ago

A medieval instrument used to determine the position of the sun:

Stels [109]

Answer: astrolabe

Explanation:

4 0

2 years ago

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago

Write a reaction equation to show HCO3 acting as an acid​

Write a reaction equation to show HCO3 acting as an acid