Ask question

Ask question

All categories

2 years ago

9

What else is produced when sodium carbonate decomposes?

2 answers:

kompoz [17]2 years ago

3 0

Answer: What else is produced when sodium carbonate decomposes?

Explanation:

frosja888 [35]2 years ago

3 0

It seems like the answer is "CO2".

You might be interested in

H. The length of the shadow is different in evening and in the day. Justify

Vinvika [58]

the shadows are exactly the same length in the morning as they are in the evening.

is so obvious it’s that when the sun is low you get long shadows and when the sun is up in the sky like in the noon the shadow is shorter.

3 0

2 years ago

Read 2 more answers

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago

A 50 kg student climbs 3m to the top of a set of stairs. Calculate the change in the student’s gravitational potential energy fr

erastova [34]

Gpe = mgh
gpe = 50*10*3
gpe = 1500 J

7 0

3 years ago

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Brums [2.3K]

The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

Initial velocity, U = 22 m/s

Deceleration, d = 1.4 $m/s^2$

Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

<em>Final velocity, V = 18.17 m/s</em>

Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

Read more: brainly.com/question/8898885

8 0

2 years ago

SOMEONE HELP ME PLSS

lara31 [8.8K]

1. 2500/60 joules/sec
2. 2,500Nm

7 0

3 years ago

Read 2 more answers