1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mamont248 [21]
1 year ago
11

5.

Chemistry
1 answer:
KonstantinChe [14]1 year ago
5 0

Answer:

1 2 4 3

Explanation:

These all have to change shape before new rocks are created

You might be interested in
Which of these choices best describes a pedigree?
mafiozo [28]
A diagram of family relationships that includes several generations. hope this helps!
7 0
2 years ago
Una piedra es lanzada horizontalmente desde la azote de un edificio de 30 metros de altura, con una velocidad de 12 m/s. Hallar
drek231 [11]

Answer: La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

Explanation:

Ok, tenemos 2 sistemas de ecuaciones para este problema.

Primero, el vertical:

La aceleración es la aceleración gravitatoria, entonces:

a = -9.8m/s^2

para la velocidad podemos integrar sobre el tiempo y obtenemos

v = (-9.8m/s^2)*t + v0

donde v0 es la velocidad inicial, pero la velocidad inicial es solo horizontal, entonces v0 = 0.

Para la posición integramos de vuelta:

p = (1/2)*(-9.8m/s^2)*t^2 + p0

donde p0 es la posición inicial, en este caso, 30m

p = (-4.9m/s^2)*t^2 + 30m

la piedra va a llegar al piso cuando la posición vertical sea igual a 0m

p = 0m =  (-4.9m/s^2)*t^2 + 30m

de aca podemos despejar el tiempo que la piedra tarda en llegar al suelo.

t = √(30/4.9) segundos = 2.47 s.

Ahora, las ecuaciones para el movimiento horizontal son:

Aceleración nula, pues no hay ninguna fuerza actuando en la dirección horizontal.

a = 0

para la velocidad, integramos sobre el tiempo:

v = 0*t + v0 = v0

donde v0 es la velocidad inicial, en este caso, es 12m/s

v = 12m/s

para la posición integramos de vuelta:

p = 12m/s*t + p0

en este caso podemos asumir que estamos inicialmente en el punto x = x0, asi que la posición inicial es p0 = x0.

p = 12m/s*t + x0

entonces, si queremos calcular la distancia entre la base del edificio y el punto donde cae la piedra, tenemos que calcular:

D = p(2.47s) - p(0s)

D = 12m/s*2.47s + x0 - (12m/s*0s + x0)

D = 29.64m

La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

3 0
3 years ago
Which of the following are examples of strong electrolytes?
mihalych1998 [28]
B. Strong acids and Strong bases 
Electrolytes are acids bases and salts
5 0
3 years ago
At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound
melisa1 [442]

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>

<em />

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g

4 0
3 years ago
What is a cool 1 week or less science fair project for grade 9?
Leno4ka [110]
Make a fizzing volcano with facts and pictures of a volcano
3 0
3 years ago
Read 2 more answers
Other questions:
  • What are some chemical formulas for compounds?
    14·1 answer
  • the coldest temperature ever recorded in Virginia was 3 degrees below zero how would you represent this temperature as an intege
    12·2 answers
  • 29. Which alcohol combines with carboxylic acid to produce the ester called ethyl butanoate?
    14·1 answer
  • You have a 5-liter container with 2.00 x 1023 molecules of ammonia gas (NH3) at STP
    11·1 answer
  • What is thermodynamics ??? <br>don't explain .-. !!!​
    9·1 answer
  • Please help due tomorrow
    5·1 answer
  • What does binding energy measure
    6·2 answers
  • Choose the formula for hydroxide.<br> OH<br> HPO4<br> Ca(OH)2<br> d Ba(OH)2
    6·2 answers
  • SKJIEDHISJLKDSKSJKA HELPP
    13·1 answer
  • Jdbdjevejeveheveveebdbdhdhdhd
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!