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larisa [96]
2 years ago
5

You place 10 grams of a salt into water and want it to dissolve. All of the following will cause a salt to dissolve faster excep

t for which one?
Chemistry
1 answer:
klio [65]2 years ago
8 0

boiling, salt water will separate the salt from the water through the physical change of water from a liquid to a gas, leaving you with just salt left over.

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Which are examples of contact forces?
lesantik [10]

Answer:

Contact force is a force that requires a contact between two bodies and is ubiquitous in nature

Explanation:

The force is distributed in two categories i.e

  • Contact force
  • Non-Contact force

Contact Forces  are the forces which is requires a contact between two bodies to occur.

The Types of contact forces are given below:

  • Frictional Force
  • Tension Force
  • Normal Force
  • Air Resistance Force
  • Applied Force
  •  Spring Force

Some common daily examples of the contact force are

  1. Frictional force between the tyres of a moving vehicle and the road.
  2. Air flowing in opposite direction of a moving object.
  3. Pushing a table with hand and the friction between its base and floor.
  4. Stretching a rubber band with hands etc.

Also non-contact forces are Gravitational force, Electrical force and Magnetic force.

8 0
3 years ago
What type of galaxy is pictured?
andrew-mc [135]
A maybe I’m not sure
7 0
2 years ago
100 POINTS!!!
mixas84 [53]

1) Area of one side of each cube

The area of one side of each cube can be calculated by multiplying the length times the width:

A=L\cdot W

where

L = length

W = width

For cube A: L=1 cm, W=1 cm

So the area is: A_A = (1)(1)=1 cm^2

For cube B: L=2 cm, W=2 cm

So the area is: A_B=(2)(2)=4 cm^2

For cube C: L=3 cm, W=3 cm

So the area is: A_C=(3)(3)=9 cm^2

2) Surface area of each cube

The surface area of a cube is given by the sum of the areas of all its faces. Since a cube has 6 identical faces, this means that the total surface area is equal to 6 times the area of one face:

A=6 A_1

where

A_1 is the area of one face of the cube

For cube A, the area of one face is A_A=1 cm^2

So the surface area of cube A is

A'_A = 6A_A=6(1)=6 cm^2

For cube B, the area of one face is A_B=4 cm^2

So the surface area of cube B is

A'_B = 6A_B=6(4)=24 cm^2

For cube C, the area of one face is A_C=9 cm^2

So the surface area of cube C is

A'_C = 6A_C=6(9)=54 cm^2

3) Volume of each cube

The volume of a cube is obtained by multiplying its length, its width and its height:

V=L\cdot W \cdot H

Where:

L = length

W = width

H = height

Moreover for a cube, all the sides have equal length, so L=W=H

So the volume can be rewritten as

V=L^3

For cube A: L = 1 cm

So the volume is V_A=(1 cm)^3 = 1 cm^3

For cube B: L = 2 cm

So the volume is V_B=(2 cm)^3 = 8 cm^3

For cube C: L = 3 cm

So the volume is V_C = (3 cm)^3 = 27 cm^3

4) Ratio surface area/volume for each cube

In this part, we have to calculate the ratio between surface area and volume of each cube:

r=\frac{A}{V}

where

A is the surface area

V is the volume

For cube A, we have:

A_A = 6 cm ^2 (surface area)

V_A=1 cm^3 (volume)

So the ratio for cube A is:

r=\frac{6}{1}=6

For cube B, we have:

A_B = 24 cm ^2 (surface area)

V_B=8 cm^3 (volume)

So the ratio for cube B is:

r=\frac{24}{8}=3

For cube C, we have:

A_C = 54 cm ^2 (surface area)

V_C=27 cm^3 (volume)

So the ratio for cube C is:

r=\frac{54}{27}=2

4 0
2 years ago
A scientist wants to display the numbers of gallons of fuel that were used up
Karolina [17]

Answer:

bar graph

Explanation:

7 0
2 years ago
A una mezcla de 300g, formada con 60% P/P de Hierro y 40% P/P de Arena, se le adicionan 135g de Cobre y 2,77g de Aluminio. ¿Cuál
Svetllana [295]

Answer:

\%P/P_{hierro}=41.1\%\\\\\%P/P_{arena}=24.4\%\\\\\%P/P_{cobre}=30.8\%\\\\\%P/P_{aluminio}=0.6\%

Explanation:

¡Hola!

En este caso, dado que estamos tratando con problem sobre porcentaje peso/peso de hierro, arena, cobre y aluminio, primero debemos calcular la masa inicial de estos dos primeros en la mezcla original de acuerdo con:

m_{hierro}=300g*0.60=180g\\\\m_{arena}=300*0.40=120g

Ahora si podemos calcular la masa de la mezcla final como la suma de las masas de todos los constituyentes de la mezcla:

m_T=180g+120g+135g+2.77g=437.77g

Finalmente, podemos calcular los porcentajes P/P como se muestra a continuación:

\%P/P_{hierro}=\frac{180g}{437.77g} *100\%=41.1\%\\\\\%P/P_{arena}=\frac{120g}{437.77g} *100\%=24.4\%\\\\\%P/P_{cobre}=\frac{135g}{437.77g} *100\%=30.8\%\\\\\%P/P_{aluminio}=\frac{g}{437.77g} *100\%=0.6\%

¡Saludos!

5 0
2 years ago
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