All categories

3 years ago

The density of water is 1.00 g/cm3. What is its density in kg/m3?

1 answer:

3 0

1 g = 1 ÷ 1000 kg
= 0.001 kg

1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³

1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³

The density is 1000 kg/m³.

You might be interested in

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons. Determine the net force a

KengaRu [80]

Explanation:

Draw a free body diagram. There are two forces: weight force pulling down and air resistance pushing up.

Sum the forces in the y direction:

∑F = 50 N − (30 kg) (10 m/s²)

∑F = -250 N

3 0

3 years ago

You decide to walk part of the way around a lake. The lake is a circle with a radius of 2.0 km You start on the shore due south

AveGali [126]

Answer:

2.828 km

Explanation:

radius of lake, r = 2 km

Let O be the centre of the circular lake and you walk from A to A to D.

Displacement is the measure of shortest distance between two points.

According to the diagram , the displacement AD is

$AD^{2}=AO^{2}+OD^{2}$ (By using the Pythagorean theorem)

$AD^{2}=2^{2}+2^{2}$

AD = 2.828 km

thus, the displacement is 2.828 km.

7 0

3 years ago

Use the table below to answer the following questions. Substance Specific Heat (J/g•°C) water 4.179 aluminum 0.900 copper 0.385

lbvjy [14]

1. $-8.78 \cdot 10^5 J$

The energy lost by the water is given by:

$Q=m C_s \Delta T$

where

m = 3.0 kg = 3000 g is the mass of water

Cs = 4.179 J/g•°C is the specific heat

$\Delta T=10.0C-80.0C=-70.0 C$ is the change in temperature

Substituting,

$Q=(3000 g)(4.179 J/gC)(-70.0 C)=-8.78 \cdot 10^5 J$

2. $3.24 \cdot 10^4 J$

The energy added to the aluminium is given by:

$Q=m C_s \Delta T$

where

m = 0.30 kg = 300 g is the mass of aluminium

Cs = 0.900 J/g•°C is the specific heat

$\Delta T=150.0 C-30.0C =120.0 C$ is the change in temperature

Substituting,

$Q=(300 g)(0.900 J/gC)(120.0 C)=3.24 \cdot 10^4 J$

3a. $-5.6^{\circ}C$

The temperature change of the water is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = -232 kJ=-2.32\cdot 10^5 J$ is the heat lost by the water

$m=10.0 kg=10000 g$ is the mass of water

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{-2.32\cdot 10^5 J}{(10000g)(4.179 J/gC)}=5.6^{\circ}C$

3b. $+10.2^{\circ}C$

The temperature change of the copper is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = 1.96 kJ=1960$ is the heat added to the copper

$m= 500 g$ is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{1960 J}{(500g)(0.385 J/gC)}=10.2^{\circ}C$

4. 42.9 g

The mass of the water sample is given by

$m=\frac{Q}{C_S \Delta T}$

where

$Q=4300 J$ is the heat added

$\Delta T=39 C-15 C=24C$ is the temperature change

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$m=\frac{4300 J}{(4.179 J/gC)(24 C)}=42.9 g$

5. 115.5 J

The heat used to heat the copper is given by:

$Q=m C_s \Delta T$

where

m = 5.0 g is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

$\Delta T=80.0 C-20.0C =60.0 C$ is the change in temperature

Substituting,

$Q=(5.0 g)(0.385 J/gC)(60.0 C)=115.5 J$

6. 0.185 J/g•°C

The specific heat of iron is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q = -47 J is the heat released by the iron

m = 10.0 g is the mass of iron

$\Delta T=25.0-50.4 C=-25.4 C$ is the change in temperature

Substituting,

$C_s = \frac{-47 J}{(10.0 g)(-25.4 C)}=0.185 J/gC$

8 0

3 years ago

What can electricty from solar power be used for

Mandarinka [93]

Regular electrical needs such as heat, a/c, appliances, anything that would be used regularly with electricity. Most electrical companies use gas, coal, and other nonrenewable resources to power houses. Solar power is just a renewable way of gathering power from the sun and converting it to electricity.

7 0

3 years ago

Susie (45kg) and Troy (65 kg) are sitting 2 meters apart. They smile at each other. What is the gravitational force between them

xxTIMURxx [149]

The gravitational force between Susie and Troy is d) $4.9 \cdot 10^{-8}N$