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melisa1 [442]
3 years ago
9

a boy pulls a wagon with a force of 6 N east as another boy pushes it with a force of 4 N east. what is the net force?

Chemistry
2 answers:
Nesterboy [21]3 years ago
7 0
Sum of the forces in the x direction = 6N - 4N = 2 Newtons East


Anton [14]3 years ago
5 0

Answer: 6 plus 4 = 10 direction is north and is unbalenced

Explanation:

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Rewrite this measurement with a simpler unit, if possible. <br> 8.3 kg•m^2/ s^2
Feliz [49]

Answer: 8.3 J

Explanation:

We have the following measurement:

8.3 \frac{kg m^{2}}{s^{2}}

Rearranging the units:

8.3 \frac{kg m}{s^{2}}m

Since 1 Newton is 1 N=1 \frac{kg m}{s^{2}}:

8.3 Nm

Since 1 Joule is 1 J=1 Nm:

8.3 J This is the simplest form possible

7 0
3 years ago
Unstable traffic flow occurs when drivers accelerate and brake too often. Unstable traffic flow wastes energy. The table below g
maksim [4K]
The answer is B because of the calculator
3 0
3 years ago
Please help 8th grade science i need #2 please!
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How can one kg of iron melt more ice than 1 kg lead at 100 °C
Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{iron} = Heat obtained from the iron by the ice

ΔQ_{iron} = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{lead} = Heat obtained from the iron by the ice

ΔQ_{lead} = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of  ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

3 0
3 years ago
Calculate the concentration of the following solution in mol/dm3 0.1 moles of NaCl in 200 cm3
taurus [48]
1 cm ^{3} = 0.001 dm ^{3}. Therefore 200 cm ^{3} = 0.2 dm ^{3}. Molarity = \frac{number of moles of NaCl}{volume of the solution} =  \frac{0.1}{0.2} = 0.5 mol/dm^{3}. Hope this helps.
8 0
3 years ago
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