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3 years ago

11

In a 4th period battle, the girls were able to overcome the in Tug-of-War. Ten boys each pulled with a force of 30 N. Six girls

were able to pull the rope toward them with a net force of 60 N. What was the minimum amount of force each of the six girls applied to the rope?

1 answer:

cupoosta [38]3 years ago

4 0

Answer:

10 N

Explanation:

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Mass is the measure of the amount of ___ in an object.

Kamila [148]

Answer:

mass is the measure of the amount of weight pulled by gravity in an object B-)

5 0

3 years ago

How many moles are there in 814.504 g of H2O?

earnstyle [38]

First you need to calculate the molar mass of H2O. To do that, look at the periodic table and add up the AMUs (atomic mass units)

H = 1.01
O = 16.0

1.01 • 2 + 16.0 = 18.02

Next, use stoichiometry to convert grams into moles. To do that, divide 814.504g by the number of grams in one mole of H2O.

814.504 ÷ 18.02 = 45.2 moles

There are 45.2 moles in 814.504 grams of H2O.

6 0

3 years ago

The pressure is changed from 500 kPa to 250 kPa. What would you expect the new volume to be if the initial volume is 200 mL?

Romashka [77]

Answer:

<h3>The answer is 400 mL</h3>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

So we have

$V_2 = \frac{500000 \times 200}{250000} = \frac{100000000}{250000} = \frac{10000}{25} \\$

We have the final answer as

<h3>400 mL</h3>

Hope this helps you

4 0

3 years ago

What is this section of the periodic table called

nasty-shy [4]

Answer:Noble gases

Explanation:

8 0

3 years ago

If 60g of rust (Fe2O₁) are made in the reaction below, how many Liters of oxygen must be used?

Aleksandr [31]

Answer:

18.

Explanation:

1) according to the reaction (M(Fe)=56; M(O₂)=32; V₀=22.4):

4Fe+3O₂⇒2Fe₂O₃; - ν(O₂)=0.75*ν(Fe);

2) ν(Fe)=m(Fe)/M(Fe)=60/56≈1.071 [mol];

3) ν(O₂)=0.75*1.071=0.804 [mol];

4) V(O₂)=V₀*ν(O₂); ⇒ V(O₂)=22.4*0.804=18 [lit].

7 0

2 years ago