Ask question

Ask question

All categories

3 years ago

9

At archery practice, Vladimir tests a bow and arrow he has recently designed. He is able to accelerate a 24.0-g arrow at an aver

age rate of 70.0 m/s^2. What is the bows average force on the arrow

1 answer:

Whitepunk [10]3 years ago

7 0

Answer: A

1.68 N

Explanation:

F = ma = 0.024(70.0) = 1.68 N

You might be interested in

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

3 0

2 years ago

A plant fertilizer contains 15% by mass nitrogen (N). In a container of soluble plant food, there are 14.1 oz of fertilizer. How

masya89 [10]

We are given with the total mass of fertilizer which is 14.1 oz. This is equivalent to 399.73 grams. 15% of the total is the amount of nitrogen. Thus, the nitrogen amount is 59.96 grams.

5 0

3 years ago

1. What would happen to the position of equilibrium when the following changes are made to the reaction below? 2Hg3O (g) ↔ 6Hg (

Mademuasel [1]

What is your answer I can help you)

4 0

2 years ago

Read 2 more answers

A sample of water at 100°C is converted to steam after absorbing 820 kJ of heat. How grams of H2O are contained in the sample? A

Varvara68 [4.7K]

Since water is already at 100<span>°C all the energy is used to evaporate it.
Now we can calculate how many </span>mols of water are evaporated with 820kJ.
$N= \frac{820}{41} =20 mol$
We calculated that we got 20 mols of water evaporated. Now, all we have to do is find how many grams is a mol of water. Molar mass of water is <span>20.16 g/mol.
</span>The final answer is:
$m=20*20.16=403.2g$

7 0

3 years ago

Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio

Mariulka [41]

Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

50.48 % O = 50.48 grams

49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

7 0

3 years ago