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3 years ago

8

one ball rolls along a shelf at a steady rate. a second ball rolls off the shelf and gains speed as it falls in a curved path. w

hich must have an unbalanced force acting on it?

1 answer:

Salsk061 [2.6K]3 years ago

6 0

The ball that rolls at a steady rate

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a total of 1.4 moles of sodium nitrate is dissolved in enough water to make 2.0 liters of an aqueous solution. the gram-formula

kompoz [17]

Answer:

1.4 moles/ 2.0 L= 0.7 M

Explanation: Molarity= moles of solute/ Liters of solution

therefore just plug the numbers in and you'll find the molarity to equal. 0.7

7 0

2 years ago

Which type of bond forms when two or more atoms share electrons?.

shepuryov [24]

Answer:

Covalent

Explanation:

Covalent is the sharing of electrons and Ionic is transferring of electrons.

7 0

2 years ago

Read 2 more answers

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

4 years ago

a compound containing only sulfur and nitrogen is by mass; the molar mass is g/mol. what are the empirical and molecular formula

kolezko [41]

The chemical compound's empirical formula is NS.

The chemical compound's molecular formula is N4S4.

<h3>What does a chemical empirical formula look like?</h3>

The empirical formula of a compound that gives the proportion (ratios) of the elements in the complex but not the precise number or arrangement of atoms is known as an empirical formula.
This would be the compound's element to whole number ratio with the lowest value.

<h3>What sort of empirical formula would that be?</h3>

The chemical structure of glucose is C6H12O6. Every mole of carbon and oxygen is accompanied by two moles of hydrogen.
Glucose has the empirical formula CH2O.
Ribose has the chemical formula C5H10O5, which can be simplified to the empirical formula CH2O.

learn more about empirical formula here

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the question you are looking for is

A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molecular formulas of the compound?

3 0

2 years ago

What happens when an electron in its lowest energy level or ground state when it is by absorbs energy?

tester [92]

Explanation:

atom changes from a ground state to an excited state by taking on energy from its surroundings in a process called absorption. The electron absorbs the energy and jumps to a higher energy level. In the reverse process, emission, the electron returns to the ground state by releasing the extra energy it absorbed

8 0

3 years ago