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True or False? If friction didn’t exist, an object in motion would stay in motion forever.

nata0808 [166]

Answer:

True

Explanation:

Friction is defined as the force that opposes motion when an object is sliding over a surface.

As a result of friction, all objects moving over a surface eventually come to rest over time.If we were to successfully create a friction-less surface, an object will remain in motion forever because it will encounter no opposition to its motion.

Hence, the resistance to the motion of objects over a surface which causes the objects to come to a halt after moving over the surface for some time is called friction.

5 0

3 years ago

Time to spice things up Woahhhhhhh

miskamm [114]

Joey king shoved me in a closet and forced me to eat frozen peas

7 0

3 years ago

What is the concentration of a solution with a volume of 1.38 mL that contains 17.36

Elden [556K]

Answer:

C = 107.97 mol/L

Explanation:

Given data:

Volume of solution = 1.38 mL (1.38 mL× 1 L /1000 mL = 0.00138 L)

Mass of ammonium sulfite = 17.36 g

Concentration of solution =?

Solution:

We will calculate the number of moles of ammonium sulfite.

Number of moles = mass/molar mass

Number of moles = 17.36 g / 116.15 g/mol

Number of moles = 0.149 mol

Concentration:

C = n/V

C = concentration

n = number of moles of solute

v = volume in L

C = 0.149 mol / 0.00138 L

C = 107.97 mol/L

5 0

3 years ago

Fluoride ion is poisonous in relatively low amounts: 0.2 g of F- per 70 kg of body weight can cause death. Nevertheless, in orde

svetoff [14.1K]

Answer:

$\boxed{\text{(a) 14 L; (b) 711 kg}}$

Explanation:

(a) Litres of water

$\text{Mass of F}^{-} = 8.5 \times 10^{7}\text{ gal} \times \dfrac{\text{0.2 mg F}^{-}}{\text{1 kg}} = \textbf{14 mg F}^{-}\\\\\text{Volume of water} = \text{14 mg F}^{-} \times \dfrac{\text{1 L water}}{\text{1 mg F}^{-}} = \textbf{14 L water}\\\\\text{The person would have to consume $\boxed{\textbf{14 L}}$ of water per day}$

(b) Mass of NaF

$\text{Volume} =8.5 \times 10^{7} \text{gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} = 3.22 \times 10^{8}\text{ L}\\\\\text{Milligrams of F}^{-} = 3.22 \times 10^{8}\text{ L} \times \dfrac{\text{ 1 mg F}^{-}}{\text{1 L}} = 3.22 \times 10^{8}\text{ mg F}^{-}\\\\\text{kilograms of F}^{-} = 3.22 \times 10^{8}\text{ mg F}^{-} \times \dfrac{\text{1 mg F}^{-}}{10^{6}\text{kg F⁻}} = \text{322 kg F}^{-}$

1 kmol of NaF (41.99 kg) contains 19.00 kg of F⁻.

$\text{Kilograms of NaF}= \text{322 kg F}^{-} \times \dfrac{\text{41.99 kg NaF}}{\text{19.00 kg F}^{-}} = \text{711 kg NaF}\\\\\text{It will take } \boxed{\textbf{711 kg}} \text{ of NaF to treat the reservoir}$

7 0

3 years ago

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What do elements in block have in common?

iragen [17]

Answer:

The d-block elements are all metals and most have one or more chemically active d-orbital electrons. Because there is a relatively small difference in the energy of the different d-orbital electrons, the number of electrons participating in chemical bonding can vary.

5 0

3 years ago

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