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Yuki888 [10]
3 years ago
6

How is the kelvin scale different from the Celsius scale?

Physics
1 answer:
k0ka [10]3 years ago
5 0
Difference is between the freezing point of water and the boiling point of water.
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What is the following atmospheric property associated with?
Nadya [2.5K]

Answer:

Your answer should be Cooled Air

Explanation:

6 0
3 years ago
No links or viruses!
hjlf

Which of the following are characteristics of noble gases?

{ \bf{ \underbrace{Answer :}}}

\sf\red{B. \:They're\: inert.} ✅

  • An inert gas is one that does not undergo chemical reactions. The noble gases have complete outer shells, so they have no tendency to lose, gain, or share electrons. This is why they are said to be inert.

\sf\purple{D.\: They \:don't \:react\: with\: other\: elements.}✅

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7 0
3 years ago
You work at a garden store for the summer, and you lift a 14 kg bag of fertilizer with a force of 227 N.
nika2105 [10]

Answer:

(a) Acceleration of the bag will be a=16.214m/sec^2  

(B) Weight of the bag will be 137.2 N

Explanation:

We have given mass of the bag m = 14 kg

Force with which bag is lifted = 227 N

(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration

So 227=14\times a

a=16.214m/sec^2

(b) Acceleration due to gravity g=9.8m/sec^2

We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity

So weight W=mg=14\times 9.8=137.2N

So weight of the bag will be 137.2 N

8 0
3 years ago
What are the units of measurement for length
Sonbull [250]
Inches, feet, yards, meters
8 0
3 years ago
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Consider a snooker ball, with a mass of 0.07kg, rolling along a table at a speed of 4.1m/s. It hits a side cushion of the snooke
zavuch27 [327]

Answer: - 3.4 \times 10^{-3} N

Explanation:

From Newton's law of motion, Force is equal to rate of change of momentum.

The initial velocity of the ball, u = 4.1 m/s

mass of the ball, m = 0.07 kg

During contact, velocity = 0

Time of contact, t = 72 ms = 72 × 10⁻³ s

F = \frac{m(v-u)}{t} = \frac{0.07 kg \times (0-4.1 m/s)}{72 \times 10^{-3}s} = - 3.4 \times10^{-3} N

Thus, net force applied to the ball during this collision is - 3.4 \times 10^{-3} N

4 0
3 years ago
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