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2 years ago

A) Khi nào một vật được coi là chuyển động ? Cho ví dụ ?

Physics

1 answer:

Lady bird [3.3K]2 years ago

3 0

Answer: ) Khi nào một vật được coi là chuyển động ? Cho ví dụ ?

b) Khi nào một vật coi là đứng yên? Cho ví dụ ?

c) Theo em câu phát biểu “ Khi khoảng cách từ vật đến vật mốc không thay đổi theo thời gian thì vật đứng yên so với vật mốc”. Có luôn đúng không, vì sao?

Explanation:

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If you walk 3 kilometers in 30 minutes , what is the average speed in kilometers per hour?

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Answer:

6 km/h

Explanation:

V avg = ∆x/∆t = 3km / 30 min ×(60min/1h) = 3 km× 2 /h = 6 km/h

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2 years ago

A child is holding a cup of hot tea. His mother, a physicist, warns him not to start running with the tea. According to Newton's

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If he stops running the tea is still going to be moving so it will spill on him.

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3 years ago

Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

vesna_86 [32]

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

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3 years ago

To meet the minimum required instrument flight experience to act as a pilot in command of an aircraft under IFR, you must have l

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I think its a tbh bc it seems to be the best answer out of a b c and d

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3 years ago

An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane travel

Licemer1 [7]

Answer:

$v=4.44\frac{m}{s}$

Explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero) and accelerates at a constant rate, position can be described like this: $x=v_{0}t +\frac{1}{2} at^{2}$ where x is the position, t is the time a is the acceleration and $v_{0}$ is initial velocity. In this way acceleration can be found. $a=\frac{2(x-v_{0}t) }{t^{2} } =\frac{2(1.11m-0)}{1s^{2} } =2.22\frac{m}{s^{2} }$ .

Now we are able to found velocity at any time with the formula: $v=v_{0} +at = 0\frac{m}{s} +(2.22\frac{m}{s^{2}}.2s)=4.44\frac{m}{s}$

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3 years ago