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yaroslaw [1]
3 years ago
13

The magnitude of a vector can never be less than the magnitude of one of its components. Group of answer choices True False

Physics
1 answer:
pogonyaev3 years ago
6 0

Answer:

True

Explanation:

As the formula for magnitude of a vector v and their components v_1, v_2, v_3, v_4,..., v_n is

v^2 = v_1^2 + v_2^2 + ... +v_n^2

Since v_1^2, v_2^2, v_3^2, v_4^2,..., v_n^2 \geq 0, this means the sum of them, v^2, is always greater or equal to v_1^2, v_2^2, v_3^2, v_4^2,..., v_n^2

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The answer is divergent boundaries.
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Basket ball has gone through some changes and hence got into shape as what we play or watch nowadays. Overall rules have the same fundamental principles as set in 1891 by the founder of this game.

December 21st, in 1891, the introductory basketball game was played in Springfield, Massachusetts. It was first brought into shape by a Canadian-by birth, Dr. James Naismith. The basic idea of the new game was to keep the sports loving students in shape during the winters or in between the outdoor game seasons.

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Initially, basket ball with the 13 rules was played with 2 peach baskets setup as goals, which now a days are the baskets on a high poles although modified but with same basic idea. In the very first game played in Springfield, the set of players were able to score a single point only, in the whole game.

By comparing Naismith’s basketball and basketball of today we see a characteristic of the original game which had no dribbling, instead a player had to pass the ball to a another player of his team ,from his the very spot where he caught it. The second thing which has now changed is the fouls, in the original game if any team made 3 consecutive foul play ,the oponent received a goal in reward. Although this scoring technique doesn't exist in the modern basketball. Rather nowadays, if any team makes five fouls in one quarter, the offending team is in the penalty ,and shoot free throws are awarded against them.

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5 0
3 years ago
A car is traveling at a velocity of 22 m/s when the driver puts on the brakes
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The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

  • Initial velocity, U = 22 m/s
  • Deceleration, d = 1.4 m/s^2
  • Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

V^2 = U^2 + 2dS

Substituting the values into the formula, we have;

V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}

<em>Final velocity, V = 18.17 m/s</em>

Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

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8 0
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Answer:

but where is the question ?

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<em>hope</em><em> it</em><em> </em><em>works</em><em> out</em>

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