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Pani-rosa [81]
3 years ago
11

Velocity is a description of an object’s blank

Physics
2 answers:
anzhelika [568]3 years ago
8 0

Velocity is the description of an object's motion from one point to another.

It tells the speed with which the trip is completed, and the direction from the start-point to the end-point.  It tells nothing about any twists, turns, loops, bends, or stops along the way.

Alisiya [41]3 years ago
3 0
Change in displacement per unit of time (Ex. 25m/s East)
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A shopping mall is putting up lights for the holidays. When they plug the holiday lights in they do not light up.
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Answer:

Explanation:

Remark

Anybody who has dealt  with Christmas Tree lights knows what can be done. You must take a bulb you know to be good and try each bulb in turn until the whole string lights up. If two bulbs are dead, give up and go buy another string. Two bulbs create a lot of combinations.

The circuit is  a series circuit.

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1 year ago
Why would two poles of different magnets repel one another?
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A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the directio
LuckyWell [14K]

Answer:

U = √Rg/sin2θ

Explanation:

Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.

Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.

According to the range formula,

R = U²sin2θ/g

Cross multiplying we have;

Rg = U²sin2θ

Dividing both sides by sin2θ, we have;

U² = Rg/sin2θ

Taking the square root of both sides we have;

√U² = √Rg/sin2θ

U = √Rg/sin2θ

Therefore, his average speed if he is to meet the ball just before it hits the ground is √Rg/sin2θ

3 0
3 years ago
Which of Newton's laws of motion describes the motion of an object that has a net<br> force of ON?
algol [13]

Newton's first and second laws of motion both do, but I think the one you're looking for is: <em>The First Law of Motion</em>.  That description is a little more direct.

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3 0
2 years ago
A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

       Speed of mailbag after 3 seconds = 26.4 m/s

       Package is 44.1 meter below helicopter

<u>Explanation:</u>

a)  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

   Initial velocity = 3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

   v = 3+9.8*3 = 32.4 m/s

  Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m

  Distance traveled by package  = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

  Speed of mailbag after 3 seconds = 26.4 m/s

 Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0
3 years ago
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