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3 years ago

PLEASE HELP ME!!!!!!!!

Physics

2 answers:

Ray Of Light [21]3 years ago

5 0

Answer:

I believe C

Explanation:

i never learned this sorry

DerKrebs [107]3 years ago

3 0

The answer is C) if I’m correct

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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

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4 years ago

If a ball with an original velocity of zero is dropped from a tall structure and it takes 7 seconds to hit the ground, what velo

Colt1911 [192]

The velocity of the ball when it reaches the ground is equal to B. 68.6 m/s. This value was obtained from the formula Vf = Vi + at. Vf is the final velocity. Vi is the initial velocity. The acceleration is "a", while the time of travel is "t". The solution is:

<span>Vf = Vi + at
</span>Vf = 0 + (-9.8 m/s^2) (7 s)
Vf = -68.6 m/s

The negative sign denotes the direction of the ball.

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4 years ago

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What is the displacement for the couple's hiking trip? Give your answer as an ordered pair, with magnitude first, followed by a

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Answer:

Explanation:

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3 years ago

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A cord of mass 0.65 kg is stretched between two supports 8.0 m apart. If the tension in the cord is 120 N, how long will it take

DedPeter [7]

The time taken by the pulse to travel from one support to the other is 0.208 s.

<h3>Given:</h3>

The mass of the cord is m = 0.65 kg.

The distance between the supports is, d = 8.0 m.

The tension in the cord is T = 120 N.

The time taken by the pulse to travel from one support to the other is given as,

$v=\frac{d}{t}$

$t=\frac{d}{v}$

Here, v is the linear velocity of a pulse. Its value is,

$v=\sqrt{\frac{T d }{m} }$

$v=\sqrt{\frac{120 * 8}{0.65} }$

$v= 38.43 m/s$

Then,

$t=\frac{8}{38.43}$

$t=0.208 s$

Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.

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2 years ago

A 5 kg ball takes 13.3 seconds for one revolution around the circle. What's the magnitude of the angular velocity of this motion

Tcecarenko [31]

Answer: 0.47 rad/sec

Explanation:

By definition, the angular velocity is the rate of change of the angle traveled with time, so we can state the following:

ω = ∆θ/ ∆t

Now, we are told that in 13.3 sec, the ball completes one revolution around the circle, which means that, by definition of angle, it has rotated 2 π rad (an arc of 2πr over the radius r), so we can find ω as follows:

ω = 2 π / 13.3 rad/sec = 0.47 rad/sec

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3 years ago

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