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natulia [17]
3 years ago
15

A park ranger wants to shoot a monkey hanging from a branch of a tree with a tranquilizing dart. The ranger aims directly at the

monkey, not realizing that the dart will follow a parabolic path and thus will fall below the monkey. The monkey, however, sees the dart leave the gun and lets go of the branch to avoid being hit. Will the monkey be hit anyway? Does the velocity of the dart affect your answer, assuming that it is great enough to travel the horizontal distance to the tree before hitting the ground?
Physics
2 answers:
blondinia [14]3 years ago
8 0
 i belive it will hit the money
Brrunno [24]3 years ago
7 0
The monkey will be hit because the force of gravity accelerates all objects to the center of the earth at equal rates. Because of this, the dart and monkey will fall the same distance in the same amount of time. Since the park ranger aimed directly at the monkey, the dart will still hit the monkey due to the fact that it falls to an equal point as the monkey.


The velocity of the dart doesn't not affect the answer.
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Most earthquakes occur in areas close to where tectonic plates meet. There are earthquakes in San Francisco. What can be conclud
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B. To have an earthquake there must be a fault line (where two or more tectonic plates meet) so if San Fran. has earthquakes they’re on a fault line.
5 0
3 years ago
Read 2 more answers
you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0
3 years ago
Một vật giao động điều hoà với chu kì T = 3,14s và biên độ A=1m tại thời điểm vật qua VTCB có giá trị là
zhenek [66]
W=2pi/T=>w=2
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5 0
2 years ago
While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.
lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

6 0
3 years ago
The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, how many oz of caffeine is left after one half life? * Y
xeze [42]

Answer:

The amount of caffeine left after one half life of 5 hours is 15 oz.

Explanation:

Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.

The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.

So that:

After one half life of 5 hours, the value of caffeine that would be left is;

                                    \frac{30}{2} = 15 oz

The amount of caffeine left after one half life of 5 hours is 15 oz.

8 0
4 years ago
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