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antiseptic1488 [7]
3 years ago
5

An injured monkey sits perched on a tree branch 9 m above the ground, while a wildlife veterinarian is kneeling down in the bush

es 90.0 m away attempting to subdue the monkey with a tranquilizer gun. The vet knows that the moment the gun fires, the monkey will be frightened and fall down from the branch. At what angle up from the ground must the veterinarian aim the gun so that the tranquilizer dart will hit the falling monkey? Given the angle, what is minimum speed at which the tranquilizer dart must leave the gun to still hit the monkey?
Physics
1 answer:
Yakvenalex [24]3 years ago
7 0

Answer:

The hunter should aim directly at the perched monkey because the tranquilizer dart will fall away from the line sight at the same rate that the monkey falls from its perch.

Tan theta = 9 / 90 = .1      so theta = 5.71 deg

The time for the monkey to reach the ground is

t = (2 h / g)^1/2 = (18 / 9.8)^1/2 = 1.36 sec

So the horizontal speed of the dart must be at least

Vx = 90 m / 1.36 sec = 66.4 m/s

Vx = V cos theta

V = 66.4 m/s / cos 5.71 = 66.7 m/s

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arlik [135]

We know that:

                             P_{1}V_{1}= P_{2}V_{2}               .......(1)

Given at P_{1}=30cm^3 , V_{1}=10^5Pa

a)

When P_{2}=2*10^5Pa , V_2=?

Using equation (1)

                                    30*10^5=2*10^5*V_2

                                     V_2=30/2=15cm^3

b)

When P_2=5*10^5Pa, V_2=?

using equation (1)

                                     30*10^5=5*10^5*V_2

                                    V_2=30/5=6cm^3

5 0
3 years ago
A ball is dropped from the top of a tall building h=5m about how long does it take for the ball to hit the ground.
pishuonlain [190]
D=-5m
a(gravity)=-9.8m/s^2
vi= 0m/s
t=?
use equation d=vi*t+0.5a*t^2
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rearrange the equation to say t^2=d/0.5a
t^2= -5/-4.9
t^2=1.02
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Fill in the blanks to complete each statement about energy in Earth’s crust. (use lower case wording only)
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Answer:

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8 0
3 years ago
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Which of the following is an example of a chemical change?
bija089 [108]

HELLO THERE

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D. A gas when vinegar and baking soda are mixed

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3 years ago
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A 2.9 × 103 kg car accelerates from rest under the action of two forces. One is a forward force of 1148 N provided by traction b
Lana71 [14]

Answer:

The car must travel 51.34 m for its speed to reach 2.7 m/s

Explanation:

Mass of car = 2.9 x 10³ kg

Forward force = 1148 N

Resistive force = 941 N

Total force = 1148 - 941 = 207 N

We know

            Force = Mass x Acceleration

              207 = 2.9 x 10³ x Acceleration

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Now we have equation of motion, v² = u² + 2as

      Initial velocity, u = 0 m/s

      Final velocity, v = 2.7 m/s

      Acceleration, a = 0.071 m/s²  

Substituting

        v² = u² + 2as

       2.7² = 0² + 2 x 0.071 x s

        s = 51.34 m

The car must travel 51.34 m for its speed to reach 2.7 m/s

5 0
4 years ago
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