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kirill [66]
2 years ago
6

Una flecha tiene una rapidez de lanzamiento inicial de 18 m/s. Si debe dar en un blanco a 31 m de distancia, que está a la misma

altura, ¿Con qué ángulo debería proyectarse?
Physics
1 answer:
maks197457 [2]2 years ago
4 0

Answer:

34.8 and 55.2º

Explanation:

This is a projectile launching exercise, as we are told that the range of the arrow must be equal to its range and = 31 m let's use the equation

         

The scope equation is

         R = v₀² sin 2θ /g

         sin 2 θ = R g / v₀²

         sin 2 θ = 31 9.8 / 18²

         2 θ = sin⁻¹ 0.93765

          θ = 34.8º

At the launch of projectiles we have two complementary angles with the same range in this case 34.8 and (90-34.8) = 55.2º

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Technician a says that using a pressure transducer and lab scope is a similar process to using a vacuum gauge. technician b says
Phantasy [73]

Answer: Both Technician A and B

Explanation:

There is a similar process in using a pressure transducer and lab scope to using a vacuum gauge.

And also, the pressure transducer can be used to tie any issues to individual cylinders if paired with a second trace consisting of the ignition pattern. Therefore, both Technician A and B are correct.

7 0
2 years ago
A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi
Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

h = ut + \frac{1}{2}\times a \times t^2

h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2

h = 313.6 m

Hence, the depth of the well is 313.6 m

4 0
2 years ago
Sandra's target heart rate zone is 135bpm—172bpm. Marissa's target heart rate zone is 143bpm—176bpm. They stop playing basketbal
Feliz [49]

Answer: Neither Sandra nor Marissa will be in her THR zone.


Explanation:


1) Actual pulse of both Sandra and Marissa : 144 bpm


2) Decrease of 20 bpm ⇒ 144 bpm - 20 bpm = 124 bpm


3) Sandra's TRH is in the range 135 - 172 bpm.


Since 124 < 135, she will be below the range.


4) Marissa's TRH range is 143 - 176 bpm.


Since, 124 < 143, she is below the range


In conlusion, neither Sandra nor Marissa will be in her THR zone.

6 0
2 years ago
Read 2 more answers
4. Going back to the dog whistle in question 1, what is the minimum riding speed needed to be able to hear the whistle? Remember
jeyben [28]

Answer:

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7  m/s

Explanation:

The Doppler shift equation is given as follows;

f' = \dfrac{v - v_o}{v + v_s} \times f

Where:

f' = Required observed frequency = 20.0 kHz

f = Real frequency = 21.0 kHz

v = Sound wave velocity = 330 m/s

v_o = Observer velocity = X m/s

v_s = Source velocity = 0 m/s (Assuming the source is stationary)

Which gives;

20 = \dfrac{330- v_o}{330+0} \times 21

330 - v_o = (20/21)*330

v_o = 330 - (20/21)*330 = 15.7 m/s

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7  m/s.

8 0
2 years ago
A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui
raketka [301]

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

8 0
2 years ago
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