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tensa zangetsu [6.8K]
3 years ago
14

A rescue plane flying horizontally at 72.6 m / s spots a survivor in the ocean 182 m directly below and releases an emergency ki

t with a parachute. Because of the shape of the parachute, it experiences insignificant horizontal air resistance. If the kit descends with a constant vertical acceleration of 5.82 m / s 2 , how far away from the survivor will it hit the waves?
Physics
1 answer:
Mila [183]3 years ago
3 0

Answer:

547 m

Explanation:

From law of motion

s = ut + ½at²

Where "t" is Time taken to reach Earth

s= distance= 182 m

a= vertical acceleration = 5.82 m / s 2

U= initial velocity in vertical position = 0

182= ½ × 5.82t²

t²=( 2× 182)/ 5.82

= 364/5.82

= 62.54

t= √62.54

t= 7.908s

horizontal distance travelled = speed x time

Horizontal speed= 72.6 m / s

horizontal distance travelled =72.6× 7.908

= 547 m

Hence, the survivor will it hit the waves at 547 m away

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50m

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A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r
katen-ka-za [31]

Answer:

  • k = 167.33 N/m

Explanation:

  • The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
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        where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =  

        4.92*10⁷ m.

        Solving for f, we get the frequency of the radio waves:

        f = 6.1 Hz

  • Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by  a fixed relationship between the spring constant k and the mass m, as follows:

       \omega_{o}^{2} =\frac{k}{m}  (1)

  • Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

       \omega = 2*\pi *f (2)    

  • We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
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3 years ago
Which option correctly describes a model of methanol. URGENT
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The last one.

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You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength
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Answer:

Period of the signal.

Explanation:

So, this question is all about a concept in physics or astronomy which is called or known as Radiation Astronomy and Galactic Nuclei that are active. This concept talks most about Quasars; a powerful radiating object which derives its power from black holes.

When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.

Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.

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3 years ago
E14. A ball rolls off a table with a horizontal velocity of 5 m/s. If
Shkiper50 [21]

a) Vertical velocity: 5.9 m/s

b) Horizontal velocity: 5 m/s

Explanation:

a)

The motion of the ball is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction)

- A uniformly accelerated motion (constant acceleration) along the vertical direction

Here we want to find the vertical component of the ball's velocity. This can be done by using the suvat equation for the vertical motion:

v_y = u_y +gt

where:

v_y is the vertical velocity at time t

u_y=0 is the initial vertical velocity (zero because the ball has been thrown horizontally)

g=10 m/s^2 is the acceleration of gravity (here we take downward as positive direction)

Substituting t = 0.6 s, which is the total time of flight, we find the vertical velocity of the ball just before it hits the ground:

v_y=0+(9.8)(0.6)=5.9 m/s

b)

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However, there are no forces acting in the horizontal direction on the ball (if we neglect the air resistance): this means that the acceleration of the ball in the horizontal direction is zero.

As a consequence, this also means that the horizontal component of the ball's velocity is constant during the motion.

Since the ball was thrown from the table with an initial horizontal velocity of 5 m/s, this means that the horizontal velocity of the ball just before it hits the floor is still

v_x =  5 m/s

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