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coldgirl [10]
3 years ago
12

A 62 kg skydiver moving at terminal speed falls 50 m in 1 s. What power is the skydiver expending on the air?

Physics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

P = 30380 W

Explanation:

given,

mass of the skydiver, m = 62 Kg

distance, s = 50 m

time, t = 1 s

we know,

Power= \dfrac{work\ done}{time}

Power= \dfrac{F.s}{t}

F = m g

Power= \dfrac{m g.s}{t}

inserting all the values

P = \dfrac{62\times 9.8\times 50}{1}

P = 30380 W

hence, Power the skydiver expending on the air is 30380 W

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A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.
pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0
3 years ago
The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are
Kisachek [45]

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

  • Sin(34°) = <em>y</em> / 10.8

Then we <u>solve for </u><u><em>y</em></u>:

  • 0.559 = <em>y</em> / 10.8
  • <em>y </em>= 6.0

And for the horizontal component, we use the formula:

  • Cos(34°) = <em>x</em> / 10.8

Then we <u>solve for </u><u><em>x</em></u><u>:</u>

  • 0.829 = <em>x</em> / 10.8
  • <em>y </em>= 9.0

So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".

3 0
3 years ago
A loudspeaker in a parked car is producing sound whose frequency is 20 510 Hz. A healthy young person with normal hearing is sta
Vsevolod [243]

Answer:

The car must be moving away from the person.

Explanation:

From Doppler's Effect, we know that when a sound source moves towards a stationary observer, the apparent frequency of that sound increases. While the apparent frequency decreases if the source moves away from the stationary observer.

The audible range of frequencies for a human ear is 20 Hz to 20000 Hz. Therefore, in order for the sound of a loud speaker to be audible for the person, the frequency must decrease below 20000 Hz.

<u>Due to this reason, the car must be moving away from the person.</u>

4 0
3 years ago
Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl
Vera_Pavlovna [14]

Answer:

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of  acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = (\frac{(m₂-m₁)}({m₁+m₂} * g = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

6 0
3 years ago
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov
son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

Where f_e is the eyepiece focal length of the microscope

  Now  making f_e the subject  of the formula

         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

         f_e  = 0.027 \ m

     

5 0
3 years ago
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