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4 years ago

13

Identify the characteristic of gravity that is incorrect,

2 answers:

34kurt4 years ago

7 0

You can that gravity is not real

lapo4ka [179]4 years ago

3 0

Answer:It makes astronauts weightless.

Explanation:

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lutik1710 [3]

The power of the engine is 320 W.

<u>Explanation:</u>

Power may be defined as the rate of doing work (or) work done per unit time. One unit of energy is used to do the one unit of work.

Power = Work done / Time taken

Given, Force = 80 N, height = 5 m , final velocity = 4 m/s

To calculate the power, we must know the time taken.

To find the time, use the distance and speed formula which is given by

Time = Distance / speed

Here distance = 5 m and speed = 4 m/s

Time = 5 / 4 = 1.25 s.

Now, Power = work done / time

= (F * d) / t = (80 * 5) / 1.25

Power = 320 W.

The standard unit of power is watt (W) which is joule per second.

4 0

3 years ago

Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

GalinKa [24]

Wave D has the same wavelength as wave A, but the amplitude is lower. The answer is Wave D.

7 0

3 years ago

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

5 0

3 years ago

Which of the following describes the importance of cyanobacteria to early Earth atmosphere formation?

Dmitrij [34]

I do believe it is a

6 0

3 years ago

A luge run is straight for the first 10 m. If the velocity of the luger when he passes the start line is +25 m/s and his acceler

dangina [55]

V^2 =U^2 +2AS
V^2 = 25 ^2 + 2x9.5x10
V^2 = 625 + 1900 = 2525
V = 50.25

3 0

3 years ago