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4 years ago

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What makes up the biosphere?

1 answer:

algol [13]4 years ago

6 0

Composed of the parts of the earth where life exists. It includes the dark enviroment of the oceans deep trenches,the rainforests and high mountaintops. Bacteria,protozoa and up to 30 million species of animals, plants and fungi are included in the biosphere.

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The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

Neporo4naja [7]

Answer:

162.8 K

Explanation:

initial current = io

final current, i = io/8

Let the potential difference is V.

coefficient of resistivity, α = 43 x 10^-3 /K

Let the resistance is R and the final resistance is Ro.

The resistance varies with temperature

R = Ro ( 1 + α ΔT)

V/i = V/io (1 + α ΔT )

8 = 1 + 43 x 10^-3 x ΔT

7 = 43 x 10^-3 x ΔT

ΔT = 162.8 K

Thus, the rise in temperature is 162.8 K.

5 0

3 years ago

Why does a rock weigh less in water than it does on land?

Cloud [144]

Water is more dense then air so it sorta holds the rock as it sinks

7 0

3 years ago

A ball dropped from a window strikes the ground 2.76 seconds later. How high is the window above the ground

sweet [91]

Answer:

37.33m

Explanation:

Using the equation of motion

S = ut + 1/2gt^2

Time t = 2.76secs

g = 9.8m/s^2

S = 0 + 1/2(9.8)(2.76)^2

S = 4.9*7.6176

S = 37.33

Hence the window is 37.33m above the ground

7 0

3 years ago

A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming

PtichkaEL [24]

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

$u=v-at$

a=-g

$u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s$

The velocity of the shell at launch and 5.4 s after the launch is given by :

$v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s$

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

6 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