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3 years ago

Which sphere of Earth do humans exist in?

Physics

2 answers:

Oksana_A [137]3 years ago

5 0

The answer would be Biosphere

PolarNik [594]3 years ago

4 0

Biosphere is the answer, hope i helped :))

You might be interested in

A bowling ball with a mass of 7.0 kg, generates 56.0 kgᐧm/s units of momentum. What

Misha Larkins [42]

Answer:

8 m/s

Explanation:

$p = mv \\ 56 = (7.0)(v) \\ v = 8.0 \: m {s}^{ - 1}$

7 0

3 years ago

Integrate your expressions for dEx and dEy from θ=0 to θ=π. The results will be the x-component and y-component of the electric

Nimfa-mama [501]

Answer:

hello your question is incomplete below is the missing part

Ex = 0

Ey = $\frac{-2kQ}{\pi a^2}$

Explanation:

Attached below is a detailed solution showing the integration of the expression dEx and dEy from ∅ = 0 to ∅ =π

Ex = 0

Ey = $\frac{-2kQ}{\pi a^2}$

6 0

3 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

$W=K_f -K_i$

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

$K_i = 66,120 J$ is the initial kinetic energy of the car

$K_f$ is the final kinetic energy

Solving,

$K_f = K_i + W=66,120+(-36,733)=29387 J$

The final kinetic energy of the car can be written as

$K_f = \frac{1}{2}mv^2$

where

m = 661 kg is its mass

v is its final speed

Solving for v,

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s$

4 0

4 years ago

In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6

stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

r₂= 7.5 cm

head upto water level is to rise = 60 + 5

= 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q

π r₁² v₁ = π r₂² v₂ = 0.015

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{7.5^2}{5^2} v_2$

$v_1= 2.25 v_2$

$v_2 = \dfrac{0.015}{\pi r_2^2}$

$v_2 = \dfrac{0.015}{\pi 0.075^2}$

v₂ = 0.848 m/s

v₁ = 1.908 m/s

Applying Bernoulli's equation

$P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h$

$P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65$

$P_p= 635539.32 Pa$

P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

6 0

4 years ago

What is the thinnest film (but not zero) of MgF2 (n = 1.38) on glass that produces a strong reflection for orange light with a w

Art [367]

The concept required to solve this problem is related to thin film interference.

The path length difference between two waves one is reflected from top surface of the film and the bottom surface of the film is equal to twice the thickness of the film:

$\gamma = 2t$

Where,

$\gamma =$ Path length difference

t = thickness

At the same time we have that the constructive interference condition for a thin film interference for strongly reflected rays is

$\gamma = m\lambda$

Where

$\gamma =$ Path length difference

$\lambda =$ wavelength

m = Any integer (order of the equation) which represent the number of repetition of the spectrum

Equation both expression we have

$2t = m\lambda$

Re-arrange to find the thickness we have

$t = m\frac{\lambda}{2n}$

Our values are given as

m = 1

$\lambda$ = 597nm

n = 1.38

Replacing,

$t = (1)\frac{597}{2(1.38)}$

$t = 216.3nm$

Therefore the thinnest thickness of the film is 216.3nm

6 0

3 years ago