Explain the difference between the justice system laws and scientific laws

According to the periodic table, the average atomic mass of helium is

vfiekz [6]

Helium has an atomic mass of 4.00 atomic mass units.

8 0

2 years ago

The James Webb Space Telescope will be the largest telescope ever put into space and will replace the Hubble Space Telescope. It

Nikolay [14]

Answer:

B : It will avoid the distortion from the atmosphere and give scientists more accurate data

Explanation:

5 0

2 years ago

wo plates with area 6.00×10−3 m2 are separated by a distance of 1.50×10−4 m . If a charge of 5.40×10−8 C is moved from one plate

liq [111]

Answer:

V = 152.542 volts

Explanation:

Given data:

area of plates $6.00\times 10^{-3} m^2$

distance between the plates is $1.50\times 10^{-4} m$

charge = $5.40\times 10^{-8} c$

we know that capacitance is given as

$C = \frac{\epsilon A}{d}$

$C = \frac{8.85\times 10^{-12} 6\times 10^{-3}}{1.50\times 10^{-4}}$

$C = 3.54\times 10^{-10} F$

potential difference is given as

$V =\frac{Q}{C} = \frac{5.40\times 10^{-8}}{3.54\times 10^{-10}}$

V = 152.542 volts

3 0

3 years ago

An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t

mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a) the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F $_L$ = C $_L$ $\frac{1}{2}$ ρV²A = W

we substitute

0.45 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F $_D$ = T = C $_D$ $\frac{1}{2}$ ρV²A

we substitute

= 0.065 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0

2 years ago

An aurora can be found in the _____.

Tomtit [17]

I believe the correct answer is d. thermosphere.

7 0

2 years ago

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