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allochka39001 [22]
2 years ago
13

Phases of the moon. Help!???????

Physics
1 answer:
Veseljchak [2.6K]2 years ago
6 0

Answer:

The Lunar Month.

New Moon.

Waxing Crescent Moon.

First Quarter Moon.

Waxing Gibbous Moon.

Full Moon.

Waning Gibbous Moon.

Third Quarter Moon.

Explanation:

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Consider the electronic elements that are cooled by forced convection in Problem 6.31. The cooling system is designed and tested
Stella [2.4K]

Answer:

surface temperature of the chip located 120 mm Ts=42.5°C

surface temperature of the chip in Mexico Ts=46.9°C

Explanation:

from the energy balance equation we have to:

q=E=30W

from Newton´s law:

Ts=Tα+(q/(h*A)), where A=l^2

N=h/k=0.04*(Vl/V)^0.85*Pr^1/3

data given:

l=0.12 m

v=10 m/s

k=0.0269 W/(m*K)

Pr=0.703

Replacing:

h=0.04*(0.0269/0.12)*(10*0.12)/((16*69x10^-6))^0.85*(0.703^1/3) = 107 W/m^2*K

The surface temperature at sea level is equal to:

Ts=25+(30x10^-3/107*0.004^2)=42.5°C

h=0.04*(0.0269/0.12)*((10*0.12)/(21*81x10^-6))^0.85*(0.705^1/3)=85.32 W/(m*K)

the surface temperature at Mexico City is equal to:

Ts=25+(30x10^-3/85.32*0.004^2)=46.9°C

8 0
3 years ago
Which of the following is a mood disorder?
gulaghasi [49]
Bipolar disorder is the answer
5 0
3 years ago
Read 2 more answers
A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?
Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

H_m = \frac{u^2sin^2 \theta}{2g}

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta  = sin^{-1}(0.2123)\\\\\theta  = 12.26^0

Therefore, the angle of projection is 12.26⁰.

6 0
2 years ago
A 10 kg turkey, He kicks the 0.5 kg ball with a force of 50N for 0.2 seconds and the ball flies straight away horizontally from
Harman [31]

Answer:

a. 20m/s

b.50N

c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.

Explanation:

a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s

Hence, the velocity of the ball after the kick is 20m/s

b.The force felt by the turkey:

#Applying Newton's 3rd Law of motion, opposite and equal reaction:

-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.

c. Using the law of momentum conservation:

-Due to ther external forces exerted on the turkey, it remains stationery.

-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.

-Momentum is not conserved due to these external forces.

5 0
3 years ago
Which two elements have the same number of valence electrons?
UNO [17]

Answer:

The answer is A. C and O..

4 0
3 years ago
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