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kvv77 [185]
1 year ago
12

Which planet has the most extreme temperature difference between day and night?

Physics
1 answer:
Softa [21]1 year ago
4 0

Answer:

Explanation:

Mercury, the first planet from the Sun, endures drastic temperature changes from day to night. During the day, the planet is incredibly near to the Sun, with temperatures reaching 430°C.

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3. Ultraviolet can kill these. <br><br>8words and 2nd letter starts with an A​
Karo-lina-s [1.5K]

Answer:

Bacteria

Explanation:

UV kills bacteria and its 8 letters

7 0
3 years ago
What is the area of 12 1/2 and 17 1/5
jonny [76]
215

I am assuming that those are the width and length. So to find area all you have to do is multiply the two because lw = a. 12 1/2*17 1/5 = 215

25/2*86/5 = 2150/10 = 215
3 0
3 years ago
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min
jekas [21]

Answer:

a)  1.301 kg/s

b) 0.001301 m³/s

c) V₁ = 6.505 m/s, V₂ = 1.626 m/s

d) 118.93 kPa

Explanation:

Given:

The number of cans  = 220

The volume of can, V = 0.355 L = 0.355 × 10⁻³ m³

time = 1 minute = 60 seconds

gauge pressure at point 2, P₂ = 152 kPa

b) Thus, the volume flow rate, Q = Volume/ time

Q = (220 × 0.355 × 10⁻³)/60 = 0.001301 m³/s

a) mass flow rate = Volume flow rate × density

since it is mostly water, thus density of the drink = 1000 kg/m³

thus,

mass flow rate = 0.001301 m³/s × 1000 kg/m³ = 1.301 kg/s

c) Given:

Cross section at point 1 = 2.0 cm² = 2 × 10 ⁻⁴ m²

Cross section at point 2 = 8.0 cm² = 8 × 10 ⁻⁴ m²

also,

Q = Area × Velocity

thus, for point 1

0.001301 m³/s = 2 × 10 ⁻⁴ m² × velocity at point 1 (V₁)

or

V₁ = 6.505 m/s

for point 2

0.001301 m³/s = 8 × 10 ⁻⁴ m² × velocity at point 1 (V₂)

or

V₂ = 1.626 m/s

d) Applying the Bernoulli's theorem between the points 1 and 2 we have

P_1+\rho gV_1 + \frac{\rho V_1^2}{2}=P_2+\rho gV_2 + \frac{\rho V_2^2}{2}

or

P_1=P_2+\rho\timesg(y_2-y_1)+\frac{\rho}{2}(V_2^2-V_1^2))

on substituting the values in the above equation, we get

P_1=152+1000\times 9.8(1.35)+\frac{1000}{2}(1.626^2-6.505^2))

it is given that point 1 is above point 2 thus, y₂ -y₁ is negative

or

P_1=118.93\ kPa

thus, gauge pressure at point 1 is 118.93 kPa

8 0
3 years ago
Please help!!!!!!!!!
denpristay [2]

What help do you want? If you tell, I may help you.

7 0
3 years ago
Compare the relative strengths of the nuclear force and the electric force
Serggg [28]

Answer:

To establish this relationship we must examine the potentials that these forces create. The electrical potential is described by

        Ve = k q / r

The potential for strong nuclear force is

       Vn (r) = - gs / 4pir exp (-mrc / h)

Where gs is the stacking constant and r the distance between the nucleons,

We can compare these potentials where the force is derived from the relationship

       E = -dU / dr

       F = q E

Explanation:

6 0
3 years ago
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