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Lynna [10]
2 years ago
12

Helppppp!!!!!! inportant need help with b​

Physics
1 answer:
KengaRu [80]2 years ago
8 0
I believe the answer is either 1.6 or 9.8
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4. With a diameter that's 11 times larger than Earth's, _______ is the largest planet.
Komok [63]

The answer is B.Jupiter

7 0
3 years ago
Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer
tekilochka [14]

Answer: 0.01 m

Explanation: The formulae for capillarity rise or fall is given below as

h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

h = 2×0.51×cos 50/0.0005×9.8×13,300

h = 0.6556/65.17

h = 0.01 m

8 0
3 years ago
Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024
myrzilka [38]
Given: Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

            G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma;  F = GMeMsat/r²     Centripetal acceleration ac = V²/r

               but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²    

GMe = 4π²r³/T²

T² = 4π²r³/GMe  

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds    or around   1.4 Hour

3 0
3 years ago
The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur
Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

           P₂ is the pressure in the hose

           ρ is the density of the fluid

           h₁ is the initial height

           h₂ is the final height

           v₁ is the initial velocity of the fluid

           v₂ is the final velocity of the fluid  and

           g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

    = (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

    = 5.92 × 10⁵ Pa

    = 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

#SPJ4

6 0
1 year ago
The graph to the right shows the change in Canada‘s harvest of Atlantic cod from 1950-2004 what year shows the clearest evidence
Ann [662]

The correct answer is C. 1995

Explanation:

The graph shows the changes in the harvest of Atlantic cod. In general, this graph illustrates how the peak occurred in the 1980s but then there was a sudden and sharp decline in 1995. Indeed, 1995 is the year with the lowest number of harvested cod as in this year there were approximately least than 10 thousand metric tonnes of cods. Also, this year shows the collapse of fishing stocks or that the population of this fish collapsed, which made it impossible to harvest as many fish as in previous years. According to this, the year that shows the collapse of fishing stocks is 1995.

3 0
2 years ago
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