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CaHeK987 [17]
3 years ago
11

When does an object falling vertically through the air reach terminal velocity?

Physics
1 answer:
zlopas [31]3 years ago
7 0

Answer:

a. when the acceleration of the objects become negative

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You are a pirate working for dread pirate roberts. you are in charge of a cannon that exerts a force 10000 n on a cannon ball wh
KATRIN_1 [288]
<span>Answer: Let m = mass of cannon Then 10000 = ma a = 10000/m v^2 = u^2 + 2as v^2 = 0 + 2as 84^2 = 2(2.21)(10000/m) 84^2 m = 4.42(10000) m = 6.264172336 = 6.26 kg Part 2 Range = u^2sin(2x38)/g = 84^2sin(76)/9.8 = 698.6129229 = 698.6 m</span>
7 0
3 years ago
Is space expanding within clusters of galaxies? why or why not?
GarryVolchara [31]

No, since their gravity is powerful enough to keep them together even while the universe expands as a whole. Space is not expanding within clusters of galaxies.

<h3>What is a galaxy?</h3>

A galaxy is a massive clump of gas, dust, and billions of stars and their solar systems bound together by gravity.

No, since their gravity is powerful enough to keep them together even while the universe expands as an entire.

Hence,space is not expanding within clusters of galaxies.

To learn more about the galaxy, refer to the link;

brainly.com/question/2905713

#SPJ1

6 0
2 years ago
What accounts for the two precipitation peaks in mbandaka?
slava [35]

The two precipitation peaks in Mbandaka during March to April and September to November is due to the intertropical convergence zone.

Intertropical convergence zone is a narrow zone located near the equator. It is where the northern and southern air masses intersect which results to low atmospheric pressure. Due to the intertropical convergence zone’s meeting of air masses, often times the air pressure are lower which will results to colder air, or even rainfall during the period of March to April, and most especially September to November in Mbandaka.

<span>Since Mbandaka is located at the cented of Tumba-Ngiri-Maindombe area, which is named as a Wetland of International importance, there is really a bigger chance that this place experience above 60mm precipitation in a year, temperatures averaging from 23 – 26 degrees Celsius.</span>

7 0
3 years ago
Where does the
GREYUIT [131]

Answer:

it comes from your knowledge and the information you have to get the reason why that is the answer so you are putting together things that you already know what the new information you have

5 0
2 years ago
What is the value of the composite constant (Gme,/r2e) to be multiplied by the mass of the object mo, in equation below:
Sedbober [7]

To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.

The equation for the gravitational force between the Earth and the object on the surface of the Earth is

F_g = \frac{Gm_em_o}{r^2_e}

Where,

G = Universal gravitational constant

m_e = Mass of Earth

r_e= Distance between object and center of earth

m_o= Mass of Object

The equation for the gravitational pulling force on the object due to gravitational acceleration is

F_g = m_o g

Equation the two expression we have

m_o g = \frac{Gm_em_o}{r_e^2}

g = \frac{Gm_e}{r_e^2}

This the acceleration due to gravity which is composite constant.

Replacing with our values we have then

g = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5.98*10^{24}kg)}{6378km(\frac{10^3m}{1km})^2}

g = 9.8m/s^2

The value of composite constant is 9.8m/s^2. Here, the composite constant is nothing but the acceleration due to gravity which is constant always.

8 0
3 years ago
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