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beks73 [17]
3 years ago
15

What is velocity Write its formula

Physics
2 answers:
fenix001 [56]3 years ago
5 0
Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation
v = Δs/Δt.
stellarik [79]3 years ago
5 0

Answer:    

Explanation:

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What is the most important element they keep in mind for the missions? of apollo
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They keep the astronauts safety as a priority. They also try to stay in budget, but mainly the astronauts safety is most important.
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In a closed system, the volume of a gas is not related to the volume of its enclosure.
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That's silly.  The volume of a gas is very intimately related
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pentagon [3]
<span>If the entropy is greater than the enthalpy, it will have more spontinaity</span>
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4 years ago
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The volume of a cantaloupe is approximated by Upper V equals four thirds pi font size decreased by 5 r cubedV= 4 3π r3. The radi
GuDViN [60]

Answer:

68.445 cm³/s

Explanation:

Given:

Volume, V = \frac{4}{3}\pi r^3

radius = 5.85 cm

Growth rate of radius = 0.5 cm/week

now

differentiating the volume with respect to time 't', we get:

\frac{dV}{dt}=\frac{d(\frac{4}{3}\pi r^3)}{dt}

or

\frac{dV}{dt}=(\frac{4}{3}\pi )3r^2\frac{dr}{dt}

now, substituting the value of r (i.e at r = 5.85cm) in the above equation, we get:

\frac{dV}{dt}=4\pi 5.85^2\times 0.5

or

\frac{dV}{dt}=68.445cm^3/s

hence, the rate of change of volume at r = 5.85cm is 68.445 cm³/s

6 0
4 years ago
g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the
Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

T^2 = \frac{4\pi^2 d^3}{2GM}

The given data are given with respect to known constants, for example the mass of the sun is

m_s = 1.989*10^{30}

The radius between the earth and the sun is given by

r = 149.6*10^9m

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

m = 8.2*1.989*10^{30}

d = 6.2*149.6*10^6

Substituting in Kepler's third law:

T^2 = \frac{4\pi^2 d^3}{2}

T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}

T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}

T = 120290789.7s

T = 120290789.7s(\frac{1year}{31536000s})

T \approx 3.8143 years

Therefore the period of this star is 3.8years

7 0
3 years ago
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