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Ghella [55]
3 years ago
13

A dripping water faucet steadily releases drops 1.0 s apart. As these drops fall, does the distance between them increase, decre

ase, or remain the same? Prove your answer.
Physics
1 answer:
ololo11 [35]3 years ago
3 0

Answer:

Distance between them increase

Explanation:

The position S of the water droplet can be determined  using equation of motion

S=ut+\frac{1}{2}  at^2

where u is the initial velocity which is zero here

t is time taken, a is acceleration due to gravity

the position of  first drop after time t is given by

S_{1}  =0 \times t+ \frac{1}{2} at^2=\frac{1}{2} at^2............(1)

the position of  next drop at same time is

S_{2}  =\frac{1}{2} a(t-1)^2 = \frac{1}{2} a(t^2+1-2t)............(2)

distance between them is S_{1} -S_{2}  is a(t-1)

from the above the difference will increase with the time

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3 0
3 years ago
A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati
Kazeer [188]

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Force

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{3.5\times 10^7}{2\times 10^6}\\\Rightarrow a=17.5\ m/s^2

Initial acceleration of the rocket is 17.5 m/s²

v=u+at\\\Rightarrow \frac{120}{3.6}=0+17.5t\\\Rightarrow t=\frac{\frac{120}{3.6}}{17.5}=1.90476\ s

Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

\Delta v=v_{e}\ln \frac{m_0}{m_f}

where,

m_0 = Initial mass of rocket with fuel

m_f = Final mass of rocket without fuel

v_e = Exhaust gas velocity

Hence, the change in velocity increases as the mass decreases which changes the acceleration

4 0
3 years ago
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