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DerKrebs [107]
3 years ago
9

Andrew walks through his garden and observes that the shapes of dewdrops are not always the same. Suppose he wants to investigat

e using the scientific method. Which questions are testable questions that he can ask to look into the reasons for the different shapes?
A) Does the shape of the dewdrop depend on the temperature of the surface?

B) Which dewdrop seems to have the most unusual shape?

C) Is the material of the surface responsible for the shape of the dewdrop?

D) Which shape of dewdrop is the most pleasing to the observer?

E) Does the shape of the dewdrop depend on the moisture in the atmosphere?
Physics
1 answer:
Over [174]3 years ago
8 0

Answer:

1

Explanation:

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A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic
olga_2 [115]

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma              

a = \frac{F}{m}      ...... (1)

acceleration (a) = \frac{dv}{dt}   ......(2)

substituting (2) into (1)

Hence, F = \frac{mdv}{dt}

\frac{dv}{dt} = \frac{F}{m}

dv = \frac{F}{m} dt

dv = \frac{1}{m}Fdt

Integrating both sides

\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt     ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt

v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}

v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|

v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |

v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |

v = \frac{1}{5} | 250 - 50 + 15 |

v = \frac{215}{5}

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

6 0
3 years ago
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