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IgorLugansk [536]
3 years ago
12

An observer sitting on shore sees a canoe traveling 5.0 m/s east, and a sailboat traveling 15.0 m/s west. What is the velocity o

f the sailboat as observed on the canoe (relative to the canoe)?
Physics
1 answer:
Dovator [93]3 years ago
3 0

Answer:

Vs/c = 20 [m/s]

Explanation:

This is a problem of relative velocities, as velocity is a vector we can use addition or subtraction of vectors for the solution.

We are asked for the velocity of the sailboat with respect to an observer located in the canoe.

V_{s/c}=V_{sailboat}-V_{canoe}\\V_{s/c}=15-(-5)\\V_{s/c}=20[m/s]

Why is the speed of the canoe negative?, it is negative because the canoe moves in the opposite direction to the sailboat.

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A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
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Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

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Reflection of the images example
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A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
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Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1.x10x^{-6} \frac{C}{m^{2}}

ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2}  \\x_{f}=0.53x^{-3}m

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If you heat soup on a stove, what happens to the movement of the soup’s particles?
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