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3 years ago

Tony walks at an average speed of 70 m/min from his home to

Physics

1 answer:

aksik [14]3 years ago

6 0

Answer:

The answer is 30 min

Explanation:

t = s/v

v = 70 m/min, s = 2100 m

t = 2100/70 = 30 min.

Hope it helps you! ＼(^ᴥ^)／

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__________ research is designed to measure the association between variables that are not manipulated by the researcher.

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Answer:

Option c. Correlational

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4 years ago

In 1970, a rocket powered car called Blue Flame achieved a maximum speed of 1.00(10 km/h (278m/s).Suppose the magnitude of the c

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Answer:

Distance traveled during this acceleration will be 6950 m

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Wear have given maximum speed tat will be equal to final speed of the car v = 278 m/sec

Constant acceleration $a=5.56m/sec^2$

As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec

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3 years ago

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

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Answer:

91.87 m/s

Explanation:

<u>Given:</u>

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

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