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Artyom0805 [142]

2 years ago

15

Two objects of equal mass are a distance of 5.0 m apart and attract each other with a gravitational force of 3.0 x 10^-7 N find

their mass.
A) 150 kg
B) 9.8 kg
C) 11.000 kg
D) 340 kg

1 answer:

Sindrei [870]2 years ago

3 0

Answer

I Think Its 150

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For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative.
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
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For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
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What are some ways to defend a castle against attack?

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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

lana66690 [7]

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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3 years ago

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Alex787 [66]

Explanation:

The given data is as follows.

Length (l) = 2.4 m

Frequency (f) = 567 Hz

Formula to calculate the speed of a transverse wave is as follows.

f = $\frac{5}{2l} \times v$

Putting the gicven values into the above formula as follows.

f = $\frac{5}{2l} \times v$

567 Hz = $\frac{5}{2 \times 2.4 m} \times v$

v = 544.32 m/s

Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.

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