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4 years ago

Please Help! Im dumb.

Physics

2 answers:

Rus_ich [418]4 years ago

8 0

The answers false I believe

My name is Ann [436]4 years ago

3 0

The answer is false hopes this helps!

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4 If someone gems to the Shore from about Kannada bhoot movies in the opposite direction .Explain, why?

dsp73

Hi Pupil Here's Your answer :::

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Answer : When a man jumps out from a boat, he pushes the boat with his feet (action) and the boat also exert an equal force on him in opposite direction reaction. As a result, the man jumps to the bank and the boat moves in the backward direction.

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Hope this helps . . . . . .

8 0

3 years ago

Rutherford discovered the nucleus of the atom by firing a particles at gold foil. An a particle has a charge of q = +2e and a ma

Readme [11.4K]

Answer: $r_0=3.037\times 10^{-14}m$

Explanation:

Given

charge on alpha particle=+2e

mass of alpha particle= $6.64\times 10^{-27}$ kg

Charge on gold nucleus=+79e

Velocity at r=1m is $1.9\times 10^{7}$

Using Energy conservation

Kinetic energy of particle will be converting to Potential energy as it approaches to nucleus

therefore

$\frac{1}{2}mv^2+U_{r=1m}=U_{closest\ to\ nucleus}$

$\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )+\frac{K\left ( 2e\right )\left ( 79e\right )}{1}=\frac{K\left ( 2e\right )\left ( 79e\right )}{r_0}$

$\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )=\frac{9\times 10^9\times 158\times \left ( 1.6\times 10^{-19}\right )}{y}\left [\frac{1}{r_0}-\frac{1}{1}\right ]$

on solving we get

$\frac{1}{r_0}=3.292\times 10^{13}$

$r_0=3.037\times 10^{-14}m$

8 0

3 years ago

A series RLC circuit has a resistance of 28 Ω and an impedance of 105 Ω. If the rms voltage applied to the circuit is 140 V, wha

sergiy2304 [10]

Answer:49.8W

Explanation:

7 0

3 years ago

Compare the magnitude of the electromagnetic and gravitational force between two electrons separated by a distance of 2.00 m. As

Lesechka [4]

First you need to know about two laws, which are:
1) Coulomb's law
2) Newton's law of gravitation

1.
According to Coulomb's law, Electric force between TWO charges is:
$F_{e} = \frac{k*q_{1}*q_{2}}{r^{2}}$ -- (A)

Where, k = 1/(4*π*epsilon_not) = $9 * 10^{9}$ $\frac{Nm^{2}}{C^{2}}$

Both $q_{1}$ and $q_{2}$ = -1.61 x $10^{-19}$ C
r = Distance between the two charges = 2.00m

Plug-in the above values in (A), you would get:

$F_{e}$ = (9 * $10^{9}$ ) (1.61 * $10^{-19}$ * 1.61 * $10^{-19}$ ) / (2*2)

$F_{e}$ = $5.83* 10^{-29}$ N

2.
According to Newton's law of gravitation:
$F_{g} = \frac{Gm_{1}m_{2}}{r^{2}}$ -- (B)

Where G = Gravitational constant = 6.674 * $10^{-11} m^{2} kg^{-1}s^{-2}$

m1 = m2 = Mass of the electron = $9.11 * 10^{-31}$ kg
r = 2.0 m

Plug-in the above values in (B), you would get:

$F_{g}$ = (6.674 * $10^{-11}$ ) ( $9.11 * 10^{-31}$ * $9.11 * 10^{-31}$ }[/tex]) / (2*2)

$F_{e}$ = $5.83* 10^{-29}$ N

$F_{e}$ = $1.38 * 10^{-71} N$

Now do Fe over Fg, you would get:
$\frac{F_{e}}{F{g}} = 4.23 * 10^{42}$

Ans: So the blanks are:
1) 5.82
2) 1.38
3) 4.23

-i

6 1

3 years ago

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Sheila lives in the seashore community of Beachview. At 6:00 a.m. one morning, she noticed that the local ocean water was at hig

Serhud [2]

The tide was high because of the sun just now coming up at noon it when the sun begins to settle causing the ocean to form at low tide .

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3 years ago

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