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2 years ago

10

Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal pr

ojection Pa of R onto axi

1 answer:

Liula [17]2 years ago

6 0

Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

$\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}$

therefore : Ra = $\frac{sin120^o * 750}{sin 35^o}$ = 1132 N

To determine the scalar component Rb

$\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}$

therefore : Rb = $\frac{sin 25^o * 750}{sin 35^o}$ = 522.6 N

To determine the orthogonal projection Pa of R onto

Pa = 750 cos $25^o$ = 679.7 N

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The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n

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</span><span> Felectric = Ftranslational
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2 years ago

Which statement about images is correct? a) A virtual image cannot be formed on a screen. b) A virtual image cannot be viewed by

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Answer:

A). A virtual image cannot be formed on a screen.

Explanation:

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2. A real image must be erect or maybe inverted.

3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.

4.A virtual image can be photographed.

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2 years ago

An electron is trapped in a square well of unknown width, L. It starts in unknown energy level, n. When it falls to level n-1 it

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Answer:

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Explanation:

Detailed explanation of answer is given in the attached files.

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A bucket of water experiencing a gravitational force of 525 N is pulled up from a water well. The net force in the y-direction i

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1 year ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

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now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

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now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

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now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

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2 years ago