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3 years ago

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At a construction site, a 68.0 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-fre

e pulley and is connected to an 85.0 kg box on a horizontal roof (see the figure (Figure 1)). The cable pulls horizontally on the box, and a 46.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown. The system is not moving.

1 answer:

Bas_tet [7]3 years ago

8 0

Answer:

Gravel Box Ms = 0.700 MK = 0.400 Concrete

Explanation:

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Which of the angles shown in this picture is the angle of incidence?<br> A<br> B

Dmitry_Shevchenko [17]

The pictures are not attached, therefore, I cannot give a specific choice.
However, I will try to help you out.

The angle of incidence is defined as the angle formed between the ray of light and the normal to the surface that the ray is falling on.

The angle of incidence can be shown in the attached image.

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A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

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4 years ago

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Assoli18 [71]

Answer:

The Roche limit for the Moon orbiting the Earth is 2.86 times radius of Earth

Explanation:

The nearest distance between the planet and its satellite at where the planets gravitational pull does not torn apart the planets satellite is known as Roche limit.

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Here $R_{P}$ is radius of planet and $D_{P}\ and\ D_{M}$ are density of planet and moon respectively.

According to the problem,

Density of Earth, $D_{P}$ = 5.5 g/cm³

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Consider $R_{E}$ be the radius of the Earth.

Substitute the suitable values in the equation (1).

$Roche\ limit=2.423\times R_{E}\times\sqrt[3]{\frac{5.5 }{3.34 } }$

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AveGali [126]

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Runner 1 has a velocity of 10 m/s west. Runner 2 has a velocity of 7 m/s east. From the frame of reference of runner 2, what is

Vsevolod [243]

Answer:

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3 years ago