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velikii [3]
2 years ago
6

A pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the

roof of a building, as shown. The rope pulls horizontally on the lower crate, and the coefficient of static friction between the lower crate and the roof is 0.666. What is the weight of the heaviest pallet of bricks that can be supported this way?
A. 400 lb
B. 350 lb
C. 250 lb
D. 266 lb

Physics
1 answer:
SashulF [63]2 years ago
5 0

Answer:

<h3><u>C</u><u>.</u><u>2</u><u>5</u><u>0</u><u> </u><u>l</u><u>b</u><u> </u><u>i</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>.</u><u>.</u><u>.</u></h3>
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Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
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Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

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What type of motion occurs when an object spins around an axis without altering its linear position?
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A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
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Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

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