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Alecsey [184]
3 years ago
11

A thin, metallic spherical shell of radius 0.227 m has a total charge of 6.03 × 10 − 6 C placed on it.

Physics
1 answer:
KATRIN_1 [288]3 years ago
3 0

Answer:

Explanation:

Given

radius r=0.227 m

Charge on surface Q=6.03\times 10^{-6} C

Point Charge inside sphere q=1.15\times 10^{-6} C

Electric Field at r=0.735 m

Treating Surface charge as Point charge and applying Gauss law

E_{total}A=\frac{q_{enclosed}}{\epsilon _0}

where A=surface area up to distance r

E_{total}=\frac{Q+q}{4\pi r^2}

E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}

E_{total}=1.194\times 10^{5} N/C

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Question 2.
boyakko [2]

The tensile stress of the wire supporting 2 kg mass is determined as 6.1 x 10⁷ N/m².

<h3>Tensile stress of the wire</h3>

The tensile stress of the wire is calculated as follows;

σ = F/A

where;

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A = πr² = πD²/4

where;

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A = 3.22 x 10⁻⁷ m²

σ = F/A = (mg)/A = (2 x 9.8)/( 3.22 x 10⁻⁷)

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A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.
Maksim231197 [3]

Answer:

v=33.764m/s

Explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

Friction Force F=3.6 N

To find

Velocity v

Solution

From the Centripetal force we know that

F_{c}=\frac{mv^{2} }{r}

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

So

F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s

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3 years ago
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