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siniylev [52]
4 years ago
7

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug

h its volume between its inner and outer surfaces. The volume charge density is?
Physics
1 answer:
Vaselesa [24]4 years ago
4 0

Answer:

1.57 * 10^{3} Q

Explanation:

The volume charge density is defined by ρ = \frac{Q}{V} (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are \frac{Coulombs}{m^{3} }, so we have to express the radius in meters:

inner radius = 4 cm * \frac{1 m}{100 cm} = 0.04m

outer radius = 6 cm * \frac{1m}{100cm}  = 0.06m

Now, we know that the volume of the sphere is calculated by the formula:

V = \frac{4}{3}\pi r^{3}, and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = \frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3}), where r_{o} is the outer radius and r_{i} is the inner radius.

Replacing the volume formula in the Equation A:

ρ = \frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}

ρ = \frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }

Replacing the values of the outer and inner radius whe have:

ρ = \frac{3Q}{4\pi (1.52 * 10^{-4})}

ρ = 1.57 * 10^{3} Q

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670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

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3 years ago
Select the correct answer. Loren has obtained a search warrant to search a suspect’s home. He is looking for a wooden bat used i
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A lead ball is dropped into a lake from a diving board 16.0 ft above the water. It hits the water with a certain velocity and th
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Answer:

138.46 ft

Explanation:

When the ball is dropped until the moment it hits the water, the ball moves in a uniform acceleration motion. Therefore, the equation that describes the movement of the ball is:

X = \frac{1}{2}*g*t^{2}  + V_{0} *t + x_0

Where X is the distance that the ball has fallen at a time t. V_0 is the initial velocity, which is 0 ft/s as the ball was simply dropped. x_0 is the initial position, we will say that this value is 0 in the position where the ball was dropped for simplicity, and it increases as the ball is falling. Now, we replace x with 16 feets and solves for t:

16 ft = \frac{1}{2} * 32.2 \frac{ft}{s^{2}} *t^{2} +0 \frac{ft}{s} * t + 0 ft

t = \sqrt{2* \frac{16 ft}{32.2 \frac{ft}{s^2}}} = 1 s

The velocity that the ball will have at the moment the ball that the ball hits the water will be:

V = V_o+g*t=0\frac{ft}{s}+32.2\frac{ft}{s^2}*1s =32.2\frac{ft}{s}

The time that will take the ball to reach the bottom from the top of the lake will be t = 5.3s - 1s = 4.3s. And as the ball will travel with constant velocity equal to 32.2 ft/s^2, the depth of the lake will be:

d = v*t = 32.2\frac{ft}{s}  * 4.3s = 138.46 ft

4 0
3 years ago
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