What are 4 physical properties for the bear

Physics

1 answer:

Pie3 years ago

3 0

Answer:

A bear normally has a short, thick neck, a rounded head, a pointed muzzle, short ears, and small eyes. Some species have round faces. Bears have poor eyesight, and most have only fair hearing.

Explanation:

Modern Bears are characterized with large body and stocky legs, a long snout, shaggy hair, plantigrade paws with five non-retractile claws and a short tail.

Grizzly bears (Ursus arctos horribilis) have concave faces, a distinctive hump on their shoulders, and long claws about two to four inches long. Both the hump and the claws are traits associated with a grizzly bear's exceptional digging ability. Grizzlies are often dark brown, but can vary from blonde to nearly black.

The brown bear has a slight hump above its shoulder, round ears, a long snout and big paws with long, curved claws that it uses for digging. Unlike the black bear, it can't climb trees. It can weigh between 350-1,500 pounds. When standing on its hind legs it can be up to 5 feet tall.

Hope this helps :)

(I didn't know which type of bear so i did brown bear and grizzly bear)

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A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$