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amid [387]
3 years ago
14

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 3,220 lb car enters an S-curve at A with a speed of 60 mi/hr with brakes applied to reduce the speed to 45 mi/hr at a uniform
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The magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

The weight is,W= 3,220 lb

The speed is,u= 60 mi/hr

The reducing speed is,v= 45 mi/hr

The distance traveled is,d= 300 ft

The radius of curvature of the path of the car at B is,R= 600 ft.

1 mile = 5280 ft

From the Newtons' equation of motion;

\rm v^2 = u^2 +2ad \\\\ \rm (45 \times \frac{5280}{3600} )^2 = (60 \times  \frac{5280}{3600}  )^2 +2a\times 300 \\\\

The tangential accelerations are;

\rm a_t = \frac{66^2 -88^2}{600} \\\\ \rm a_t =  -5.65 ft/sec^2 \\\\

The force is found as;

\rm \sum F = ma \\\\ \frac{3220}{32.2} \times -5.65 \\\\ F_T= 565 \ lb

The normal force  is;

\rm F_n = \frac{3220}{32.2} \times \frac{66^2}{600} \\\\ F_N =726 \ lb

The net or the total friction force exerted by the road on the tires at B. is found as;

\rm F = \sqrt{(565)^2+(726)^2} \\\ F = 919.946 \ lb

Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

#SPJ1

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