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icang [17]
3 years ago
15

ILL GIVE THE BEST ANSWER BRAINLIEST I PROMISE AND ILL THANK EVERY ANSWER A polar bear is able to walk on ice without breaking it

by spreading out its force. If a polar bear has a mass of 200. kg and when spread out on ice, covers an area of 2.5 m2, what is the pressure the polar bear exerts on the ice?
Physics
2 answers:
Luden [163]3 years ago
7 0

Answer:

(200KG)(9.81 m/s^2=1962

1962N/2.5m^2=784.8pa or 7.84.8 N/m^2

krek1111 [17]3 years ago
5 0

Answer:

Polar bears (Ursus maritimus) are of special interest because of their large size, white color and position as the top-level carnivore in the remote arctic environment. They occur only in the northern hemisphere nearly always in association with sea ice. They have only two colors of fur: tan and white. Polar bears were created to withstand cold temperatures and are quite adaptable. Polar bears are ferocious and the most dangerous of bears. They can lop a person's head off with one swoop of their paw.

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Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add
jok3333 [9.3K]

Answer:

When x = 2.8 cm, B_{x1} = 0.0265 T

When x = 5.5 cm, B_{x2} = 0.0209 T

when x = 7.3 cm, B_{x3} = 0.0169 T

When x = 11.0 cm, B_{x4} = 0.0103 T

Explanation:

According to Biot-Savart law,

B_{x} = \frac{N \mu_{o}IR^{2}  }{2(x^{2} +R^{2}  )^{3/2} }\\.......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

\mu_{o}  = 4\pi  * 10^{-7} N/A^{2}

When x₁ = 2.8 cm = 0.028 m

B_{x1} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.028^{2} +0.11^{2}  )^{3/2} }\\B_{x1} = 0.0265 T

When x₂ = 5.5cm = 0.055 m

B_{x2} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.055^{2} +0.11^{2}  )^{3/2} }\\B_{x2} = 0.0209 T

When x₃ = 7.3 cm = 0.073 m

B_{x3} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.073^{2} +0.11^{2}  )^{3/2} }\\B_{x3} = 0.0169 T

When X₄ = 11.0 cm = 0.11 m

B_{x4} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.11^{2} +0.11^{2}  )^{3/2} }\\B_{x4} = 0.0103 T

4 0
4 years ago
If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be
il63 [147K]

Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

This is equal to q / ε where ε is a constant called permittivity  which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .

If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .

It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors

1 ) charge q and

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3 years ago
The difference between verbal and non-verbal communication is in verbal we use words and in non-verbal we can use body language.
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Your answer would be true! :)
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With good tires and breaks, a car traveling 50 mi/hr on dry pavement can travel 400 ft when the driver reacts to something he se
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Answer:6.72 m/s^2

Explanation:

Given

initial velocity u=50 mi/hr\approx 73.33 ft/s

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using equation of

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where v=final velocity

u=initial Velocity

a=acceleration

s=displacement

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0-73.33^2=2\times a\times 400

a=-6.72 m/s^2

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What type of heat transfer is heat that comes from a heat lamp in a bathroom?
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