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3 years ago

How polar bear can survive in the polar region

Physics

1 answer:

ladessa [460]3 years ago

7 0

Answer:

Hey mate.....

Explanation:

This is ur answer...

<em>Polar bears are well adapted for survival in the Arctic. Their adaptations include: a white appearance - as camouflage from prey on the snow and ice. thick layers of fat and fur - for insulation against the cold.</em>

Hope it helps!

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Say you have a differential drive robot that has an axle length of 30cm and wheel diameter of 10cm. Find the angular velocity fo

Klio2033 [76]

Answer:

a) ω1 = 18rpm ω2 = -18rpm

b) ω1 = 102rpm ω2 = 138rpm

c) ω1 = ω2 = 3.18rpm

Explanation:

For the first case, we know that each wheel will spin in a different direction but with the same magnitude, so:

ωr = 6rpm This is the angular velocity of the robot

$\omega = \frac{\omega r * D/2}{r_{wheel}}$ where D is 30cm and rwheel is 5cm

$\omega = \frac{6 * 30/2}{5}=18rpm$ One velocity will be positive and the other will be negative:

ω1 = 18rpm ω2 = -18rpm

For part b, the formula is the same but distances change. Rcircle=100cm:

$\omega 1 = \frac{\omega r * (R_{circle} - D/2)}{r_{wheel}}$

$\omega 2 = \frac{\omega r * (R_{circle} + D/2)}{r_{wheel}}$

Replacing values, we get:

$\omega 1 = \frac{6 * (100 - 30/2)}{5}=102rpm$

$\omega 2 = \frac{\omega r * (100 + 30/2)}{5}=138rpm$

For part c, both wheels must have the same velocity:

$\omega = \frac{V_{robot}}{r_{wheel}}=20rad/min$

$\omega = 20rad/min * \frac{1rev}{2*\pi rad}=3.18rpm$

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4 years ago

We have a long wire with a circular cross section and radius a = 2.40 cm. The current density in this wire is uniform, with a to

Alika [10]

Answer:

The magnitude of the magnetic field is 7.49x10⁻⁶T

Explanation:

The magnetic field is:

$B=(\frac{\mu i}{2\pi R^{2} } )r$

Where

i = current = 3 A

R = radius = 2.4 cm = 0.024 m

r = distance = 0.72 cm = 7.2x10⁻³m

Replacing:

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3 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

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Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

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How polar bear can survive in the polar region​

How polar bear can survive in the polar region