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2 years ago

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I was in Anchorage Alaska during the Summer Solstice. The sun rose around 3am and set around 1 am the next day-- almost 24 hours

of sunlight. If the tilt of the Earth was changed to be 10 degrees, how would the length of daylight in Alaska change at the Summer Solstice

1 answer:

m_a_m_a [10]2 years ago

3 0

Answer:

Daylight hours would be shorter.

Explanation:

If there were no tilt of the axis at all, every place on the planet except the north and south poles would have 12 hours of daylight and 12 hours of night every day of the year.

At a 10° tilt, the arctic circle and antarctic circles would be a little less than half the distance from the poles as they are today.

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A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

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3 years ago

1. °C= (°F-32) X 5/9. what will be the answer?

aleksley [76]

Answer:

F=9/5(°C)+32

Explanation:

Lets take an example

Take 25°C

$\\ \sf\longmapsto \dfrac{9}{5}(25)+32$

$\\ \sf\longmapsto 9(5)+32$

$\\ \sf\longmapsto 45+32$

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Answer:

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Answer:

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Explanation:

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