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3 years ago

Why does one part of earth experience summer while another part experiences winter?

Physics

2 answers:

lana66690 [7]3 years ago

6 0

The combination of Earth's tilted axis and its revolution causes the atmosphere to be unequally heated by the sun.

Elis [28]3 years ago

4 0

Answer:

Earth revolves around sun in 1 year. It moves in nearly circular orbit. Seasons are caused on Earth due to the tilt of the earth. Earth's axis is tilted to about 23.5 degrees from the horizontal. When it moves in its orbit, sometimes norther hemisphere faces sun and sometimes it faces away from sun and then the southern hemisphere faces towards it.

So, the part which faces towards sun experiences summer while the other which is away faces winters.

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A three-way 120 V lamp bulb that contains two filaments is rated for 100-200-300 W. One filament burns out. Afterward, the bulb

ololo11 [35]

Answer:

The larger resistance value is 144 ohms and the smaller resistance value is 48 ohms.

Explanation:

Given that Resistance, R, is inversely proportional to Power, the largest value of power will yield the smallest value of resistance and vice versa.

The largest value of resistance will be:

P = V²/R

R = V²/P

R = 120²/100

R = 144 ohms

The smallest value of resistance will be:

R = 120²/300

R = 48 ohms

8 0

4 years ago

Luis wants to compare the density of three solid objects that are different shapes and sizes. What information does he need to m

tatyana61 [14]

Answer:

u need to make sure that comparison is = to shapes and then find the shapes sizes and add them

7 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$