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During a total lunar eclipse, where is the Moon located?

krok68 [10]

During a total lunar eclipse, the Moon located in the umbra. The answer is letter A. For a total lunar eclipse to occur, the Sun, Earth and Moon must be aligned in a straight line. The Earth’s umbra complete covers the Moon. The earth’s umbra is about 870,000 miles wide.

5 0

3 years ago

What is the frequency of a wave if the speed is 4300 m/s and the wavelength is 32 m?

givi [52]

4300 they didn’t look alike because DJ sinwosnbube is wow

8 0

3 years ago

the deepest point in the sea is 100m below sea level .what arr the water pressure at this depth and the total pressure due to wa

fomenos

The water pressure at this depth and the total pressure due to water and atmosphere are 10.3 x 10⁵ Pa and 11.31× 10⁵ Pa.

<h3>What is pressure?</h3>

The pressure is the amount of force applied per unit area.

Atmospheric pressure, Patm =1.01×10⁵ Pa

Density of water, ρ=1030 kg/m³

Depth, h=100 m

Pressure =ρgh

P = 1030×10×100

P = 10.3 x 10⁵ Pa.

Total pressure, P=Po +ρgh

P=1.01×10⁵ + 1030×10×100

P=11.31× 10⁵ Pa

Hence, total pressure is 11.31× 10⁵ Pa.

Learn more about pressure.

brainly.com/question/12971272

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4 0

2 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.