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arsen [322]
3 years ago
11

A ___ can only take place when the moon is in the new moon phase

Physics
1 answer:
Kisachek [45]3 years ago
5 0
An eclipse i believe is the answer?
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Why echo sound cannot be heard in small room
Art [367]
They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.
7 0
3 years ago
Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.
Ad libitum [116K]

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

\theta= arctan(\frac{v_y}{v_x})=arctan(\frac{4.0 m/s}{6.0 m/s})=arctan(0.67)=33.7^{\circ}

relative to the direction of the river.


b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

t=\frac{S_y}{v_y}=\frac{360 m}{4.0 m/s}=90 s


c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

S_x = v_x t =(6.0 m/s)(90 s)=540 m

so, Dave's landing point is 540 m downstream.


d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

7 0
4 years ago
1. To get to school, a girl walks 1 km North in 15 minutes. She
yulyashka [42]

Answer:

Average velocity of the girl is 1.25 m/s.

Explanation:

3 0
3 years ago
In which of the following ways is a puffin dependent upon salmon? Question 4 options: Puffins have a commensalistic relationship
galben [10]
<span>Puffins have a predator-prey relationship with salmon.

</span>
5 0
3 years ago
Read 2 more answers
A​ DC-9 aircraft leaves an airport from a runway whose bearing is N4343degrees°E. After flying for one half 1 2 ​mile, the pilot
bearhunter [10]

Answer:

103.4° or S76.6°E

Explanation:

The direction N43°E is perpendicular to the direction south-east when the plane turn 90° and heads in the south-east direction.

Since the distance 1/2 mile N43°E is perpendicular to the distance 1 mile south-east, we have a right angled triangle.

So, the angle θ between the aircraft's new position and old position is gotten from tanθ = 1 ÷ 1/2 = 2

θ = tan⁻¹(2) = 63.43°

So, the total angle from North to its new position is 40° + 63.43° = 103.43°

Since we need the south-east bearing, the angle from south is 180° - 103.43° = 76.57° ≅ 76.6°

So, our bearing is 103.4° or S76.6°E

3 0
4 years ago
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