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slava [35]
2 years ago
14

If Roger dropped pages instead of mongoer, you would have would have to wait longer to catch them than when you were catching ma

ngoes Why would have to wait longer? ​
Physics
1 answer:
slamgirl [31]2 years ago
8 0
I’m confused retype your question please
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Is this statement true or false concerning squall line thunderstorm development? These often form ahead of the advancing front b
Pavlova-9 [17]

Answer: The following statement is true about squall line thunderstorm development: <em><u>These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.</u></em>

<em>An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.</em>

3 0
2 years ago
PLEASE HELP ASAP WILL REWARD BRAINLIEST:))
laiz [17]

Answer:

120s^-1

Explanation:

v=12v

I=10A

and since rate is with time, therefore rate=energy/time.

H=IV

10×12=120/s

therefore the rate is 120s^-1

6 0
2 years ago
What is the difference between landforms and landmarks
Anika [276]
A land form is a natural feature of earths surface and a land mark is can be man-made or natural and it is used to mark something significant. 
7 0
3 years ago
An object is in motion when its distance from a(n) (what is in the blank?) is changing.
Ratling [72]
<span>An object is in motion when its distance from its point of origin is changing.</span>
3 0
3 years ago
The 10 kg dancer leaps into the air with an initial velocity of 5 m/s at angle of 45° from the floor. How far will she travel in
natulia [17]

Answer:

2.55 m

Explanation:

The motion of the dancer is the motion of a projectile, which consists of 2 independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The horizontal range of a projectile can be found by using the equations of motions along the two directions, and it is given by:

d=\frac{v^2 sin(2\theta)}{g}

where

v is the intial velocity

\theta is the angle of projection

g=9.8 m/s^2 is the acceleration due to gravity

For the dancer in this problem, we have:

v = 5 m/s

\theta=45^{\circ}

Therefore, the horizontal range is:

d=\frac{(5)^2(sin 2\cdot 45^{\circ})}{9.8}=2.55 m

5 0
3 years ago
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