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Kisachek [45]
3 years ago
10

Based on the weather station systems shown below, what is the most likely location of the low pressure system?

Physics
2 answers:
BigorU [14]3 years ago
5 0

Answer: Based on the weather station symbols shown below, what is the most likely location of the low pressure system?

A.Just West of City A.

kicyunya [14]3 years ago
3 0
D just east of station b
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I need help with this
Fed [463]
HEY THERE !!

It is called SUBLIMATION.

so the correct answer is (D).
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3 years ago
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a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou
zmey [24]

Answer:

1.56 J

Explanation:

The potential energy only depends on the vertical height from the ground level.

We consider the ground level to have zero P.E.

So when it is 2 m above the ground level,

P.E. =  mgh

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7 0
3 years ago
A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr
Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

v_f^2 -v_i^2 =2ad

substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

v_f=\frac{121.2 m/s}{2}=60.6 m/s

In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

3 0
3 years ago
If I travel 300 m east, then 400 m west, what is my distance &amp;<br> displacement?
ad-work [718]

Answer:100m west

Explanation:

6 0
3 years ago
An object is placed at O ona number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to
kvv77 [185]

Answer:

You have a displacement of 5 units to the right.

Explanation:

First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.

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