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Andrew [12]
4 years ago
12

In science, the inquiry process involves observing, questioning, ________, and summarizing.​

Physics
2 answers:
posledela4 years ago
8 0

Answer:

proceesing

Explanation:

Tju [1.3M]4 years ago
6 0

Answer:

predicting

Explanation:

for ap3x users

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A bicycle, a motorcycle, a sports car, and a pickup truck are traveling at the same velocity. Which vehicle requires the greates
trapecia [35]
D. Pickup truck because it's heavy and it gained more momentum
3 0
3 years ago
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an athlete runs 5.4 laps around a circular track that is 400.0 m long. If this takes 540 s, what is the average velocity of the
Slav-nsk [51]

Answer:

Average velocity is 0.296 m/s.

Average speed is 4.0  m/s.

Explanation:

Given:

Distance of the circular track is, D=400.0\ m

Number of laps ran is, n=5.4

Time taken for the run is, t=540\ s

Now, total distance covered in 5.4 laps = D_T=D\times n=400\times 5.4=2160\ m

Also, since the path is a circle, the final position of the athlete after 5.4 laps will be 0.4 of 400 m ahead of the starting point.

Distance covered in 0.4 laps is, \textrm{Displacemet}=0.4\times 400=160\ m

Therefore, the displacement of the athlete will be 160 m as the athlete is 160 m ahead of the starting point and displacement depends on the initial and final points only.

Now, average velocity is given as:

v_{avg}=\frac{\textrm{Displacemet}}{t}=\frac{160}{540}=0.296\ m/s

Average speed is the ratio of total distance covered to total time taken.

So, average speed = \frac{D_T}{t}=\frac{2160}{540}=4\ m/s

6 0
4 years ago
Pleasee help me pwease
Romashka [77]
I mean it’s alaska so i’m assuming B but i could be very wrong
8 0
3 years ago
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Compare the characteristics of 4d orbitals and 3d orbitals and compared the following sentences. Check all that apply
charle [14.2K]

Answer :

<em>(b) 4d orbitals would be larger in size than 3d orbitals</em>

<em>(e) 4d orbitals would have more nodes than 3d orbitals</em>

Explanation :

As we move away from one orbital to another, the distance between nucleus and orbital increases. So, 4d orbitals would be far to the nucleus than 3d orbitals.

Hence, 4d orbitals would be larger in size than 3d orbitals.

Number of nodes is any orbital is n - 1 where, n is principal quantum number.

So, number of orbital in 4d is 3.

And number of orbital in 3d is 2.

So, options (b) and (e) are correct.

8 0
3 years ago
Find the x-component of this
Sholpan [36]

Answer: 74.8m

Explanation:

We have the vector defined by:

r = 101m

θ = 42.2°

When we want to write this as rectangular components, we have that:

x = r*cos(θ)

y = t*sin(θ)

this is because we can construct a triangle rectangle, where the module of the vector is the hypotenuse, the x component is the adjacent cathetus and the y component is the opposite cathetus.

Then, here we have that the x component is

x = 101m*cos(42.2°) = 74.8m

5 0
3 years ago
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