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HACTEHA [7]
3 years ago
10

What is a example of analyze and interpret data?

Physics
2 answers:
Evgen [1.6K]3 years ago
8 0

Answer:

Data from a cross-sectional study or survey might need to incorporate weights or design effects in the analysis.The analysis plan should specify which variables are most

Explanation:

allochka39001 [22]3 years ago
5 0

Explanation:

For example, scientists on a ship may examine SONAR data collected in real-time to determine the shape of the seafloor (Fig 2.7 A). ... Biologists might graph the number of box jellyfish over time and compare these data to the phases of the moon to look for patterns (Fig 2.7 B).

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Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source
Alex_Xolod [135]

Answer:

The magnetic field strength  required to hold anti-protons, moving at 5.70 ✕ 10⁷ m/s in a circular path of 3.20 m in radius is 0.186 T.

Explanation:

To solve the question we note that the magnetic force on a moving charge is given by

F = q·v·B

Where

q = Charge

v = Velocity of the charge =5.70 ✕ 10⁷ m/s

B = Magnetic field

Based on Newton's second law,

Force = Mass, m × Acceleration, a = m × a

Where:

a = Acceleration

m = Mass of anti-proton = Mass of proton = 1.6726219 × 10⁻²⁷ kg

We note that for circular motion, acceleration a is given by

\alpha = \frac{v^{2} }{r} .

Where:

r = Radius = 3.20 m

Therefore, for the circular motion, force, F = \frac{m\cdot v^{2} }{r}

Equating the magnetic force equation to the circular force equation, we have

\frac{m\cdot v^{2} }{r} = q·v·B So that, we find B by making the subject of the formula as follows

B= \frac{m\times v^{2} }{r\times q \times v} . Which gives

B= \frac{m\times v }{r\times q} =  \frac{(1.6726219 \times 10^{-27}) \times (5.7\times 10^{7}) }{(3.20)\times (1.602\times 10^{-19}) } =  0.186 T

The magnetic field strength is

B = 0.186 T

4 0
3 years ago
What keeps the balanced rock balanced
pogonyaev
Hello,

Here is your answer: 

The proper answer to this question is "because of there substantial size the rock rests on another rock which keeps it balanced".

If you need anymore help feel free to ask me!

Hope this helps!
5 0
3 years ago
The sun emits energy by converting hydrogen into helium. what is this process called?
AleksAgata [21]
The answer is Nuclear Fusion. The sun emits energy by converting hydrogen into helium. Nuclear fusion does two things, it converts hydrogen into helium and it also converts mass to energy.
7 0
4 years ago
How far something moves in a specific amount of time
Svetlanka [38]
Distance is how far something moves in a specific amount of time. 
5 0
3 years ago
Read 2 more answers
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude
Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

7 0
3 years ago
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