Answer:
"Narrow the focus of research question"
Explanation:
O Narrow the focus of research question
This is good! You can still use your question, but focus in on something so you have a proper research project.
O Add another research question
Would adding another question to an already broad question help? No.
O Use the very first source you find for your project
If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question
O Change the scope of your project
You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
The measuring devices for mass is scales and Balances
The centripetal acceleration is 
Explanation:
For an object in uniform circular motion, the centripetal acceleration is given by

where
v is the speed of the object
r is the radius of the circle
The speed of the object is equal to the ratio between the length of the circumference (
) and the period of revolution (T), so it can be rewritten as

Therefore we can rewrite the acceleration as

For the particle in this problem,
r = 2.06 cm = 0.0206 m
While it makes 4 revolutions each second, so the period is

Substituting into the equation, we find the acceleration:

Learn more about centripetal acceleration:
brainly.com/question/2562955
#LearnwithBrainly
Answer:
magnetic fields is stronger at the pulls because opposites attract which is why the pull is stronger.
this was written by me.
Explanation: