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Andrew [12]
2 years ago
6

The students changed the release distance of the sphere during their investigation. Which statements best describe the purpose f

or increasing the release distance between the trials? Select all that apply.
A. to indicate balanced forces are acting on the spheres

B. to determine the types of forces acting on the spheres

C. to measure the strength of the force acting on the spheres

D. to change the direction of the force acting on the spheres

E. to observe the effect of unbalanced forces acting on the spheres
Physics
1 answer:
vodka [1.7K]2 years ago
4 0

Since unbalanced forces acts on the spheres, the aim of increasing the release distance of the spheres is to observe the effect of unbalanced forces acting on the spheres.

<h3>What is an experiment?</h3>

An experiment is a study whose objective is to study cause and effect relationships. In this case, the students had to increase the distance at which the spheres was released.

The aim of increasing the release distance of the spheres is to observe the effect of unbalanced forces acting on the spheres.

Learn more about experiments: brainly.com/question/11256472

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Find the final price of a 10-speed bike whose original price was $150.00 with a 10% discount and a 6% sales tax.
goldenfox [79]
We have given
original price =150.00
discount=10%=0.1
sales tax=6%=.06
now 
discount=150*0.1=15cost of bike after discount =150-15=135
sales tax=135*.06=8.1
final price =cost of bike after discount +sales tax
<span>final price=135+8.1
</span>final price=143.1<span />
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Answer:

7.0\cdot 10^{-13}C

Explanation:

The magnitude of the electrostatic force between two charged objects is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

r=0.070 cm =7\cdot 10^{-4} m is the distance between the charges

q_1=q_2=q since the charges are identical

F=9.0\cdot 10^{-9}N is the force between the charges

Re-arranging the equation and solving for q, we find the charge on each drop:

F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C

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3 years ago
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4 0
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Answer:

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Net Force (N)
DaniilM [7]

Explanation:

F = m a

F= 4.0×4.0

F= 16 N

......................................................................................................

F= ma

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m= 25/4.998

m= 5.002 kg

......................................................................................................

F=ma

53= 3 × a

a= 53/3

a= 17.666 m/s

5 0
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