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algol [13]
3 years ago
10

What is the opposite of the metric system?

Physics
1 answer:
Nana76 [90]3 years ago
7 0
The opposite of metric system is Imperial system
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If you rub an inflated balloon against your hair and place it against a door, by what mechanism does it stick? explain.
Snowcat [4.5K]
When you rub inflated balloon with your hair or your kitten's fur, charge is induced on all over the balloon's surface. This is called "charging by friction" because you developed charges by rubbing to bodies with each other. It will also stick on your wall you can check it out. This is because of "unlike charges attract each other". Rubbed balloon and wall possessed unlike charges which made them stick together. 
8 0
4 years ago
A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po
VladimirAG [237]

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

V4=1500/183 =9.197v

5 0
3 years ago
In a elastic lab what did you change and why?
Vilka [71]

I changed my undershorts.  The elastic on the old ones I put on that day was deteriorated, and it completely failed when I dripped lab coffee on it, causing falldown.

7 0
3 years ago
Read 2 more answers
A scientist measures the growth of a bamboo plant over time. The table below shows the results.
mihalych1998 [28]
A. 1.35 is the number in between 1.2 and 1.5.
3 0
3 years ago
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A small object carrying a charge of -2.50 nc is acted upon by a downward force of 18.0 nn when placed at a certain point in an e
Vesna [10]
Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?

B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>

<span>Solution:

A) A charge q </span>under an electric field of intensity E will experience a force F  equal to:

F=qE

In our problem we have q=-2.5 nC=-2.5\cdot 10^{-9} C and F=18 nN = 18 \cdot 10^{-9} N, so we can find the magnitude of the electric field:

E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.

B) The proton charge is equal to

e=1.6\cdot 10^{-19} C

Therefore, the magnitude of the force acting on the proton will be

F=eE=1.6\cdot 10^{-19} C \cdot 7.2 V/m=1.15 \cdot 10^{-18} N

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.

7 0
3 years ago
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