+0  
 
0
1254
14
avatar+1832 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 Aug 29, 2014

Best Answer 

 #12
avatar+330 
+10

 . . . what about 6 , 7 and 8 . . .

 

Yes, they are correct. I double checked them.

 

~~D~~

 Aug 31, 2014
 #1
avatar+32822 
+5

Questions 9 through 21 are correct.  

 

Several of the other images are too small for my jaded eyes to make out! 

 Aug 29, 2014
 #2
avatar+1832 
0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.
 Aug 30, 2014
 #3
avatar+1832 
0

are the pictures still unclear 

 Aug 30, 2014
 #4
avatar+115747 
+5

I am sorry 315, I have not had a chance to look at them and I don't know how many I could answer anyway.

I will add this thread to the other one and high light them as unanswered.

Alan may have missed it and David has not been about for a while.

Chris may have a go at some when he is back on.

Anyway that is all I can do for you at the moment. sorry.

 Aug 30, 2014
 #5
avatar+1832 
0

Okay melody thank you and I will wait 

 Aug 30, 2014
 #6
avatar+4473 
+5

1. 

 Aug 30, 2014
 #7
avatar+4473 
+5

2. Correct.

 Aug 30, 2014
 #8
avatar+4473 
+5

3. Correct.

 

4. Correct.

 Aug 30, 2014
 #9
avatar+330 
+5

Q-1) The answer is (A). Remember to subtract the length (1cm from the top and 1 cm from the bottom) already included in the other measurements.

 

Q-2 through Q-5 ) Answers are correct.

 

For Q-23 . . . . .

 

Your answer, (B), is (probably) correct. If it didn’t say “always” and/or have (D) “Any value” as an alternative, then I might be more certain.

 

 

This is an ambiguous question. (These are often found on physics tests).

 

For most equations friction work is (usually) signed as negative. Applied forces, are usually positive.

 

Technically, friction opposes motion and, hence, does negative work. However, friction is not a vector, so the negative sign does not tell us that it's moving backward.

 

Friction removes energy from the system. That is why it is (usually) signed as negative. To overcome this, energy needs to be (continuously) added to the system.

 

Nota bene: Another consideration is friction can be (0) zero. If this is the case, then it is neither positive nor negative. This really is an ambiguous question

 

~~D~~

 

P.S. Now, that I have reread the question, it is clear it refers to the previous question (Q-20). Your answer, (B), is correct.

 

It does help to read the whole question.

 Aug 31, 2014
 #10
avatar+1832 
0

Thanj you AzizHusain  and DavidQD ... 

 Aug 31, 2014
 #11
avatar+1832 
0

but what about 6 , 7 and 8 

are there correct 

 Aug 31, 2014
 #12
avatar+330 
+10
Best Answer

 . . . what about 6 , 7 and 8 . . .

 

Yes, they are correct. I double checked them.

 

~~D~~

DavidQD Aug 31, 2014
 #13
avatar+1832 
0

Now its very clear .. thank you david 

 Aug 31, 2014
 #14
avatar+115747 
+5

Thank you Alan, Aziz and David :)

 Aug 31, 2014

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