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I need help in these questions in **details **

and Please understand that I struggle a little with English

I determine the velocity is negative because the slope is negative ! I chose (b) .. right ?

xvxvxv Aug 26, 2014

#3**+10 **

**Question 5** is right.

**Question 11** There may be a 'normal' method for doing this, I am not a physics specialist.

$$magnitude\\

=\sqrt{7.6^2+5.2^2}\\

=9.20869\\

=9.2$$

$$\\tan\theta=\frac{5.2}{7.6}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.2}}}{{\mathtt{7.6}}}}\right)} = {\mathtt{34.380\: \!344\: \!723\: \!845^{\circ}}}$$

direction = 180+34.38 = 214.38 degrees

**So the answer is B** 9.2 and 214.38 degrees with the positive x axis.

If you do not understand what I have done then make sure you ask me to explain more.

Melody Aug 26, 2014

#5**+10 **

Good,

Lets try **question 16**

easterly force = 3N

Northerly force =4N

Use Pythagoras's theorem to get net force.

sqrt(9+16)=5N

Now, F=ma

5=2*a

$$acceleration =5/2=2.5m/s^2$$

.Melody Aug 26, 2014

#6**+10 **

**Question 19**

NOTE: I do not know this stuff. I am referencing this site

http://zonalandeducation.com/mstm/physics/mechanics/forces/newton/mightyFEqMA/mightyFEqMA.html

It looks pretty good.

u=30m/s, mass=2000kg, Braking force=10000N, v=0, find t

F=ma therefore a=F/m=-10000/2000=-5m/s^{2}

v=u+at

0=30-5t

5t=30

t=6 seconds

------------

d=ut+(1/2)at^{2}

d=30*6+(1/2)(-5)*36

d=180+-90

distance=90metres I think that D is the answer.

-------------

Melody Aug 26, 2014