All Answers 358150 Answers

There are 2 forces acting on the car. One is the force acting on the mass of the car from gravity (and the road supporting it). The other is centrifugal force which causes the car to move in a tangent to the angle of circular motion.

 Because there is no friction from the tires the forces will balance when

m(g) sin(theta) = m(v²)/r)cos(theta)

Rearranging and reducing this

r = (v²/g)cot(theta)

 or

r = (v²) / (g * tan(theta))

r = (16.667 m/s)² / ((9.81 m/s²) * (tan(20)))

r = (277.778 m²/s²) / ( 3.571 m/s² )

r = 77.8 m

_7UP_

thanks Alan,

I can see how that G is derived when you are talking about Earth and a small object close to the Earth's surface

BUT  why is it the same constant when you are talking about two equally sized objects that could be anywhere in the universe?

I don't get it.   

Welcome to the forum Dragontmaermaths :))

I really like your method Chris - it is so quick  

Think about this

you have 2 equal sized cakes.  they are big slab cakes.

you cut the first one into 36 equal peices and put one peice on the first plate

you cut the other one into 216 equal peices and put one peice on the second plate

Which plate has the biggest slice of cake?

...؟؟؟؟؟؟؟؟؟.......

Thanks...just missing  "A y intercept of 3 means the point (0. 3)...". That is very helpful.

A y intercept of 3 means the point (0, 3)...

Now note something.......we could find the slope arithmetically, but think about this....from (0,3) to (2, -4)...y changes by -7 and x changes by 2......and that's the definition of the slope....so...our slope is -7 / 2 

Now....using the point slope form, we have

y - y1 = m(x - x1)     where m is the slope and (x1, y1) is a point on the line....(let's use (0, 3))

So we have

y - 3 = (-7/2)(x - 0)

y - 3 = (-7/2)x      add 3 to both sides

y = (-7/2)x + 3

And there you go !!!

In my head...-98

To check: $${\mathtt{1}}{\mathtt{\,-\,}}\left({\mathtt{1}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{100}}}}\right){\mathtt{\,\times\,}}{\mathtt{100}} = -{\mathtt{98}}$$

Notice that...... √(25² * 4²)  = √[(25 * 4)² ] = √ [(100)² ]  = 100

I must have had a calculations error...

Duh it's 11 got one thats hard too you actually

How do I multiply (say the question was: $${\sqrt{{{\mathtt{25}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{2}}}}}$$)

Yes that does make more sense! Thank you! 

Ok you had me confused - Gino answered the wrong question.  Easy done :)))

I'm still getting the hang of square roots multiplying, adding, etc...

Forunately, I'm learning this before I make mistakes with it in class, so I'm better prepared (we haven't gone over how negatives work in this sense; now I'll be sure to remember it.) Thank you.

25^2=625

4^2=16

625+16=641

So what's the problem?

Mathematician.....

-aⁿ    means that the exponent only applies to the "a"...not to the "understood" -1 in front

But (-a)ⁿ  means that the exponent applies to everything in the parentheses....

Does that help???....students are often confused about this.....!!!

that's correct and it makes more sense if you see it in a different context.

eg

$$12-5^2=12-25=-13$$

you would never think of saying that this is 12+25 would you?  

So if it's negative, it needs parenthesis; otherwise the negative applies afterword.

(-3² + 4²) + (3² / 3)

(- 9 + 16) + (9 / 3)

( 7 ) + ( 3)

10

I suspect that you evaluated -3²  as (-3)² .....remember...the first is -9, but the second is +9

Hi Mathematician,

$$(-3^2+4^2)+(\frac{3^2}{3})\\
=(\textcolor[rgb]{1,0,0}{-1}*3^2+4^2)+(\frac{3^2}{3})\\
=(\textcolor[rgb]{1,0,0}{-1}*9+16)+3\\
=-9+16+3\\
=10$$

2{4^3 - 6[4 - (2-4)] - 3}    simplify

2{64 - 6[4 - (-2)] - 3}

2{64 - 6[4 + 2] - 3}

2{64 - 6[6] - 3}

2{64 - 36 - 3}

2{28 - 3}

2{25}

50

Note....parentheses, brackets and braces are all the same in these operations....we could have used any combination of them.....

$${\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{100}}$$

It is a very simple muliplication problem that most people know by heart.

The easiest way to multiply by 10 is to move the decimal point once to the right, or add a zero as the last digit if it's a non-decimal number.