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The centrifugal force pushing th box outward is given by m*v²/r where m is the mass of the box, v is the lorry speed and r is the radius of curvature.  This force is balanced by μ*m*g, where μ is the coefficient of static friction and g is the gravitational acceleration.

mv²/r = μmg

v²/r = μg

v² = μgr

v = √(μgr)

You can plug in the numbers.

Identify the forces on each block and use Newton's laws of motion.  Have a look at;

Hi Selena, 

you learn by copying - justr little things at first like making a fraction.  there is lots of code that you can play with in this thread.   :)

Normal force of surface on lower block = (2.05+4.82)*9.81N,

hence horizontal friction force = 0.257*(2.05+4.82)*9.81N.

Vertical force of upper block on lower block = 2.05*9.81N,

hence horizontal friction force = 0.257*2.05*9.81N.

Newton's 2nd law on lower block:

4.82*a = 106.6 - 0.257*(2.05+4.82)*9.81 - 0.257*2.05*9.81

a = (106.6 - 0.257*(2.05+4.82)*9.81 - 0.257*2.05*9.81)/4.82

$${\mathtt{a}} = {\frac{\left({\mathtt{106.6}}{\mathtt{\,-\,}}{\mathtt{0.257}}{\mathtt{\,\times\,}}\left({\mathtt{2.05}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4.82}}\right){\mathtt{\,\times\,}}{\mathtt{9.81}}{\mathtt{\,-\,}}{\mathtt{0.257}}{\mathtt{\,\times\,}}{\mathtt{2.05}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}{{\mathtt{4.82}}}} \Rightarrow {\mathtt{a}} = {\mathtt{17.450\: \!448\: \!879\: \!668\: \!049\: \!8}}$$

acceleration ≈ 17.45 m/s²

.

Sorry Aziz but I am going for bugzy's interpretation of the question.  

$$\\9*2\frac{1}{3}\\\\
=9*\frac{2*3+1}{3}\\\\
=9*\frac{7}{3}\\\\
=\frac{9*7}{3}\\\\
=\frac{\not9^3*7}{\not3^1}\\\\
=21$$

You write down all the forces on each mass, then resolve them parallel and perpendicular to the surface of the hill.  The perpendicular forces on each mass must balance; the net force down the hill must give rise to an acceleration (the same for both masses since they are rigidly fixed together).  

You don't say which mass is lower so I will assume mass A is lower (this will only affect the sign of the tension);

Mass A:

Perpendicular:   F_NA = m_Ag*cos(θ)                    ...(1)

where F_NA is normal force, m_A is mass of A;  g is gravitational acceleration;  θ is angle to horizontal.

Parallel:           m_A*a = m_Ag*sin(θ) - μ_AF_NA - T  ...(2)

where a is acceleration; μ_A is coefficient of friction for mass A; T is tension in the bar.

Mass B:

Perpendicular:  F_NB = m_Bg*cos(θ)   ...(3)

Parallel:           m_B*a = m_Bg*sin(θ) - μ_BF_NB + T  ...(4)

Use (1) in (2) and (3) in (4)

m_A*a = m_Ag*sin(θ) - μ_Am_Ag*cos(θ) - T     ...(5)

m_B*a = m_Bg*sin(θ) - μ_Bm_Bg*cos(θ)  + T   ...(6)

Add equations (5) and (6):

(m_A + m_B)*a = (m_A + m_B)g*sin(θ) - (μ_Am_A + μ_Bm_B)g*cos(θ)

or

a = g*sin(θ) -  (μ_Am_A + μ_Bm_B)g*cos(θ)/(m_A + m_B)   ...(7)

Substitute (7) back into (5) or (6) and rearrange to get T.  I'll leave you to do this and to plug in the numbers (you should get something close to a ≈ 3.5m/s² and T ≈ -25N)

Welcome back Aziz, I have missed you as well.  

This is worth a watch.

$$\frac{299}{3\frac{1}{4}}\\\\
=\frac{299}{\frac{13}{4}}\\\\
=299\div \frac{13}{4}}\\\\
=299\times \frac{4}{13}}\\\\
=\frac{299*4}{13}}\\\\
=\frac{23*4}{1}}\\\\
=92$$

$$\\4y^2 + 12x^2 y + 9x^4\\\\
(2y+3x^2)^2$$

Thanks Chris,

It is the easiest way of doing it.

Just multiply by 10^n  where n is the number of digits that are repeating.  do the subtraction and simplify.   

J=5B (1)

J-64=B+64   (2a)

add 64 to both sides

J=B+128       (2b)

(1)=(2b)

5B=B+128

4B=128

b=128/4

b=$32

J = 5*32 = $160

The normal force is the vertical force the ground exerts on the sled to stop it sinking through the surface!  In this case the sled is partly held up by the force on the rope, so the reaction force of the ground on the sled is

normal force = mass*g - F*sin(θ)  where g is gravitational acceleration (9.8m/s²) and θ is the angle between the rope and the ground. F*sin(θ) is the vertical component of the force applied to the sled through the rope.   If we have θ = 32° then

$${\mathtt{normalforce}} = {\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{32}}^\circ\right)} \Rightarrow {\mathtt{normalforce}} = {\mathtt{218.504\: \!036\: \!788\: \!35}}$$

or normal force ≈ 218.5N

All the numbers are multiplied by 4 so the scale factor is  4 new:1old    The scale factor is 4.

The centre of the dilation is the origin.   [ also, I assume you meant B' to be (-8,24) ]

always draw the picture when you do these types of questions.

This is the boat in the water.  The red line is the water.

Can you work it out now?

y2 + 6xy + 9x2      This is a perfect square trinomial....the key to factoring these is to take the square root of the first and last terms and separate them by the sign on the middle term....then just square the whole thing...

So we have.....

(y + 3x)

Now...just square the whole thing

(y + 3x)²

And there you are  !!!

d(t) = 3/2 sin (5πt/31) + 23

The average depth of the water, over time, is 23 feet.

The amplitue of the tide is 1.5 feet

So the maximum depth of the water is 24.5m

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

You do not need this but the period is 2pi divided by 5pi/31 = 2pi*31/5pi = 62/5 = 12.4

I think it fair to assume that we are talking tides and time is in hours.

Here's my take on this one :

Melody is correct

The amplitude is 1

The vertical shift is "up' 3 units

The horizontal shift is to the left by 5 rads

Note that the "4" tells us how many periods there are in 2pi...thus, the normal cosine graph is 'compressed" by a factor of 4 !!!

AzizHusain, it is hotter in California

You are wonderful Ghost of Web2.0calc.   You write your lay so beautifully.  

Kitty is one fortunate Princess to have such attention lavished upon her.  

Our magical realm is truly blessed.

Queen Guinevere.

18 - 24x + 8x^2   Let's start by taking out the greatest common factor (2)    .....so we have....

2(9 - 12x + 4x^2)   And the polynomial in the parentheses is a perfect square trinomial, so we have.....

2(3 - 2x) (3 - 2x)  =   2(3 - 2x)²

And that's it !!

$$\\(\sqrt3(2^7))^3\\\\
=(3^{0.5}2^7)^3\\\\
=3^{0.5*3}(2^{7*3})\\\\
=3^{1.5}*2^{21}\\\\$$

Thanks DS and Chris! 

I am doing alright. Just lots of things happening these days and yes, it is good...the usual. Right now, no grades yet since it just started. I'm glad to be back as well. Will visit from time to time!

However, the weather is crazy here...it is in the 90s (Fahrenheit) in October! I am going crazy.

You can just copy the information by highlighting the numbers and right-clicking to "copy" (or ctrl+c) and then paste by right-clicking and then "paste" (or ctrl+v) when you need that information again.