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2 years ago

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helps us adapt to our environment. It also generally with age. A. Plasticity . . . decreases B. Developmen

t . . . increases C. Plasticity . . . increases D. Development . . . decreases

2 answers:

Nana76 [90]2 years ago

8 0

Not sure this is a physics question (probably biology).
Anyway, the correct answer is A):
"Plasticity helps us to adapt to our environment. It also generally decreases with age".
Plasticity is the ability to adapt to the environment. Since this ability is linked with brain functions, and brain functions worsen with age, then plasticity decreases with age.

grin007 [14]2 years ago

4 0

Answer: A, Plasticity ... decreases

Explanation:

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Can two objects exert a force on each other without touching? example

lara [203]

Answer:

where is the example...?

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2 years ago

The curvature of the helix r(t)equals(a cosine t )iplus(a sine t )jplusbt k (a,bgreater than or equals0) is kappaequalsStartF

4vir4ik [10]

Answer:

$\kappa = \frac{1}{2 b}$

Explanation:

The equation for kappa ( κ) is

$\kappa = \frac{a}{a^2 + b^2}$

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

$\kappa (a) = \frac{a}{a^2 + b^2}$

Now, the conditions to find a maximum at $a_0$ are:

$\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0$

$\frac{d^2\kappa(a)}{da^2} \left | _{a=a_0} < 0$

Taking the first derivative:

$\frac{d}{da} \kappa = \frac{d}{da} (\frac{a}{a^2 + b^2})$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da} (\frac{1}{a^2 + b^2} )$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1) (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da} (a^2+b^2)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a (\frac{1}{(a^2 + b^2)^2} ) (2* a)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2} - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 - 2 a^2)$

$\frac{d}{da} \kappa = \frac{b^2 - a^2}{(a^2 + b^2)^2}$

This clearly will be zero when

$a^2 = b^2$

as both are greater (or equal) than zero, this implies

$a=b$

The second derivative is

$\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 - a^2}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 - a^2 ) + (b^2 - a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2 a ) + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

$\frac{d^2}{da^2} \kappa = \frac{-2 a}{(a^2 + b^2)^2} + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

We dcan skip solving the equation noting that, if a=b, then

$b^2 - a^2 = 0$

at this point, this give us only the first term

$\frac{d^2}{da^2} \kappa = \frac{- 2 a}{(a^2 + a^2)^2}$

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

$\kappa = \frac{b}{b^2 + b^2}$

$\kappa = \frac{b}{2* b^2}$

$\kappa = \frac{1}{2 b}$

3 0

3 years ago

You are spinning a rock, of mass 0.75 kg, at the end of a string of length 0.86m in a vertical circle in uniform circular motion

Nina [5.8K]

Answer:

v (minimum speed) = 2.90 m/sec.

$\\ \\ maximum speed (v)= 6.57 m/sec.\\$

Maximum value of speed will occur at lowest point of vertical circle.

Explanation:

a) What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?

Using the force balance expression at the top of the circle,

Gravitational Force + Tension force = Centrifugal force

$m*g + T = m*v^2/R$

Given that : T = 0

R = length of string = 0.86 m

mass of the spinning rock = 0.75 kg

$v = \sqrt{g*R}$

$v = \sqrt{9.81*0.86}$

v (minimum speed) = 2.90 m/sec.

b) what is the maximum speed the rock can have so that the string does not break?

Here the force balance at bottom of circle is represented by the illustration:

$T = m*g + m*v^2/R$

Given that:

maximum tension T = 45 N

maximum speed v = ??

mass m = 0.75 kg

∴

$45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\$

c)

At what point in the vertical circle does this maximum value occur?

Maximum value of speed will occur at lowest point of vertical circle.

This is so because at the lowest point; the tension in string will be maximum.

4 0

2 years ago

How much centripetal force is needed to make a body of mass 0.5kg to move in a circle of radius 50cm with speed 3ms^-1 ?

scoundrel [369]

Centripetal force is given by F= mv²/r.

Given: m = 0.5 kg, v = 3 m/s, r = 0.5 m

Putting values,

F= mv²/r = 0.5× 3²/0.5 = 9 N

7 0

2 years ago

Read 2 more answers

A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu

Katyanochek1 [597]

Answer:

a) V_f = 25.514 m/s

b) Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

V_f = V_i + a*t

V_f = 20.5 j + 0.31 i *49

V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

V_f = sqrt ( 20.5^2 + 15.19^2)

V_f = sqrt(650.9861)

V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

tan(Q) = (V_y / V_x)

Q = arctan (20.5 / 15.19)

Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

4 0

3 years ago