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3 years ago

Use your calculator to evaluate -3.7 meter/second-13.9 meter/second 21.4 second-72 second

Physics

1 answer:

cluponka [151]3 years ago

6 0

Answer :

(-3.7 meter/second) - (13.9 meter/second) = -17.6 meter/second

(21.4 second) - (72 second) = -50.6 second

Explanation :

(1) As we are given the expression :

(-3.7 meter/second) - (13.9 meter/second)

Now we have to evaluate this expression, we get:

⇒ -17.6 meter/second

(2) As we are given the expression :

(21.4 second) - (72 second)

Now we have to evaluate this expression, we get:

⇒ -50.6 second

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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

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

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

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

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3 years ago

A 195 g glider is moving at 5.3 m/s on a frictionless air track. It then collides with a stationary 295 g glider.

lord [1]

Answer:For a perfectly elastic collision, the final velocities of the carts will each be 1/2 the velocity of the initial velocity of the moving cart. For a perfectly inelastic collision, the final velocity of the cart system will be 1/2 the initial velocity of the moving cart.

Explanation:

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Par 1/2

BaLLatris [955]

part 1

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mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

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part 2

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convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

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Which is present when iodine changes from brown to blue or purple?

vladimir1956 [14]

If iodine is added to a starch solution, they react with each other and the iodine darkens to an almost pitch black.

however, if iodine is added to a solution containing no starch, it will show up only as an extremely pale brown. almost colorless and hardly visible.

when following the changes in some inorganic oxidation reduction reactions, iodine may be used as an indicator to follow the changes of iodide ion and iodine element. soluble starch solution is added. only iodine element in the presence of iodide ion will give the characteristic blue black color. neither iodine element alone nor iodide ions alone will give the color result.

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State Pascal's principle of transmission of pressure

bulgar [2K]

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