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Y_Kistochka [10]
3 years ago
7

A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i

nstant the first ball is released. determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. (use h and g as appropriate in your equation.)
Physics
1 answer:
Darya [45]3 years ago
8 0

The position of the first ball is

y_1=h-\dfrac g2t^2

while the position of the second ball, thrown with initial velocity v, is

y_2=vt-\dfrac g2t^2

The time it takes for the first ball to reach the halfway point satisfies

\dfrac h2=h-\dfrac g2t^2

\implies\dfrac h2=\dfrac g2t^2

\implies t=\sqrt{\dfrac hg}

We want the second ball to reach the same height at the same time, so that

\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2

\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)

\implies h=v\sqrt{\dfrac hg}

\implies v=\sqrt{hg}

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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 equals​0) 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
A certain shade of blue has a frequency of 7.24 × 1014 Hz. What is the energy of exactly one photon of this light?
ANEK [815]
E = hf

E = 6.63* 10 ⁻³⁴ * 7.24* 10¹⁴

<span>E = 4.80012 × 10⁻¹⁹ J</span>
4 0
3 years ago
Read 2 more answers
A 2 kg ball is dropped above the surface of Planet X. If the gravitational field strength at the surface of Planet X is 5 N/kg,
Trava [24]

Given data:

* The mass of the ball is 2 kg.

* The gravitational field strength at the surface of planet X is 5 N/kg.

Solution:

The weight of the ball on the planet X is,

W=ma

where m is the mass of ball, a is the gravitational field strength,

Substituting the known values,

\begin{gathered} W=2\times5 \\ W=10\text{ N} \end{gathered}

Thus, the weight of the ball on the surface of planet X is 10 N.

3 0
1 year ago
Mechanical energy is conserved in the presence of which of the following types of forces?magnetic
Tju [1.3M]
<h2>Answer: electrostatic and gravitational force </h2><h2 />

Mechanical energy remains constant (conserved) if only <u>conservative forces</u> act on the particles.  

In this sense, the following forces are conservative:  

-Gravitational  

-Elastic

-Electrostatics  

While the Friction Force and the Magnetic Force are not conservative.

According to this, mechanical energy is conserved in the presence of electrostatic and gravitational forces.

7 0
3 years ago
In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times
Sergeeva-Olga [200]

Answer:

0 m/s

Explanation:

Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.

In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

Average velocity = displacement/time

v=0/30

v = 0 m/s

Hence, the correct option is (1).

6 0
3 years ago
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