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14 hours ago

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Wahadło matematyczne wykonuje 10 pełnych wahnięć w ciągu 20 sekund. Oblicz częstotliwość wahań tego wahadła. Jaką drogę pokona k

ulka tego wahadła w tym czasie, jeśli wiadomo, że amplituda jego drgań wynosi 3 cm?

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Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to

Sav [3153]

Arginine is classified as a basic amino acid since it has two amino groups alongside a single acid group. At a low pH level, all ionizable groups are protonated. As the pH rises slightly, the acid group loses its proton. When the pH increases further, one of the amino groups also loses a proton. At considerably high pH levels, none of the ionizable groups remain protonated.

Pkas

<span> <span><span> <span> pka1 = 1.82 </span> <span> pka2 = 8.99 </span> <span> pka3 = 12.48 </span> </span> </span></span> Thus, 9.20 is above the second pKa and below the third pKa. This indicates that the acid has already lost its proton, as has one of the amino groups, while the second amino group remains protonated. When an acid is not protonated, it carries a negative charge. An unprotonated amino group is neutral, whereas when protonated, the amino group bears a positive charge. Therefore, this amino acid exhibits one positive charge (from one of the amino groups) and one negative charge (from the acid), resulting in an overall neutral charge.

4 0

1 month ago

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

Ostrovityanka [3204]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Utilizing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Now let’s solve for t.

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

The computed velocity magnitude is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The ball's angle is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

3 months ago

Use the formula t = (0.25) s1/2 to find the time t in seconds it will take a stone to drop a distance s of 200 feet. Round your

inna [3103]

Answer:

The duration, t = 3.53 seconds

Explanation:

The following information is provided:

The equation to calculate the time t is expressed as:

$t=(0.25)s^{1/2}$ ...... (1)

Where

s denotes the distance in feet

We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet

Substituting the value of s into equation (1) yields:

$t=(0.25)\times (200)^{1/2}$

t = 3.53 seconds

Thus, the time taken by the object is 3.53 seconds, which provides the required answer.

4 0

3 months ago