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3 years ago

A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.4 m/s. The drag force is of the form bv^2 What is the value of b?

Physics

1 answer:

julia-pushkina [17]3 years ago

6 0

Drag Force = bv^2 = ma; a = g = 9.81 m/s^2

b = mg/v^2 = (0.0023×9.81)/(9.4^2)

b = 0.000255

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3 years ago

4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The

GalinKa [24]

Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time $t$ , the horizontal position $x$ and vertical position $y$ of the ball are given respectively by

$x = v_i \cos(\theta_i) t$

$y = v_i \sin(\theta_i) t - \dfrac g2 t^2$

and the horizontal velocity $v_x$ and vertical velocity $v_y$ are

$v_x = v_i \cos(\theta_i)$

$v_y = v_i \sin(\theta_i) - gt$

The ball reaches its maximum height with $v_y=0$ . At this point, the ball has zero vertical velocity. This happens when

$v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g$

which means

$y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g$

At the same time, the ball will have traveled half its horizontal range, so

$x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g$

Solve for $v_i$ and $\theta_i$ :

$\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0$

Since $0^\circ$ , we cannot have $\sin(\theta_i)=0$ , so we're left with (e)

$3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}$

Now,

$\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}$

$\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}$

so it follows that (d)

$R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}$

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

$v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}$

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

$v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}$

Then with $v_y=0$ , the ball's speed $v$ is

$v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}$

Finally, in the work leading up to part (e), we showed the time to maximum height is

$t = \dfrac{v_i \sin(\theta_i)}g$

but this is just half the total time the ball spends in the air. The total airtime is then

$2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}$

and the ball is in the air over the interval (a)

$\boxed{0 < t < 2\sqrt{\frac R{3g}}}$

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Answer:

c) wafers of crystalline silicon treated with metals that absorb solar radiation and generate electricity

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Photovoltaic solar cells are used in solar panels. Most of the photovoltaic cells are made from crystalline silicon.

It has a property using which it converts the light (solar radiation) falling on it to electrical energy. This process occurs due to the photovoltaic effect.

It is a way generating energy without consuming fossil fuels. Thus it reduces the carbon emission and thus plays a major role in environment protection.

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3 years ago

A cylinder measuring wide and high is filled with gas. The piston is pushed down with a steady force measured to be . Calculate

jekas [21]

Explanation:

Let us assume that the given cylinder is 2.6 cm wide and its height is 3.1 cm. And, when piston is pushed down then the steady force is equal to 15 N.

Now, radius of the cylinder will be as follows.

r = $\frac{diameter}{2}$

= $\frac{2.6 cm}{2}$

= 1.3 cm

or, = 0.013 m (as 1 m = 100 cm)

As, area of cylinder = $\pi \times r^{2}$

= $3.414 \times (0.013 m)^{2}$

= $5.77 \times 10^{-4} m^{2}$

Relation between pressure and force is as follows.

Pressure = $\frac{Force}{Area}$

= $\frac{15 N}{5.77 \times 10^{-4} m^{2}}$

= 25996 $N/m^{2}$

Since, 1 $N/m^{2}$ = 1 Pa (as 1 kPa = 1000 Pa)

Therefore, P = 25996 $N/m^{2}$

= 25.99 kPa

= 26 kPa (approx)

Thus, we can conclude that pressure of the gas inside the cylinder is 26 kPa.

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3 years ago

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denis23 [38]

Answer:

the weight of the large stone is greater than a small one

Explanation:

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3 years ago