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3 years ago

Which ball (if either) has the greatest speed at the moment of impact

Physics

2 answers:

Ierofanga [76]3 years ago

5 0

Are there any options??

I would have to say metal of course but without options I can't assume anything

Alenkinab [10]3 years ago

4 0

Whichever ball is placed higher, but if the balls are at the same level then both experience the same acceleration thus same speed.

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A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

$\Omega = \frac{Mgd}{I\omega}$

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

$\omega$ = Angular velocity

Replacing the value for moment of inertia

$\Omega= \frac{MgR}{MR^2 \omega}$

$\Omega = \frac{g}{R\omega}$

The value for our angular velocity is not in SI, then

$\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})$

$\omega = 104.7rad/s$

Replacing our values we have that

$\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}$

$\Omega = 1.17rad/s$

The precession frequency is

$\Omega = \frac{2\pi rad}{T}$

$T = \frac{2\pi rad}{\Omega}$

$T = \frac{2\pi}{1.17}$

$T = 5.4 s$

Therefore the precession period is 5.4s

7 0

2 years ago

What will most likely occur if sulfur forms an ionic bond with another element?

patriot [66]

If sulfur forms an ionic bond with another element it will more then likely create a chemical reaction of some sort.

6 0

3 years ago

Read 2 more answers

what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

Approximately $5.11 \times 10^{-19}\; {\rm J}$ .

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

Look up Planck's constant: $h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}$ .

Apply the Rydberg formula to find the wavelength $\lambda$ (in vacuum) of the photon in question:

$\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}$ .

The frequency of that photon would be:

$\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}$ .

Combine this expression with the Rydberg formula to find the frequency of this photon:

$\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}$ .

Apply the Einstein-Planck equation to find the energy of this photon:

$\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}$ .

(Rounded to three significant figures.)

6 0

2 years ago

A cat falls from a table of height 1.3 m. What is the impact speed of the cat? ( i want the answer NOT THE FORMULA)

telo118 [61]

The impact speed will be
v^2 = 2*9.8*1.3
v^2 = 25.48
v= 5.04 m/s

4 0

3 years ago

Read 2 more answers

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.