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2 years ago

A 69 kg zebra is traveling 7 m/s east. What is the zebra’s momentum? kg-m/s

Physics

1 answer:

Hitman42 [59]2 years ago

6 0

Linear momentum is the product of mass and velocity. In this case, it is simply:
$69\times 7 = 483 kg\cdot m/s$

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Which of these experiments would make use of qualitative data?

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"A study of the different amounts of time it takes water to evaporate completely"

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2 years ago

An atom has seven protons, eight neutrons, and six electrons. What type of electrical charge does it possess?

Crank

Answer:

positive

Explanation:

No. of positive charge= 7

No.of negative charge = 6

so 7 - 6 = 1 positive charge

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2 years ago

All waves transmit energy. Only one type of wave does not require a medium to transmit energy. That is a _____wave?

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2 years ago

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the force shown in figure 7-15 moves an object from x=0 to x=0.75 m. How much work is done by the force?

Phantasy [73]

Work is force multiplied by the distance the force moves the object

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2 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

2 years ago