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3 years ago

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What the unit of work?

1 answer:

ra1l [238]3 years ago

5 0

Answer:

yes

Explanation:

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What is the Octet rule

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the octate rule says that electron after completing all the valency of the element doesnot goes as 9 but continues next round after completing the next octat then the first shell completes the all electron valency .thanks if you dont understand then please comment l will try to solve your confusion

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3 years ago

Read 2 more answers

A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta

Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, $\theta = 54^{\circ}$

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, $R = \frac{v^{2}sin2\theta}{g}$

$R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m$

Time of flight, t' = $\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s$

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

$s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m$

Distance traveled vertically:

$s_{y} = vcos\theta - \frac{1}{2}gt^{2}$

$s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m$

Now,

$s_{x} = \frac{mx + m'x'}{M}$

$10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}$

x' = 3.535 m

Similarly,

$s_{y} = \frac{my + m'y'}{M}$

$4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m$

The location of the other fragment is at (3.535, 1.162)

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4 years ago

There is evidence that elephants communicate via infrasound, generating rumbling vocalizations as low as 14 Hz that can travel u

amid [387]

Answer:

$L = 96.2 dB$

Explanation:

As we know that the level of intensity is given as

$\beta = 10Log(\frac{I}{I_o})$

here we know

$\beta = 104 dB$

$104 = 10Log(\frac{I}{10^{-12}})$

$I = 0.025 W/m^2$

now the sound is travelling in all possible directions so we have

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$\frac{0.025}{I_2} = \frac{10^2}{4.1^2}$

$I_2 = 4.2 \times 10^{-3} W/m^2$

now for level of sound we have

$L = 10Log(\frac{4.2 \times 10^{-3}}{10^{-12}})$

$L = 96.2 dB$

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3 years ago