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3 years ago

WDYK? Can you remember how ionic bonds were defined from your physical science class?

1 answer:

5 0

Yes, ionic bond is formed by the transfer of electrons from and an atom of metallic element to the atom of an non metallic element

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When the moon is waxing, it's lighted part appears

kvv77 [185]

To increase it's size

Waxing is the opposite of waning, which is to decrease.

7 0

3 years ago

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What is the wave speed of a 6 m wave with a frequency of 20 Hz?

andrew-mc [135]

Answer: Wave speed= frequency x wavelength

=20 x 3

=60 m/s

Explanation:

8 0

3 years ago

Which of the following measurements has two significant digits?

myrzilka [38]

0.022 has 2 digits because you would count from the left starting with the first nonzero number

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3 years ago

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A ship's wheel has a moment of inertia of 0.930 kilogram·meters squared. The inner radius of the ring is 26 centimeters, and the

Vikki [24]

We can use the formula of the moment of inertia given by:

$r\cdot F=I\alpha$

Where:

r = Distance from the point about which the torque is being measured to the point where the force is applied

F = Force

I = Moment of inertia

α = Angular acceleration

So:

$\begin{gathered} r\cdot F=(-0.26\times314+290\times0.32)=92.8-81.64=11.16 \\ I=0.930 \\ so,_{\text{ }}solve_{\text{ }}for_{\text{ }}\alpha: \\ \alpha=\frac{r\cdot F}{I} \\ \alpha=\frac{11.16}{0.930} \\ \alpha=\frac{12rad}{s^2} \end{gathered}$

Answer:

12 rad/s²

8 0

1 year ago

Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the

Scrat [10]

Answer: $1.8\°$

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d=3.4(10)^{-5}m$ is the width of the slit

$\lambda=540 nm=540(10)^{-9}m$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the second-order diffraction angle is given when $n=2$ , hence equation (1) becomes:

$dsin\theta_{2}=2\lambda$ (2)

Now we have to find the value of $\theta_{2}$ :

$sin\theta_{2}=\frac{2\lambda}{d}$ (3)

Then:

$\theta_{2}=arcsin(\frac{2\lambda}{d})$ (4)

$\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})$ (5)

Finally:

$\theta_{2}=1.8\°$ (6)

5 0

3 years ago