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2 years ago

Which statement describes ionic compounds?

Physics

1 answer:

sladkih [1.3K]2 years ago

8 0

Answer:the answer is b

Explanation:

This is because in an ionic bond negative and positive bonds are present

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580 nm light shines on a double slit

lesantik [10]

Answer:

Angular position of the first minimum is 0.27 degree

Explanation:

As we know by the formula of single slit diffraction pattern

$a sin\theta = m\lambda$

here we have

a = d = 0.000125 m

for m = 1

$\lambda = 580 nm$

now we have

$0.000125 sin\theta = 580 \times 10^{-9}$

$sin\theta = 4.64 \times 10^{-3}$

$\theta = 0.27 degree$

6 0

3 years ago

1) There are many positive and negative aspects associated with nuclear power. Which is a negative aspect associated with nuclea

kherson [118]

Nuclear power emits radiation.

5 0

3 years ago

What two elements are named after Dimitri Mendeleev?

bezimeni [28]

I'm pretty sure there is only one element named after Mendeleev: Mendelevium.

8 0

3 years ago

Read 2 more answers

A cat chases a mouse across a 1.2 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the

Drupady [299]

The cat has two directions of motions:
The horizontal motion = Dx = 2.2 m
The vertical motion = Dy = -1.3 m (negative sign indicates that the cat is falling)
a = 9.8 m/sec^2
Vy = zero (since you are not moving up)

From the laws of motion:
Dy = Vyt + 0.5ayt^2
-1.3 = 0(t) + 0.5(-9.8)t^2
t = 0.52s

Then, again using the laws of motion (but for the horizontal direction this time)
Dx = Vxt
2.2 = Vx0.52 
Vx = 2.2/0.52 
= 4.23 m/s

Therefore the cat's speed when it slid off the table is 4.23 m/s horizontally.

7 0

3 years ago

How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight

Natalka [10]

By using Ohm's law, we can find what should be the resistance of the wire, R:
$R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega$

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
$r=35 mm=0.35 \cdot 10^{-3} m$
So the area is
$A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2$

And by using the resistivity of the Aluminum, $\rho=2.65 \cdot 10^{-8} \Omega m$ , we can use the relationship between resistance R and resistivity:
$R= \frac{\rho L}{A}$
to find L, the length of the wire:
$L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m$

4 0

3 years ago