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3 years ago

Do non metals have luster

1 answer:

4 0

Yes, It probably true. Because the Luster it has a nonmetal, is going to be found in periodic table.

Hope it helped you.

-Charlie

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Due to efficiency considerations related to its bow wake, the supersonic transport aircraft must maintain a cruising speed that

givi [52]

Answer:

decreases.

Explanation:

When the aircraft is flies from the warm air into the colder air then its speed will be decreases.

as we know that

we know mach number is constant

so that here Mach number M is expressed as

M = $\frac{u}{v}$ .............................1

here u is Local flow velocity with respect to the boundarie and v is the speed of sound in the medium

If the aircraft flies from hot air to cold air, the speed of sound in the medium will decrease. But the Mach number remains constant. Therefore, the local flow velocity relative to the boundaries also decreases.

7 0

3 years ago

Please dont delete this question......

Juli2301 [7.4K]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

You are given the melting points of three unknown substances and are asked to predict which one is an ionic compound. You would

nikitadnepr [17]

The answer is <span>highest</span>

4 0

3 years ago

A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

6 0

3 years ago

Projectile motion in two dimensions cannot be determined by breaking the problem into two connected one-dimensional problems.

andriy [413]

The answer I believe is False

8 0

3 years ago