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valentina_108 [34]
3 years ago
5

Is carbon a metal, nonmetal, or semi-metal

Physics
1 answer:
Vlada [557]3 years ago
5 0

Answer: Non metal.

Explanation: Carbon has 4 electrons in it’s valence shell which makes it a metalloid but commonly it is considered as a non metal.

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What is the best unit to use when measuring the mass of a mineral sample?
Ksivusya [100]
Measuring density: Measure the mass (in grams) of each mineral sample available to you. The mass of each sample is measured using a balance or electronic scale. Record mass on a chart.
8 0
3 years ago
A light ray incident on a block of glass makes an incident angle of 50.0° with the normal to the surface. The refracted ray in t
Vladimir [108]

Answer:

The index of refraction of the glass is 1.3

Explanation:

Given data:

i = incident angle = 50°

r = refracted angle = 36.1°

The index of refraction according Snell´s law is:

n=\frac{1*sini}{sinr} =\frac{1*sin50}{sin36.1} =1.3

4 0
4 years ago
4. Calculate the total resistance for two 180 ohm resistors connected in<br> parallel
solmaris [256]

Answer:

90 ohms

Explanation:

1/r = 1/180 + 1/180

1/r= 2/180

take the reciprocal of 2/180 which is 180/2 and its 90 ohms

3 0
3 years ago
A substance is round in one container and has a volume of one liter. In a much larger container, the substance is rectangular bu
UkoKoshka [18]
Answer: liquid

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Liquids also have a set volume, but can flow to take the shape of the bottom of their container.
8 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
3 years ago
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