All categories

2 years ago

What are most elements on the periodic table?

Physics

2 answers:

Inessa [10]2 years ago

6 0

Answer:

Is metals, Non metals, Metalloids

The periodic table on the left separates elements into three groups : the metals (green in the table), nonmetal (orange), and metalloids (blue) . most elements are metal. they are usually shiny,very dense , and only melt at high temperatures .

Molodets [167]2 years ago

5 0

Answer:

Metals, nonmetals and metalloids.

Explanation:

You might be interested in

(e)

Hatshy [7]

Answer:

≈933.3kg/m^3

Explanation:

Density=Mass/Volume

11200kg/12.0= 933.3333kg/m^3

5 0

3 years ago

A boat moves with a speed of +2.5 m/s in a direction 25° north of east. If the mass of the boat is 15,000 kg, what is the moment

nadya68 [22]

Momentum - mass in motion
P=MV
P=(15,000 kg)(2.5 m/s)
P=37 500 kg x m/s to the north
Hope this helps

5 0

2 years ago

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each othe

deff fn [24]

Answer:

$\theta_2 - \theta_1 = 156.93 degree$

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

so we have

$y = A sin\omega t$

$y = \frac{A}{5}$

so now we have

$\frac{A}{5} = A sin\omega t$

now we have

$\theta_1 = 11.53 degree$

so the phase other particle in opposite direction is given as

$\theta_2 = 180 - 11.53 = 168.46 degree$

so we have phase difference given as

$\theta_2 - \theta_1 = 168.46 - 11.53$

$\theta_2 - \theta_1 = 156.93 degree$

7 0

2 years ago

How can accuracy be limited?

lozanna [386]

Answer:

accuracy refers to the deviation of a measurement from a standard or true value of the quantity being measured

5 0

2 years ago

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,

Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

$-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2$

now we know that

$v_0 = 660 m/s$

$M = 5.98 \times 10^{24} kg$

$m = 5 \times 10^9 kg$

$r = 4 \times 10^9 m$

$R = 6.37 \times 10^6 m$

now from above formula

$GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

now we have

$2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2$

now plug in all data

$2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2$

$v = 7934.2 m/s$

5 0

2 years ago