1. Use the periodic table to determine which of the following groups of elements would have properties most similar to arsenic (As)


A . Boron (B) , silicon (Si) , and germanium (Ge)

B. Nitrogen (N), Phosphorous (P) , and antimony (Sb)

C. Germanium (Ge) , selenium (Se) , and bromide (Br)

D. Gallium (Ga) , germanium (Ge) , and selenium (Se)

Answer: microwaves, infrared light, ultraviolet light, X-rays and gamma-rays

Hopes this helps!

Red light

Orange light

Green light

Violet light

Laser light

Bright light

Dim light

Sun light

Answer:

hiiiiiiiiiiiiiiiiii

mm

Answer:

It is called the angle of refraction

Answer:

Option D

Explanation:

Taking 2020 unit test for edu right now.

Answer: solute: sugar, coffee powder, salt, cocoa powder, liquid detergent, monosodium glutamate (vetsin).

solvent: cow's milk, water, fruit extract

Explanation:

A solution is a mixture that is made up of two or more substances including solute and solvent.

The solute is one which can be dissolved in a liquid part that is solvent. The solute can be liquid or solid. Solutes are sugar, salt, coffee powder, cocoa powder, liquid detergent, monosodium glutamate.

The solvent is one which dissolves in it solute particles or molecules. Solvent are cow's milk, water, and fruit extract.

Answer:

Matter in its gas state has the most thermal energy than when it is a solid or a liquid. Because gasses have more thermal energy than others, they move differently than others.

Explanation:

;P

Answer:

Julie's car.

Explanation:

The acceleration of an object is equal to the rate of change of velocity. It can be given by :

[tex]a=\dfrac{v-u}{t}[/tex]

Jodi's car moves at a constant 60 mph for 10 s.

Julie's car transitions from a speed of 20 mph to a speed of 30 mph for the same time interval i.e. 10 s.

For Jodi's car, the change in velocity is equal to 0. As a result, its acceleration is equal to 0 m/s².

For Julie's car,

Initial speed, u = 20 mph = 8.9408 m/s

Final speed, v = 30 mph = 13.4112 m/s

Acceleration,

[tex]a=\dfrac{13.4112-8.9408}{10}\\\\a=0.447\ m/s^2[/tex]

It is clear that Julie's car has the greater acceleration than the Jodi's car.

Answer:

Yes, the value of g affected by the radius.

Explanation:

The formula for the force of gravity of 2 objects is

[tex]F_{gravity} = G\frac{m_{1}m_{2}}{r^2}[/tex], where m1 and m2 are the masses of the 2 objects, r is the radius, and G is the gravitational constant, which is approximately [tex]6.67 \cdot 10^{-11}[/tex].

Therefore, as the radius if bigger, the force of gravity is going to be smaller exponentially.  

Answer: A

Explanation:

The guy above me was right

Motion is caused by the forces acting on a body. A body will move in the direction of in which the net force is greater. Therefore, objects 1 and 2 will not move but 3 and 4 will accelerate upward.

From the table ; the upward force of 18N is lesser than the force maintained by the objects at rest. Hence, objects 1 and 2 will not move.

However, objects 3 and 4 have forces 16N and 8N respectively which are lesser than the upward force, hence, the body will move upward since the upward force is greater.

Therefore, objects 1 and 2 will not move but 3 and 4 will accelerate upward.

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When the object falls through a viscous fluid with terminal velocity the total downward force becomes equal to total upward force

[tex]\\ \rm\rightarrowtail mg=U+F_v[/tex]

[tex]\\ \rm\rightarrowtail mg=m_1g+6\pi \eta rv_T[/tex]

[tex]\\ \rm\rightarrowtail (m-m_1)g=6\pi \eta rv_T[/tex]

[tex]\\ \rm\rightarrowtail v\rho g=6\pi \eta rv_T[/tex]

[tex]\\ \rm\rightarrowtail \dfrac{4}{3}\pi r^3(\rho-\sigma)g=6\pi\eta rv_T[/tex]

[tex]\\ \rm\rightarrowtail v_T=\dfrac{4gr^2(\rho-\sigma)}{18\eta}[/tex]

[tex]\\ \rm\rightarrowtail v_T=\dfrac{2}{9}\dfrac{gr^2(\rho-\sigma)}{\eta}[/tex]

Answer:

Explanation:

Waves and wave functions are direct ways to graph vibrational motion. Specifically, for cases dealing with vibrating strings. If you were to move a tensioned string up and down with your arm, you would produce a wave with a given frequency, period, and amplitude.

Answer:

d = 3.6 m

Explanation:

The formula for Work: W = F * d

Where F is force in Newtons (N) and d is the distance travelled in meters (m). In order to do work, you need to force an object throughout a distance.

That's why pushing a wall (that doesn't move) is not considered work, because you can't move the wall to a distance.

So what we know:

F = 5,000 N

W = 18,000 J

d = W / F (according to the formula, right?)

d = 18,000 / 5,000 = 3.6 m

Answer:

El plomo, el aluminio y la plumavit caerán al mismo tiempo.

Explanation:

En cinemática, la caída libre es un movimiento dónde solamente influye la gravedad. En este movimiento se desprecia el rozamiento del cuerpo con el aire, es decir, se estudia en el vacío.

La caída libre es un movimiento uniformemente acelerado de un cuerpo dónde solamente influye la gravedad y se estudia en el vacío, es decir, se debe despreciar  el rozamiento del cuerpo con el aire. Entonces a este movimiento se denomina caída libre ideal. Y en esas condiciones, la aceleración que adquiriría el cuerpo sería debida exclusivamente a la gravedad, siendo independiente de su masa.

Entonces, el plomo, el aluminio y la plumavit caerán al mismo tiempo ya que adquirirían la misma aceleración, g, que es la aceleración de la gravedad.

For a block of solid copper sits on a flat, level table. Copper has a density of 8. 94 ✕ 103 kg/m3, the volume of the block  is mathematically given as

V = 2.28x10^3

Generally, the equation for the volume of the block is mathematically given as

V = m/p

V = 20.4/8.94x10^3

V = 0.00228

V = 2.28x10^3

In conclusion, the volume of the block is

V = 2.28x10^3

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Answer:

Approximately [tex]11121\; {\rm kg}[/tex] (rounded to the nearest whole number as required, assuming that planet Earth is a uniform sphere.)

Explanation:

Let [tex]G[/tex] denote the gravitational constant. [tex]G \approx 6.6743 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}[/tex].

Let [tex]M[/tex] denote the mass of planet Earth. Let [tex]m[/tex] denote the mass of the telescope. Let [tex]r[/tex] denote the distance between the telescope and the center of planet Earth.

Note the unit conversion for the distance [tex]r[/tex]:

[tex]\begin{aligned}r &= 6940 \; {\rm km} \times \frac{10^{3}\; {\rm m}}{1\; {\rm km}} \\ &= 6.940 \times 10^{6}\; {\rm m}\end{aligned}[/tex].

By Newton's Law of Universal Gravitation, the magnitude of the gravitational force between planet Earth and this telescope would be:

[tex]\begin{aligned}W &= \frac{G\, M\, m}{r^{2}}\end{aligned}[/tex].

Rearrange this equation to find the mass [tex]m[/tex] telescope in terms of [tex]G[/tex], [tex]M[/tex], and [tex]r[/tex]:

[tex]\begin{aligned}G\, M\, m &= W\, r^{2}\end{aligned}[/tex].

[tex]\begin{aligned}m &= \frac{W\, r^{2}}{G\, M}\end{aligned}[/tex].

Substitute in the value of [tex]G[/tex], [tex]M[/tex], and [tex]r[/tex]:

[tex]\begin{aligned}m &= \frac{W\, r^{2}}{G\, M} \\ &\approx \frac{9.21 \times 10^{4}\; {\rm N} \times (6.940 \times 10^{6}\; {\rm m})^{2}}{6.6743 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}} \times 5.98 \times 10^{24}\; {\rm kg}} \\ &\approx 11121\; {\rm kg}\end{aligned}[/tex].

Answer:

11,121 kg

Explanation:

Given parameters:

Height of Everest  = 8848m

Acceleration due to gravity on earth  = 9.8m/s²

Radius of the earth = 6400km = 6400000m

Unknown:

Acceleration due to gravity at the peak of Everest  =?

Solution:

According to Newton's law of universal gravitation, we have that acceleration due to gravity is proportional to the altitude.

 To solve this problem, use the mathematical expression below;

        g[tex]_{e}[/tex]   = g(1 - [tex]\frac{2h}{R}[/tex])

where g[tex]_{e}[/tex]  is gravity on top of Everest

            g is the gravity on the earth surface

            h is the height of the mountain

            R is the radius of the earth

      g[tex]_{e}[/tex]   = 9.8 (1 -[tex]\frac{2 x 8848}{6400000}[/tex] )

      g[tex]_{e}[/tex]  = 9.8 (1-0.002765)   = 9.77m/s²

The acceleration due to on top of Everest is  9.77m/s²

Answer:

m = 2.34 kg

Explanation:

The given data shows the relation between the potential energy and the height of an object. The formula for the potential energy is given by :

E = mgh

g is the acceleration due to gravity

We have,

E = 98.61 J, h = 4.3 m

Putting all the values to find m. So,

[tex]m=\dfrac{E}{gh}\\\\m=\dfrac{98.61}{9.8\times 4.3}\\\\m=2.34\ kg[/tex]

So, the mass of the object is 2.34 kg.

Answer:

1350 meters

Explanation:

this is easy

15 meters per second

90 seconds

90 * 15 = 1350

Answer:

-56.9 m/s

Explanation:

Given:

Δy = -165 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-165 m)

v = -56.9 m/s

The food package will strike the ground at 11 degrees below the horizontal.

The time for the food package to hit the ground is calculated as follows;

h = vt + ¹/₂gt²

let the initial velocity be horizontal

4900 = 0(t) + (0.5 x 9.8)t²

4900 = 4.9t²

t² = 4900/4.9

t² = 1,000

t = √1,000

t = 31.62 s

vf(y) = vo + gt

vf(y) = 0 + (31.62 x 9.8)

vf(y) = 309.88 m/s

The horizontal component of the speed will be constant, while vertical component will change

[tex]tan(\theta ) = \frac{V_y}{V_x} \\\\\theta = tan^{-1} (\frac{V_y}{V_x})\\\\\theta = tan^{-1} (\frac{309.88}{58.1} )\\\\\theta = 79^0[/tex]

Angle below the horizontal = 90 - 79 = 11⁰

Thus, the food package will strike the ground at 11 degrees below the horizontal.

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Kepler's Law (3)

T² ≈ r³

The square of the orbital period ≈ the cube the distance to the sun

The farther away the planet is, the bigger the radius, the bigger the period

Answer: From What i Seeing it is Option B

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state the effect of momentarily touching the conductor with a finger while the charged rod is still near the conductor

[tex]\small\color{blue}{⊱─━━━━━━━━━⊱༻●༺⊰━━━━━━━━━─⊰}[/tex]

If you have an insulator that's charged up with a negative charge, and you touch a conductor, the negatively charged excess electrons will move to the conductor until the electrons would rather not move anywhere at all. When this happens, we say that the system has reached equilibrium

[tex]\small\color{blue}{⊱─━━━━━━━━━⊱༻●༺⊰━━━━━━━━━─⊰}[/tex]

-PoliteCat

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[tex]\large\boxed{\text{Together We Go Far!}}[/tex]