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Sunny_sXe [5.5K]
2 years ago
5

1. Compared to most other metals, what properties do the alkali metals have? А low melting points and high densities B low melti

ng points and low densities C high melting point and high densities D high melting points and low densities​
Physics
1 answer:
ivann1987 [24]2 years ago
3 0

Answer:

B. low melting points and low densities

Explanation:

Alkali metals are any of the monovalent elements found in Group IA of the periodic table. They readily lose their one valence electron to form ionic compounds with nonmetals. Examples of alkali metal are Potassium (K), Lithium (L), and Sodium (Na).

Compared to most other metals, the chemical properties that alkali metals have are low melting points (28.5°C) and low densities that is typically less than 1 grams per cubic centimeters.

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Is snowman chemical or physical changes??
frosja888 [35]
Physical. You are only moving the matter (snow) into a different shape. Hope this helps!

3 0
3 years ago
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Which language style would be most appropriate for the given situation?
polet [3.4K]

Answer:

informal language

Explanation:

you do not need to be formal! you are not at a business conference. technical is not needed either since you are not discussing the intricacies of your job or some computer language.

5 0
2 years ago
Karla Ayala pulls a sled on an icy road (dangerous!). Because of Karla's pull, the tension force is 151 N, and the rope makes a
skelet666 [1.2K]

Answer:

W = 1418.9 J = 1.418 KJ

Explanation:

In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:

W = F.d

W = Fd Cosθ

where,

W = Work Done = ?

F = Force = 151 N

d = distance covered = 10 m

θ = Angle with horizontal = 20°

Therefore,

W = (151 N)(10 m) Cos 20°

<u>W = 1418.9 J = 1.418 KJ</u>

6 0
3 years ago
(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r
Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface U_1=\frac{GM_em}{R_e}

Gravitational Potential Energy at height h is U_2=\frac{GM_em}{R_e+h}

Energy required to lift the satellite E_1=U_1-U_2

E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

h=\frac{R_e}{2}

h=3.19\times 10^6\ m

(b)For greater height E_1  is greater than E_2

thus energy to lift the satellite is more than orbiting around earth

4 0
2 years ago
The specific heat capacity of sea water is 4100 J/Kg°C and the boiling point of 100.6 °C. (i) Calculate the energy required to r
maw [93]

Answer:

334.314 (kJ)

Explanation:

1) the formula for the required energy is: Q=c*m(Bp-t), where c - 4100 J/kg*C; m - 0.9 kg; Bp - 100.6 C; t - 10 C.

2) according to the formula above:

Q=4100*0.9*(100.6-10)=41*9*906=334314 (J).

6 0
2 years ago
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