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sveta [45]
2 years ago
6

What are valence electrons?

Physics
1 answer:
SashulF [63]2 years ago
8 0

Answer:

The correct answer is D. Electrons in an atom that can bond with other atoms.

Explanation:

For those of you that need it still

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The gravitational force between two objects:
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C. Gravitational force between two objects of masses m1/m2

is defined by Gm1xm2/r^2, where G is a universal constant, and r is the distance between the masses. This shows you why B is wrong; it would increase. Also, A is not a good choice, as shown. In the case of B, it would increase by say, 2/1 to 4/1 by doubling ones mass. In C's case the bottom would decrease on the fraction alot (square) if they were moved closer, thus the force would be greater.

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Why is most space exploration accomplished with spacecraft that do not have astronauts on board
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The demand for your hand-made skateboards, in weekly sales, is q = −3p + 700 if the selling price is $p. You are prepared to sup
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you should sell your skateboards at $240

Explanation:

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q (sales) = q (supply)

-3p + 700 = 2p - 500

700 + 500 = 2p + 3p

1200 = 5p

1200/5 = p

$240 = p

3 0
3 years ago
Newton's three laws of motion allow scientists to calculate the following properties of moving objects. speed density path posit
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3 years ago
A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe
Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet =  327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf  = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

     = 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

   = 0.00353 × 128 × (327-40)

   = 0.00353 × 128 × 287

   = 129.68 J

Quantity of heat required to melt the bullet:

Q = mH_f

Q = 0.00353 × 24500

   = 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

                                                                      = 216.17 J

3 0
3 years ago
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