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serg [7]
3 years ago
14

A teacher wants to perform a classroom demonstration that illustrates both chemical and physical changes. Which would be the bes

t demonstration that she could use?
Physics
2 answers:
goblinko [34]3 years ago
4 0
Putting something in acid, as the object would chemically be broken down and physically break down.
Alex17521 [72]3 years ago
3 0
She can put chalk in vinegar as the vinegar will disintegrate the chalk chemically demonstrating chemical changes. But for physical changes she can break the chalk into small pieces by smashing it with something or her hand.
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Which of the following describes the electron sharing between hydrogen and fluorine A.Hydrogen and fluorine share one electron w
atroni [7]
Hydrogen has one electron in its outermost shell, while fluorine has seven electron in its outermost shell, hence both hydrogen and fluorine needs a single electron to complete its outermost shell. 
That's why there is a single bond between hydrogen and fluorine.
Hence both hydrogen and fluorine share one electron with each other, so option "A" is correct.
5 0
3 years ago
What happens when you brake earphones in to smaller and smaller pieces
nikklg [1K]
They dont work because you crushed them duh
5 0
3 years ago
Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo
Helga [31]

Answer:3.23 m/s^2

Explanation:

Given

\mu_s =0.330

Frictional Force is balanced by force due to car acceleration

Frictional force F_s

F_s=ma_{min}

\mu_sN=ma_{min}

\mu_s\cdot mg=ma_{min}

a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2

6 0
2 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
A wheelchair ramp is 5.2 m long and 0.8 m high. Calculate the ramp’s mechanical advantage
Arlecino [84]

Answer:

5

Explanation:

62

3 0
3 years ago
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