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3 years ago

Which one of the following is the general equation for a single-displacement reaction?

Physics

1 answer:

azamat3 years ago

8 0

Answer: A. AB + C → AC + B

Explanation:

A single displacement reaction is also known as single replacement reaction. A reaction in which an element replaces another element from the compound is known as single displacement reaction.

AB + C → AC + B

Here, element C replaces another element B from compound AB.

Thus, option A represents is a general equation for a single displacement reaction

Option B represents a double displacement reaction where one element of a compound replaces another element of another compound and vice-versa, option C is a synthesis reaction where two reactants or more combine to form one product. Option D is a decomposition reaction in which multiple products are formed from a single reactant.

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(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

Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (

RideAnS [48]

Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

4 0

3 years ago

In 8.5 s a fisherman winds 2.4 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be constant as an approximation)

SSSSS [86.1K]

Answer:

9.412 rad/s.

Explanation:

Velocity is the rate of change of an object's position.

V = x/t

Where x is the distance in m

= 2.4 m

t is time taken in s

= 8.5 s

V = 2.4/8.5

= 0.2824 m/s.

Equating linear velocity and angular velocity,

V = ω*r

Where,

ω Is the angular speed in rad/s

r is the radius of the circle in m

= 3 cm

= 3cm * 1m/100 cm = 0.03 m

ω = V/r

= 0.2824/0.03

= 9.412 rad/s.

4 0

4 years ago

The gas within a cylinder of an engine undergoes a net change in volume of 1.50 × 10-3 m3 when it does work at a constant press

Lemur [1.5K]

Answer:

work =p×v =3.27×10^5×1.5×10^-3 =490.5 joule

efficiency =w/q in

:. qin= w/efficiency =490.5/0.225=2180 joule

qout =q in - work =1689.5 joule

q out is work given as heat

4 0

4 years ago

If the mass of the cement is 15 000 kg, calculate the density of this cement sample in kgm-3

MrRa [10]

Explanation:

Hi there!

Mass of the container (m) = 15000 kg

Also from the figure;

Volume of the container (V) = l*b*h

{ From figure length = 2m , breadth = 1.1m, height = 2.5m}

So,

V = (2*1.1*2.5) m³

= 5.5m³

Then;

Density (d) = m/v

= 15000/5.5

= 2727.27 kgm³

Therefore, the density is 2727.27 kgm³.

Hope it helps!

5 0

2 years ago