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Alenkasestr [34]
2 years ago
6

Chemical equations should be balanced so that they demonstrate the law of conservation of mass. Which of the following statement

s about balanced chemical equations is correct?
A
The products must contain the same numbers and types of atoms as the reactants.

B
The products must contain the same arrangement of atoms as found in the reactants.

C
The products must contain the same numbers but not the same types of atoms as the reactants.

D
The products must contain the same types but not the same numbers of atoms as the reactants.
Physics
1 answer:
faust18 [17]2 years ago
7 0

Answer:

The products must contain the same numbers and types of atoms

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What area of the earth contains semi-solid rock and lava
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Answer:

mantle

Explanation:

Below the crust lies a layer of very hot, almost solid rock called the mantle. Beneath the mantle lies the core. The outer core is a liquid mix of iron and nickel, but the inner core is solid metal. Sometimes, hot molten rock, called magma, bursts through Earth's surface in the form of a volcano.

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What is the y axis on a graph?
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^^^^^^^^^^^^^^^^^^^^^^^^^^^ is correct
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2 years ago
HELP PLS!<br> (LOOK AT THE PICTURE)
Arturiano [62]
The answer is b !!!! Hope it helps
4 0
2 years ago
A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0
2 years ago
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