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Degger [83]
3 years ago
13

Identify the type of chemical reaction that is described.

Physics
1 answer:
Anastasy [175]3 years ago
6 0

Water breaks down into  hydrogen and oxygen: It is a decomposition reaction as a single substance decomposes to give two products.

H_2O \rightarrow H_2 +\frac {1}{2}O_2

Leaves make starch using  chlorophyll and carbon dioxide: Synthesis reaction: as the synthesis reaction involves two or more than two reactants which join together to result into a single main product along with the formation of simple by products.

6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2

Food burns in oxygen gas  and releases a lot of energy: Combustion: Combustion process involves the use of oxygen to give products along with release of energy.

C_6H_{12}O_6 + 6O_2\rightarrow 6CO_2 + 6H_2O

Adding vinegar (acid) to baking soda (alkali) gives  a product that is neither  acidic nor alkaline: Neutralization: acetic acid in vinegar reacts with soda (base) to give salt (neutral) and .

CH_3COOH + NaHCO_3\rightarrow CH_3COONa +H_2O + CO_2





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A=0.80

Explanation:

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An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

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