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tatiyna
2 years ago
11

During WW2, battleships would shoot their 16-inch main guns to project an 800km projectile at the enemy as far as 22 miles aways

. If the projectile leaves the muzzle moving at +850m/s, what is the momentum of that projectile?
Chemistry
1 answer:
Nesterboy [21]2 years ago
3 0

The momentum of the projectile is 6.8 * 10⁵ kgm/s

<h3>Momentum and projectiles</h3>

Momentum of an object is the product of the mass of that object and its velocity.

  • Momentum, p = m*v

where m is mass of object

v is velocity of object

A projectile is an object launched into space and allowed to fall freely under the force of gravity.

The momentum of a projectile is conserved in the horizontal direction since no net force acts on the projectile in the horizontal direction.

using the formula for momentum;

p = m*v

Assuming the mass of the projectile to be 800 kg

Velocity of the projectile = 850 m/s

p = 800 kg * 850 m/s

p = 6.8 * 10⁵ kgm/s

Therefore, the momentum of the projectile is 6.8 * 10⁵ kgm/s

Learn more about momentum and projectiles at: https://brainly.in/question/201038

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Actually, there are four kinds of reptile motion: 

Concertina - vermiform. Circular muscles around the snake squeeze the front of the snake's body out long, then the latter half is pulled forward. 

Rectilinear crawling - Belly scutes are moved forward individually in a wave-like motion. 

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Lateral undulation - Most common form of movement. Snake presses on alternating pressure points to force body forward (or backward)

(taken from a user on Yahoo from Correct Answers)
3 0
3 years ago
The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri
Nimfa-mama [501]

Answer: The expression for equilibrium constant is \frac{[NH_3]^2}{[H_2]^3[N_2]}

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as k_c

For a general reaction:

aA+bB\rightleftharpoons cC+dD

The equilibrium constant is written as:

k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}

Chemical reaction for the formation of ammonia is:

N_2+3H_3\rightleftharpoons 2NH_3

k_c=1.6\times 10^2

Expression for k_c is:

k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}

1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}

8 0
3 years ago
Give the chemical symbol of an element in the third period (row) of the periodic table with four 3p electrons.
Ann [662]

<u>Answer: </u>The chemical symbol of the element is Sulfur.

<u>Explanation:</u>

The element which is present in third period of the periodic table having four 3p electrons is Sulfur. Sulfur is the 16th element of the periodic table which has 6 valance electrons.

The electronic configuration of this element is:  [Ne]3s^23p^4

This element is considered as a non-metal because it will accept electrons in order to attain stable electronic configuration.

Hence, the chemical symbol of the element is Sulfur.

7 0
3 years ago
In your video respond to the following question/promt. Make sure you state a clear claim and explain with
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Answer:

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Explanation:

6 0
2 years ago
Read 2 more answers
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
3 years ago
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