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Iteru [2.4K]
2 years ago
7

Why was John Paul Jones's victory important in the war? His victory increased France's confidence in the skill of the Continenta

l navy. His victory against a large warship dealt a blow to the Spanish navy. His victory against a superior British foe inspired the American troops.
Physics
2 answers:
Sati [7]2 years ago
7 0

Answer:

c

Explanation:

IRINA_888 [86]2 years ago
4 0

Answer:

C. His victory against a superior British foe inspired the American troops.

Explanation:

John Paul Jones is considered the hero in the Revolutionary War. He is known as the Father of the US Navy.

In the Revolutionary War, Jones sided with the American colonists against the British and took hold of naval ships. In 1779, when the British warship <em>Serapis</em> was in conflict with the American warship <em>Bon Homme Richard, </em>Jones plugged the American warship with the Britisher's warship and tossed a grenade into the opponent warship. Thus when Jones was victorious in the war, this boosted the American spirits for the war.

Therefore, option C is the correct answer.

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Why doesn’t the ball roll on forever after being kicked at a soccer game?
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If you have two uncertainties, and they are from two different sources and contribute to the uncertainty of a measurement, what
Darya [45]

The propagation errors we can find the uncertainty of a given magnitude is the sum of the uncertainties of each magnitude.

                           Δm = ∑  | \frac{dm}{dx_i} | \ \Delta x_i

Physical quantities are precise values ​​of a variable, but all measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is used the propagation errors to use the variation of each parameter, keeping the others constant and taking the worst of the  cases, all the errors add up.

If m is the calculated quantity, x_i the measured values ​​and Δx_i the uncertainty of each value, the total uncertainty is

                      Δm = ∑  | \frac{dm}{dx_i } | \ \Delta x_i    | dm / dx_i | Dx_i

               

for instance:

If the magnitude is  a average of two magnitudes measured each with a different error

                     m = \frac{m_1+m_2}{2}

                     Δm = | \frac{dm}{dx_1} |  Δx₁ + | \frac{dm}{dx_2} | Δx₂

                     \frac{dm}{dx_1} = ½

                     \frac{dm}{dx_2} = ½

                     Δm = \frac{1}{2} Δx₁ + ½ Δx₂

                     Δm = Δx₁ + Δx₂

In conclusion, using the propagation errors we can find the uncertainty of a given quantity is the sum of the uncertainties of each measured quantity.

Learn more about propagation errors here:

brainly.com/question/17175455

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