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aalyn [17]
4 years ago
5

What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2.40×104

m/s2? What is the magnitude of the force exerted on the ship by the artillery shell?
Physics
2 answers:
ladessa [460]4 years ago
7 0

<u>Answer:</u> The force exerted on the artillery shell is 2.64\times 10^6N  and the magnitude of force exerted on the ship by artillery shell is 2.64\times 10^6N

<u>Explanation:</u>

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

F=ma

where,

F = force exerted on the artillery shell

m = mass of the artillery shell = 1100 kg

a = acceleration of the artillery shell = 2.40\times 10^4m/s^2

Putting values in above equation, we get:

F=1100kg\times 2.40\times 10^4m/s^2\\\\F=2.64\times 10^6N

Now, according to Newton's third law, every action has an equal and opposite reaction.

So, the force exerted on the artillery shell will be equal to the force exerted on the ship by artillery shell acting in opposite direction.

Hence, the force exerted on the artillery shell is 2.64\times 10^6N  and the magnitude of force exerted on the ship by artillery shell is 2.64\times 10^6N

pickupchik [31]4 years ago
6 0

Answer:

Force exerted, F = 2.64 × 10⁷ Newton

Explanation:

It is given that,

Mass of the artillery shell, m = 1100 kg

It is accelerated at, a=2.4\times 10^4\ m/s^2

We need to find the magnitude of force exerted on the ship by the artillery shell. It can be determined using Newton's second law of motion :

F = ma

F=1100\ kg\times 2.4\times 10^4\ m/s^2

F = 26400000 Newton

or

F = 2.64 × 10⁷ Newton

So, the force exerted on the ship by the artillery shell is 2.64 × 10⁷ Newton.

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sergejj [24]

Answer:

Weight\ loss=1.6321N

Explanation:

From the question we are told that:

Weight W=85.9kg

Altitude h= 6.33 km

Let

Radius of Earth r=6380km

Gravity g=9.8m/s^2

Generally the equation for Gravity at altitude is mathematically given by

 g_s=9.8(\frac{6380}{6380+6.33})^2

 g_s=9.781m/s^2

Therefore

Weight at sea level

 W_s=9.8*85.9

 W_s=841.82N

Weight at 6.33 altitude

 W_a=9.781*85.9

 W_a=840.2N

Therefore

 Weight loss=W_s-W_b

 Weight loss=841.82-840.2

 Weight loss=1.6321N

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3 years ago
Line segment Q R , Line segment R S and Line segment S Q are midsegments of Î"WXY. Triangle R Q S is inside triangle X Y W. Poin
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The perimeter of ΔWXY is : ( D ) 14.5 cm

<u>Calculating the </u><u>perimeter </u><u>of ΔWXY</u>

QR = WY / 2

RS = XW / 2

QS = XY / 2

Given that : QR = 2.93 cm ,  RS = 2.04 cm,  QS = 2.28 cm

Therefore

Perimeter of ΔWXY = ∑ WY + XW + XY  

                                 = 2SR + 2QS + 2QR

                                 = 2(2.04) + 2(2.28) + 2(2.93)

                                 = 14.5 cm

Hence we can conclude that the perimeter of ΔWXY = 14.5 cm

learn more about perimeter calculations : brainly.com/question/24744445

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A teacher explains that a scientist named Boyle did many experiments to determine what the relationship is between the volume an
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An object with a resistance of 28 Ω has 76 V applied to it. How much electric current is going through this object? Answer in un
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Answer : The electric current of a circuit is, 2.8 A.

Explanation :

Using Ohm's law :

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V = voltage of circuit = 76 volts = 76 V

I = current flowing in a circuit = ?

Now put all the given values in the above formula, we get :

I=\frac{V}{R}

I=\frac{76V}{28\Omega }

I=2.8A

Therefore, the electric current of a circuit is, 2.8 A

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