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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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A proton and an electron are placed in an electric field. Which undergoes the greater acceleration?
iren2701 [21]

Newton's 2nd law:

Fnet = ma

Fnet is the net force acting on an object, m is the object's mass, and a is the acceleration.

The electric force on a charged object is given by

Fe = Eq

Fe is the electric force, E is the electric field at the point where the object is, and q is the object's charge.

We can assume, if the only force acting on the proton and electron is the electric force due to the electric field, that for both particles, Fnet = Fe

Fe = Eq

Eq = ma

a = Eq/m

We will also assume that the electric field acting on the proton and electron are the same. The proton and electron also have the same magnitude of charge (1.6×10⁻¹⁹C). What makes the difference in their acceleration is their masses. A quick Google search will provide the following values:

mass of proton = 1.67×10⁻²⁷kg

mass of electron = 9.11×10⁻³¹kg

The acceleration of an object is inversely proportional to its mass, so the electron will experience a greater acceleration than the proton.

6 0
4 years ago
Suppose your hair grows at the rate of 1/55 inch per day. Find the rate at which it grows in nanometers per second. Because the
qwelly [4]

Answer:5.35nm

Explanation:

Consider that 1 inch is = 0.0254m

we have,

1m= 1x10^9 nm  

While:

0.0254m = 2.54x10^7nm  

1/55 (2.54x10^7) = 4.6181 x 10^5nm  

1 day= 24 hrs  

= (24x60) when calculating in min  

= (24x60x60) calculating in seconds we have:

= 8.64x10⁴sec  

In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm

Therefore, the amount by which the hair grows in 1 second  will be;

= (4.6181 x 10^5)/(8.64x10^4)  

= 5.35nm  

The rate of growth will be 5.35nm

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7 0
4 years ago
A truck traveling at a constant speed of 24 m/s passes a more slowly moving car. The instant the truck passes the car, the car b
Ymorist [56]

Answer:

7.15 m/s

Explanation:

We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.

The truck moves at constant speed, we can use the equation for position under constant speed:

Xt = X0 + v*t

The car is accelerating with constant acceleration, we can use this equation

Xc = X0 + V0*t + 1/2*a*t^2

We know that both vehicles will meet again at x = 578

Replacing this in the equation of the truck:

578 = 24 * t

We get the time when the car passes the truck

t = 578 / 24 = 24.08 s

Before replacing the values on the car equation, we rearrange it:

Xc = X0 + V0*t + 1/2*a*t^2

V0*t = Xc - 1/2*a*t^2

V0 = (Xc - 1/2*a*t^2)/t

Now we replace

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Answer:

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