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Vika [28.1K]
2 years ago
13

What is the mass of a cannonball if a force of 2,500 N gives the cannonball an acceleration of 200 m/s2? Question 5 options: 15.

0 kg 12. 5 kg 20 kg 25 kg.
Physics
1 answer:
djverab [1.8K]2 years ago
8 0

The mass of the given cannonball is 12.5 kg. Option B is correct.

<h3>What is Acceleration?</h3>

It is defined as the change in the direction and speed of the object. It is a vector quantity.

From, Newton's second law of motion

F = ma

Where,

F - force = 2,500 N

m - mass

a - acceleration =  200 m/s²


Put the values in the formula,

2500 = m \times 200 \\\\&#10;m = \dfrac {2500}{200}\\\\&#10;m = 12.5\rm \  kg &#10;

Therefore, the mass of the given cannonball is 12.5 kg.

Learn more about Acceleration:

brainly.com/question/2437624

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The center-seeking change in velocity of an object moving in a circle is the centripetal acceleration.

So, by Newton's laws, we know that an object moving with a given velocity will remain in constant motion with a constant velocity until we apply an acceleration.

So we define acceleration as the rate of change of the velocity, also remember that velocity is a vector (has magnitude and direction), so, if there is a change the direction of the velocity, we have an acceleration that causes that.

In circular motion, the velocity vector is always perpendicular to the radius of the circle, and it can only be possible if the velocity direction is changing constantly. This will happen because of something called centripetal acceleration.

This acceleration points radially inwards (to the center of the circle) so is also perpendicular to the velocity of the moving object, and this is what causes the constant change in the direction of the velocity of the moving object.

Just to give an example, if you have a string with a mass on one end, and with your hand, you rotate the mass (from the string), the tension of the string would be the centripetal acceleration.

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Let two objects of equal mass m collide. Object 1 has initial velocityv, directed to the right, and object 2 isinitially station
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<span>Our equation 1 would be
m*v=M*V1+m*V2
v=V1+V2
 v-V1=V2 

the equation 2 would look like this
 </span>V^2=V1^2+V2^2  
V^2-V1^2=V2^2
(V-V1)*(V+V1)=V2^2Dividing with the 1
V+V1=V2 
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A solid uniform ball of mass 1.0 kg and radius 1.0 cm starts from rest and rolls down a 1.0-m high ramp. there is enough frictio
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S GP A projectile of mass m moves to the right with a speed vi (Fig. P11.51a). The projectile strikes and sticks to the end of a
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What is the kinetic energy of the system after the collision?

K_f=\frac{3}{2} \frac{m^{2}v_i^{2}  }{(M+3m)}

How this is calculated?

Given:

Initial speed=v_i

mass of rod=M

Let, Initial kinetic energy =K_i

Final kinetic energy=K_f

Moment of inertia =I

What is the moment of inertia?

I=(I_p)_0+(I_{rod})_0\\I=m(\frac{d}{2})^{2}  +\frac{Md^{2} }{12} \\I=\frac{(M+3m)d^{2} }{12}

What is the angular momentum?

By conservation of angular momentum,

L_i=L_f

mv_i\frac{d}{2}=\frac{(M+3m)d^{2}\omega }{12}  \\\omega=\frac{6mv_i^{2} }{d(M+3m)}

We know that, the final kinetic energy is given by,

K_f=I\omega^{2}\\K_f=\frac{1}{2} *\frac{(M+3m)d^{2} }{12} *\frac{36m^{2}v_i^{2}}{d^{2}(M+3m)^{2}}\\ K_f=\frac{3}{2} \frac{m^{2}v_i^{2}  }{(M+3m)}

What is the kinetic energy?

  • In physics, the kinetic energy of an object is the energy that it possesses due to its motion.
  • It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
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To know more about kinetic energy, refer:

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