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Brums [2.3K]
3 years ago
12

According to Newtons second law of motion, which is equal to the acceleration of an object?A: net force + massB: net force x mas

sC: net force - massD: net force divides mass
Physics
2 answers:
Assoli18 [71]3 years ago
5 0

D: net force divides mass

Application of F = ma.

mart [117]3 years ago
4 0

Answer:

The acceleration of an object is equal to the net force divided by the mass of the object.

(D) is correct option.

Explanation:

According to Newton's second law of motion

The net force is equal to the product of the mass of the object and the acceleration of the object.

In mathematically terms,

F = ma

The acceleration is

a = \dfrac{F}{m}

Where, a = acceleration

F = net force

m = mass

Hence, The acceleration of an object is equal to the net force divided by the mass of the object.

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Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

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initial instant. before the crash

         p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

         p_f = (m + M) v

the moment is preserved

          p₀ = p_f

          m v₀ = (m + M) v

          v = \frac{m}{m+M} \ v_o

for when n balls have collided

          v = \frac{m}{n \ m + M}  v₀

Now let's analyze the case of the bouncing ball (elastic)

     

initial instant

        p₀ = m v₀

final moment

        p_f = m v_{1f} + M v_{2f}

        p₀ = p_f

        m v₀ = m v_{1f} + M v_{2f}

       m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

        K₀ = K_f

        ½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

        v₁ = v_{1f}            v₂ = v_{2f}

        m (v₀² - v₁²) = M v₂²

let's use the identity

         (a² - b²) = (a + b) (a-b)

we write our equations

         m (v₀ - v₁) = M v₂                       (1)

         m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

         v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

          m (v₀ - v₁) = M (v₀ + v₁)

          v₀ (m -M) = (m + M) v₁

          v₁ = \frac{m-M}{m + M}   v₀

we substitute in equation 1 to find v₂

            \frac{M}{m}  v₂ = v₀ -  \frac{m-M}{m+M}   v₀

            v₂ = \frac{m}{M}  ( 1 - \frac{m-M}{m+M} ) \ v_o

            v₂ = \frac{m}{M}  ( \frac{2M}{m+M} ) \ \ v_o

            v₂ = \frac{2m}{m +M}  \ v_o  

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

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