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2 years ago

Find the mass if the force is 18 n and the acceleration is 2 m/s/s

Physics

1 answer:

Mazyrski [523]2 years ago

4 0

Answer:

<h2>The answer is 9 kg</h2>

Explanation:

The mass of an object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{18}{2} \\$

We have the final answer as

Hope this helps you

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A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it i

lianna [129]

Answer:

<em>A) the moment of inertia of the system decreases and the angular speed increases. </em>

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

$I_{1} w_{1} = I_{2} w_{2}$ ....1

where $I_{1}$ and $I_{2}$ are the initial and final moment of inertia respectively.

and $w_{1}$ and $w_{2}$ are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

$I = mr^{2}$ ....2

where $m$ is the mass of the rotating body,

and $r$ is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from $I_{1}$ to $I_{2}$ will cause the angular speed of the system to increase from $w_{1}$ to $w_{2}$ .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

7 0

3 years ago

A basketball player drops a 0.4-kg basketball vertically so that it is traveling at 5.7 m/s when it reaches the floor. The ball

Alik [6]

Answer:

(a) p = 3.4 kg-m/s (b) 37.78 N.

Explanation:

Mass of a basketball, m = 0.4 kg

Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)

It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)

(a) Change in momentum = final momentum - initial momentum

p = m(v-u)

p = 0.4 (2.8-(-5.7))

p = 3.4 kg-m/s

(b) Impulse = change in momentum

Ft = 3.4

We have, t = 0.09 s

$F=\dfrac{3.4}{0.09}\\\\F=37.78\ N$

Hence, this is the required solution.

4 0

3 years ago

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$