All categories

2 years ago

What do you like about engaging to different reereational activities

Physics

1 answer:

Novay_Z [31]2 years ago

6 0

Answer:

What I like the most about engaging to different recreational activities is that these contribute to my entertainment and relax after working.

That is, recreational activities allow individuals to abstract from their daily responsibilities, encouraging distraction and concentration of the individual on issues of less importance, allowing them to relax.

Therefore, in my case I love that different recreational activities, such as sports or social gatherings, allow me to abstract from my responsibilities and allow me to enjoy moments of fun.

You might be interested in

A massless rod of length L has a small mass m fastened at its center and another mass m fastened at one end. On the opposite end

konstantin123 [22]

Answer:

onservation of energy

U top = K bottom

(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2

So 2m*g*L = 1/2*1.25*m*L^2*?^2

So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)

3 0

2 years ago

Part complete during a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s

Inga [223]

The average force applied to the ball= 106.7 N

Explanation:

Force is given by

f= ΔP/t

ΔP= change in momentum= m Vf- m Vi

m= mass =0.2 kg

Vf= final velocity= 12 m/s

Vi=initial velocity= -20 m/s ( negative because it is going towards the wall which is treated as negative axis)

t= time= 60 ms= 0.06 s

now ΔP= 0.2 [ 12-(-20)]

ΔP=0.2 (32)=6.4 kg m/s

now force F= ΔP/t

F= 6.4/0.06

F=106.7 N

8 0

2 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:

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