Ask question

Ask question

All categories

3 years ago

9

A 22.0-kg child is riding a playground merrygo-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he i

s 1.25 m from its center?

1 answer:

Tems11 [23]3 years ago

6 0

Answer:

F=480.491 N

Explanation:

Given that

mass ,m = 22 kg

Angular speed ω = 40 rev/min

$\omega=\dfrac{2\pi \times 40}{60}\ rad/s$

ω =4.18 rad/s

The radius r= 1.25 m

We know that centripetal force is given as

F=m ω² r

Now by putting the values in the above equation we get

$F=22\times 4.18^2\times 1.25\ N$

F=480.491 N

Therefore the centripetal force on the child will be 480.491 N.

You might be interested in

A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc

german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = $\frac{1}{2} kx^2$ and the kinetic energy gained by the box is given by KE = $\frac{1}{2} mv^2$

Since work done by the spring = kinetic energy gained by the box we have

$\frac{1}{2} mv^2$ = $\frac{1}{2} kx^2$ therefore we have v = $\sqrt{\frac{kx^2}{m} }$ = $x\sqrt{\frac{k}{m} }$ = $0.15\sqrt{\frac{300}{0.2} }$ = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = $\frac{v^2}{2g}$ = $\frac{5.81^{2} }{2*9.81}$ = 1.72 m

6 0

3 years ago

What is the formula to calculate the thickness of a test tube?

Maslowich

Answer:

Having the inside dimensions (ID) and the outside dimensions (OD) will allow you to figure out the wall thickness on tubing. You would need to subtract the ID from the OD and then divide by two. This number is the wall thickness.

Explanation:

4 0

3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago

A skateboarder is skating back and forth on the halfpipe as seen below. As he skates his energy transforms from potential energy

egoroff_w [7]

Answer:

Friction and air resistance cause some of his kinetic energy to be “lost”. This makes him slow down.

Explanation:

The law of conservation of energy states that in absence of frictional forces, the mechanical energy of an object (given by the sum of its kinetic and potential energy) is conserved. In such a situation, the skateboarder would never stop his motion, because potential energy is continuously converted into kinetic energy and vice-versa, but the total energy remains the same so he would never stop.

In a real world, however, this is not true. In fact, in a real world some frictional force are present, in particular:

- friction: this force is due to the contact between the skateboard and the surface of the halfpipe, and its direction is always opposite to the motion of the skateboarder

- Air resistance: this force is due to the resistance opposed by the molecules of air that the skateboarder meets during his motion, and its direction is also opposite to the motion of the skateboarder

This two forces are said to be non-conservative forces, which means that they cause some of the mechanical energy of the skateboarder to be "lost", in the sense that it is dissipated as heat and it is no longer available for the skateboarder.

Therefore, the correct option is

Friction and air resistance cause some of his kinetic energy to be “lost”. This makes him slow down.

7 0

4 years ago

A disk-shaped merry-go-round of radius 2.63 m and mass 152 kg rotates freely with an angular speed of 0.526 rev/s. A 51.7 kg per

zalisa [80]

Explanation:

Given

radius $r=2.63 m$

$N=0.526 rev/s$

$\omega =3.30 rad/s$

mass disc $M=152 kg$

mass of person $m=51.7 kg$

velocity of Person $v=2.76 m/s$

moment of inertia $I=Mr^2$

$I=0.5\times 152\times 2.63^2=827.64 kg-m^2$

Initial angular momentum

$L_i=I\omega +mvr$

$L_i=827.64\times 3.30+51.7\times 2.76\times 2.63$

$L_i=2731.212+375.27=3106.48 Js$

Final Moment inertia

$I_f=0.5Mr^2+mr^2$

$I_f=(152\cdot 0.5+51.7)\cdot (2.63)^2=1185.243 kg-m^2$

final angular momentum

$L_f=I_f\omega _f$

Conserving angular momentum

$L_i=L_f$

$3106.48=1408.97\times \omega _f$

$\omega _f=2.62 rad/s$

4 0

3 years ago