17. If a good is a luxury, what type of response will it cause?

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

4 0

3 years ago

high school physics, no need detail explain, just give the answer, but you have to make sure thank you

Goshia [24]

Answer:many questions add point

6 0

3 years ago

What can you say about the speed of light in diamond?

Inessa05 [86]

So n=c/v, n= index, c=speed of light and v= speed of light in diamond. 2.42=c/v so v=c/2.42, c≈3x108 m/sec so v=1.24x108 m/sec.
Hope this helps.

7 0

3 years ago

Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

$\Delta W = \Delta PE + \Delta KE$

Where,

PE = Potential Energy

KE = Kinetic Energy

$\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2$

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

$P = \frac{\Delta W}{\Delta t}$

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

The rate of mass flow is,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

Where,

$\rho_w$ = Density of water

A = Area of the hose $\rightarrow A=\pi r^2$

The given radius is 0.83cm or $0.83 * 10^{-2}$ m, so the Area would be

$A = \pi (0.83*10^{-2})^2$

$A = 0.0002164m^2$

We have then that,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

$\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)$

$\frac{\Delta m}{\Delta t} = 1.16856kg/s$

Final the power of the pump would be,

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

$P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)$

$P = 57.1192W$

Therefore the power of the pump is 57.11W

6 0

3 years ago

Joaquin has hockey practice at the same ice rink where Alton has curling practice. One day, they were both playing on the ice wi

joja [24]

the puck recoils in each case.

larger mass stone gives puck greater recoil, smaller stone, smaller recoil

8 0

3 years ago

Read 2 more answers