A.

Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i

Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

$L = L_i_n_i + L_3_s$

$L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt$

$L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s$

Moment if inertia is

$I = 2ml^2$

$I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2$

Angular speed

ω = L/I

$= 38 / 5.75\\\\=6.61$

The speed of each ball is

V = ωL

$= 6.61\times0.5\\\\= 3.31m/s$

7 0

3 years ago

Which of the following could classify as a non-essential need according to the hierarchy of needs? A. healthy food B. human cont

sergejj [24]

D. According to the hierarchy of needs the body and mind must be taken care of first and foremost.

4 0

3 years ago

When the starter motor on a car is engaged, there is a 310 A current in the wires between the battery and the motor. Suppose the

Blababa [14]

Answer:

Diameter will be 351.42 mm

Explanation:

We have given current flowing in the copper wire i = 310 A

Voltage drop across the wire V = 0.55 volt

We know that resistance is given by $R=\frac{V}{i}=\frac{0.55}{310}=0.00177\Omega$

Length of the copper wire l = 1 m

Resisitivity of the copper wire $\rho =1.72\times 10^{-8}\Omega m$

We know that resistance $R=\frac{\rho l}{A}$

$0.00177=\frac{1.72\times 10^{-8}\times 1}{A}$

$A=969.4545\times 10^{-8}m^2$

As area $A=\pi r^2$

$3.14\times r^2=969.4545\times 10^{-8}$

$r=17.57\times 10^{-4}m$

So diameter $d=17.57\times 10^{-4}\times 2=35.142\times \times 10^{-4}m$ = 351.42 mm

3 0

3 years ago

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

3 years ago

Select the correct answer.

Simora [160]

Answer:

A. 2.36 Newtons

Explanation:

F = GmM/d²

F = 6.673 x 10⁻¹¹(1)(5.98 x 10²⁴) / (1.3 x 10⁷)²

F = 2.36121...

Very poor question design.

mass of box... 1 significant digit

distance... 2 significant digits

mass of earth... 3 significant digits

value of G... 4 significant digits

Answer precision to 3 significant digits is not justifiable

7 0

3 years ago