What's being transferred between a cell phone and a cell phone tower through radio waves?

A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about

Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[ $0^{2} -11^{2}$ ]

x-x₀=6.05* $10^{-2}$ m

=6.05 cm

3 0

2 years ago

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ????0 . Find the minimu

Reptile [31]

Answer:

Explanation:

Threshold frequency = 4.17 x 10¹⁴ Hz .

minimum energy required = hν where h is plank's constant and ν is frequency .

E = 6.6 x 10⁻³⁴ x 4.17 x 10¹⁴

= 27.52 x 10⁻²⁰ J .

wavelength of radiation falling = 245 x 10⁻⁹ m

Energy of this radiation = hc / λ

c is velocity of light and λ is wavelength of radiation .

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 245 x 10⁻⁹

= .08081 x 10⁻¹⁷ J

= 80.81 x 10⁻²⁰ J

kinetic energy of electrons ejected = energy of falling radiation - threshold energy

= 80.81 x 10⁻²⁰ - 27.52 x 10⁻²⁰

= 53.29 x 10⁻²⁰ J .

4 0

2 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$