This nutrient can be hidden in food and has twice as many calories as other nutrients

Give a calculate answer to show that the two values (English system and metric system) for the Planck Constant are equivalent.

Nataly_w [17]

Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

Converting this in to metric system .

We have 1 eV = 1.6 x 10⁻¹⁹ J

Converting

4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

So Given values of Planck Constant are equivalent in English system and metric system.

7 0

2 years ago

A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

k = Dielectric constant = 40

$\epsilon_0$ = Electric constant = $8.854\times 10^{-12}\ \text{F/m}$

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$

Number of electron is given by

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}$

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0

2 years ago

Incident rays parallel to the axis of a concave mirror reflect parallel to the axis.

coldgirl [10]

No they don't. Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.

3 0

2 years ago

Show all work.

lys-0071 [83]

The new gravitation force at the new location is 40 N

Explanation:

The weight of the astronaut is given by the equation

$F=mg$ (1)

where

m is the mass of the astronaut

g is the acceleration of gravity

The acceleration of gravity at a certain distance $r$ from the centre of the Earth is given by

$g=\frac{GM}{r^2}$

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

$F=\frac{GMm}{r^2}$

When the astronaut is on the Earth's surface, $r=R$ (where R is the Earth's radius), so his weight is

$F=\frac{GMm}{R^2}=640 N$

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

$r'=3R+R=4R$