How many laws does newton have?

In order to effectively radiate a radio signal, an antenna must be at least one-half the wavelength of transmission in length, i

Virty [35]

The equation that relates Wavelength & frequency of electromagnetic waves is : Velocity (c) = Wavelength (λ) * frequency (f) ------ (1) Electromagnetic wave velocity (c) = Speed of Light = 3 * 10^8 m/s From (1), Wavelength , λ = c / f λ= (3*10^8)/(980*10^3) λ = 306.12 m Minimum height of the antenna for effective transmission = λ * 0.5 = 306.12*0.5 Answer : 153 m (rounded off)

6 0

3 years ago

2N2 Hy 03 2N₂²_0₂ +42 H₂O

Sholpan [36]

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4 0

3 years ago

Which would melt first, germanium with a melting point of 1210 k or gold with a melting point of 1064oc?

iren2701 [21]

Germanium To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15 1064 + 273.15 = 1337.15 K So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.

8 0

3 years ago

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0

3 years ago

Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.

kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, $4.67\times 10^7$

(b) The number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

Explanation :

For part (a) :

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{3.82\text{ days}}$

$k=1.81\times 10^{-1}\text{ days}^{-1}$

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.81\times 10^{-1}\text{ days}^{-1}$

t = time passed by the sample = 12 days

a = initially number of radon atoms = $4.1\times 10^8$

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

$12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}$

$a-x=4.67\times 10^7$

Thus, the number of radon atoms will remain after 12 days is, $4.67\times 10^7$

For part (b) :

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = $(4.1\times 10^8)-(4.67\times 10^7)$

The number of radon nuclei have decayed = $3.6\times 10^8$

Thus, the number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

5 0

3 years ago