What would you have loved to press the pause button on so you could go deeper

The logarithm of x, written log(x), tells you the power to which you would raise 10 to get x. So, if y=log(x), then x=10^y. It i

fomenos

To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.

From the definition of logarithm we know that,

$Log_{10}(10) = 1$

In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,

$log_{10}(1,000,000)$

We can express this also as,

$log_{10}(10^6)$

By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.

So this can be expressed as

$6*log_{10}(10)$

Since the definition of the base logarithm 10 of 10 is equal to 1 then

$6*1=6$

The value of the given logarithm is equal to 6

8 0

3 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

MatroZZZ [7]

Answer:

The wavelength is $\lambda_2 = 534 *10^{-9} \ m$

Explanation:

From the question we are told that

The wavelength of the first light is $\lambda _ 1 = 587 \ nm$

The order of the first light that is being considered is $m_1 = 10$

The order of the second light that is being considered is $m_2 = 11$

Generally the distance between the fringes for the first light is mathematically represented as

$y_1 = \frac{ m_1 * \lambda_1 * D}{d}$

Here D is the distance from the screen

and d is the distance of separation of the slit.

For the second light the distance between the fringes is mathematically represented as

$y_2 = \frac{ m_2 * \lambda_2 * D}{d}$

Now given that both of the light are passed through the same double slit

$\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1$

=> $\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1$

=> $\lambda_2 = \frac{m_1 * \lambda_1}{m_2}$

=> $\lambda_2 = \frac{10 * 587 *10^{-9}}{11}$

=> $\lambda_2 = 534 *10^{-9} \ m$

4 0

3 years ago

The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point

Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth, R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

w = 2π / T

w = 2π / 24*3600

w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

v1 = R*w

v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

π/2 ........... s

x ............ 1/5 s

x = π/2*5 = 18°

- The radius of the earth R' at point where θ = 18° from the equator is:

R' = R*cos(18)

R' = (6.37 * 10 ^6)*cos(18)

R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

v2 = R'*w

v2 = (6058230.0088)*(7.27 * 10^-5)

v2 = 440.433 m/s

5 0

3 years ago

The largest graduated cylinder in my lab holds 2 L and has an inner diamter (the part that holds the water) of 8 cm. When it is

mestny [16]

Answer:

3924 Pa

Explanation:

Volume of cylinder = 2 L = 0.002 m^3 (1000 L = 1 m^3)

diameter of the inner cylinder = 8 cm = 0.08 m (100 cm = 1 m)

radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m

area of the inner cylinder = $\pi r^{2}$

where $\pi$ = 3.142,

and r = radius = 0.04 m

area of inner cylinder = 3.142 x $0.04^{2}$ = 0.005 m^2

height h of the water in this cylinder = volume/area

h = 0.002/0.005 = 0.4 m

pressure at the bottom of the cylinder due to the height of water = pgh

where

p = density of water = 1000 kg/m^3

g = acceleration due to gravity = 9.81 m/s^2

h = height of water within this cylinder = 0.4 m

pressure = 1000 x 9.81 x 0.4 = 3924 Pa

3 0

3 years ago

. You are in an airplane flying over the ocean. How could you most accurately determine your distance from the water?

pshichka [43]

Answer:

i dont nobhhjkkkkkkuii

Explanation:

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8 0

1 year ago