An earth scientist who studies hydrosphere

Throw two balls from the same height at the same time at an initial speed of 20 m/s. One is thrown vertically down, while the ot

labwork [276]

The time difference between their landing is 2.04 seconds.

<h3>Time of difference of the two balls</h3>

The ball thrown vertical upwards will take double of the time taken by the ball thrown vertically downwards.

Time difference, = 2t - t = t

t = √(2h/g)

where;

h is the height of fall
g is acceleration due to gravity

Apply the principle of conservation of energy;

¹/₂mv² = mgh

h = v²/2g

where;

v is speed of the ball

h = (20²)/(2 x 9.8)

h = 20.41 m

<h3>Time of motion</h3>

t = √(2 x 20.41 / 9.8)

t = 2.04 s

Thus, the time difference between their landing is 2.04 seconds.

Learn more about time of motion here: brainly.com/question/2364404

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6 0

2 years ago

Before the cells in a person's body can use the food that the person eats, the food must be A. chemically broken down into sugar

larisa86 [58]

A chemically broken down into sugar

6 0

3 years ago

Read 2 more answers

What is the total displacement of a child who walks 4m south 2m north 5m south and 5 m north?

seropon [69]

Your answer is going to be 2m south

3 0

3 years ago

Coherent light with wavelength 599 nm passes through two very narrow slits with separation of 20 μm, and the interference patter

goblinko [34]

Answer:

134.77 mm

Explanation:

Wave length of light λ = 599 x 10⁻⁹ m

Slit separation d = 20 x 10⁻⁶ m

Screen distance D = 3 m

Distance of second dark fringe from centre

= 1.5 x λ D / d

Putting the values given above

distance = $\frac{1.5\times599\times10^{-9}\times 3}{20\times10^{-6}}$

= 134.77 x 10⁻³ m

= 134.77 mm.

7 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

2 years ago