How the relative motion and relative position of the Sun, Earth and Moon affect the seasons, phases of the moon and eclipses?

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

Read 2 more answers

A wave traveling in water has a frequency of 250 Hz and a wavelength of 6.0 N. What is the speed of the wave?

enyata [817]

The speed of a wave is determined by the product of the frequency and the wavelength; we already have the wavelength and the frequency, so all we need to do is multiply them by each other and use our proper unit of measure.

Velocity (speed) = Frequency x Wavelength
V = 250 x 6
V = 1500

Your answer is 1500 m/s.

I hope this helps!

5 0

3 years ago

Calculate the acceleration of a student riding his bicycle in a straight line that speeds up from 4 m/s to 6 m/s in 5 seconds.

solong [7]

Acceleration is equal to velocity final minus velocity initial divided by time. 6m/s minus 4m/s divided by 5 seconds is 0.4m/s^2.