Looks like a cosine function graph. The wave pattern is transversal waves . The faster the amplitude the higher the wave. The force of the drop hitting the water pushes the water down and out causing waves If the water hits from a higher amplitude the waves raise bigger . When you increase the frequency of the water drops the waves move faster but no bigger. When you increase the frequency of the water drops the wave ripples faster .

Explanation:

7 0

3 years ago

A solid cylinder is mounted above the ground with its axis of rotation oriented horizontally. A rope is wound around the cylinde

Romashka [77]

Answer:

(a)10.5 rad/s2

(b) 20.9 rev

Explanation:

As the block of mass 53 kg is falling and pulling on the rope. The tension force on the rope must be equal to the gravity acting on the block according to Newton's 3rd law

T = mg = 53*9.81 = 519.93 N

Since this tension force would rotate the cylinder freely without any friction. The torque created by this tension force is

To = TR = 519.93 * 0.36 = 187.17 Nm

This solid cylinder would have a moment of inertia around it's rotating axis of:

$I = \frac{mR^2}{2} = \frac{275 * 0.36^2}{2} = 17.82kgm^2$

(a)We can use Newton's 2nd law to calculate the angular acceleration exerted by such torque on the solid cylinder

$\alpha = \frac{To}{I} = \frac{187.17}{17.82} = 10.5 rad/s^2$

(b) With such constant angular acceleration, the angle it would make after 5s is

$\theta = \frac{\alphat^2}{2} = \frac{10.5*5^2}{2} = 131.3 rad$

Since each revolution equals to $2\pi rad$ of angle, we can calculate the number of revolution it makes

$\frac{\theta}{2\pi} = \frac{131.3}{6.28} \approx 20.9 rev$

(c) Assume the thickness of the rope is negligible (and its wounded radius is unchanging), we can calculate the rope length unwinded after rotating 131.3rad

$\theta R = 131.3*0.36 = 47.27 m$

3 0

3 years ago

A car is sitting still. It accelerates to a constant speed then it decelerates again to zero speed. While the car is acceleratin

Kobotan [32]

Answer:

in the acceleration process the quantity α and w must increase

the deceleration process the alpha quantity must constant a direction opposite to the angular velocity

Explanation:

Acceleration and angular velocity are related to linear

v = w xr

a = αx r

The bold letters indicate vectors and the cross is a vector product, therefore if

we can see that the relationship between linear and angular variables is direct

therefore in the acceleration process the quantity α and w must increase as well as their linear counterparts

in the deceleration process the alpha quantity must constant as the linear acceleration and must have a direction opposite to the angular velocity

5 0

4 years ago

Calculate the distance moved by a runner who runs with a speed of 5 km/h for a period of 1.5 hours.

FrozenT [24]

Answer:

7.5 km

Explanation:

h5 per hour means that he traveled 5 km in 1 our. And then half of the hour, which means half an hour 5 km which is 2.5.

5 + 2.5 = 7.5

or just 1.5 x 5 = 7.5

8 0

3 years ago