Answer:
Explanation:
Given,
- Work done by the rope 900 m/s.
- Angle of inclination of the slope =

- Initial speed of the skier = v = 1.0 m/s
- Length of the inclined surface = d = 8.0 m
part (a)
The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.
part (b)
- Initial speed of the skier = v = 1.0 m/s.
Rate of the work done by the rope is power of the rope.

Part (c)
- Initial speed of the skier = v = 2.0 m/s.
Rate of the work done by the rope is power of the rope.

Explanation:
En un balde de 25 cm de altura se hace un orificio a 15 cm del suelo.calcula la velocidad de salida del agua
Answer:
592.92 x 10³ Pa
Explanation:
Mole of ammonia required = 10 g / 17 =0 .588 moles
We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .
From the relation
PV = nRT
P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )
P = 592.92 x 10³ Pa
Answer:
a) 
b) 
c) 
d) 
Explanation:
<u>Given equation of pressure variation:</u>
![\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]](https://tex.z-dn.net/?f=%5CDelta%20P%3D%20%281.78%5C%20Pa%29%5C%20sin%5C%20%5B%280.888%5C%20m%5E%7B-1%7D%29x-%28500%5C%20s%5E%7B-1%7D%29t%5D)
We have the standard equation of periodic oscillations:

<em>By comparing, we deduce:</em>
(a)
amplitude:

(b)
angular frequency:


∴Frequency of oscillations:


(c)
wavelength is given by:



(d)
Speed of the wave is gives by:



Acceleration is measured in meters per second square.