Mass of 25 kg weight is 25/9.81 slug of mass.
Answer:
(a) Workdone = -27601.9J
(b) Average required power = 1314.4W
Explanation:
Mass of hoop,m =40kg
Radius of hoop, r=0.810m
Initial angular velocity Winitial=438rev/min
Wfinal=0
t= 21.0s
Rotation inertia of the hoop around its central axis I= mr²
I= 40 ×0.810²
I=26.24kg.m²
The change in kinetic energy =K. E final - K. E initail
Change in K. E =1/2I(Wfinal² -Winitial²)
Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]
Change in K. E= -27601.9J
(a) Change in Kinetic energy = Workdone
W= 27601.9J( since work is done on hook)
(b) average required power = W/t
=27601.9/21 =1314.4W
Answer:
m = 28.7[kg]
Explanation:
To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.
where:
Ek = kinetic energy = 1800 [J]
m = mass [kg]
v = 11.2 [m/s]
![1800=\frac{1}{2}*m*(11.2)^{2}\\m = 28.7[kg]](https://tex.z-dn.net/?f=1800%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2A%2811.2%29%5E%7B2%7D%5C%5Cm%20%3D%2028.7%5Bkg%5D)
Answer:
1. bending of light in gravitational fields.
2. effect of gravitational redshift.
3. perihelion precission of mecury.
Explanation:
1 bending of light in gravitational fields, we can think of it like this:
by noting the change in position s of stars as they pass near the sun on the celetial sphere, so since the sun creates a gravitational field even the star thats not in our line of side(behind the sun) can be seen because its light is bent.
2. effects of gravitational redshift:
this says that if you are in the gravitational field, your clock moves slower when it is seen by a distant observer.
3. perihelion precission of mecury:
according to Newtonian physics a two body system consisting of a lone orbiting the spherical mass would trace out an ellipse with the center of mass of the system as the focus but mercury deviates from that precission. then according to Einstein, the change in orientation of the orbital ellipsewithin its orbital plane is the effect of gravitation being mediated by the curvature of space-time.
Answer:
Average acceleration on first part of the chunk is given as
Average acceleration on second part of the chunk is given as
Explanation:
By momentum conservation along x direction we will have
so we have
also by energy conservation
by solving above equation we will have
Average acceleration on first part of the chunk is given as
Average acceleration on second part of the chunk is given as
