Potential Energy = mass * Hight * acceleration of gravity
PE=hmg
PE = 1.5 * .2 * 9.81
PE = 2.943
it lost .6 so 2.943 - .6 = 2.343
now your new energy is 2.343 so solve for height
2.343 = mhg
2.334 = .2 * h * 9.81
h = 1.194
the ball after the bounce only went up 1.194m
Answer:
E=0
Explanation:
Electric field due to each thin sheet of charge=\sigma/2\varepsilon
let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .
In the region between the plates,the electric field due to each plate is in same direction,
E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)
E=\sigma/\varepsilon
in the region outside the plates, the field due to the plates is in opposite directions
E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)
E=-\sigma/2\varepsilon+\sigma/2\varepsilon
E=0
Answer:
3.5 N
Explanation:
Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm
- Tension of the string attached at the 0cm end is 0 as moment arm is 0
- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise
- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise
- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise
- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise
- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.
Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm
Total counter-clockwise moment = 100T
For this to balance, 100 T = 350
so T = 350 / 100 = 3.5 N
Answer:
Kathmandu
Explanation:
As the altitude get higher, the gravitational pull of the earth on the object increases, therefore, the mass is higher up above.
Any two-dimensional vector in cartesian (x,y) coordinates can be broken down into individual horizontal and vertical components using trigonometry. If a train goes up a hill with 15 degree incline at a speed of 22 m/s, the horizontal component is 22cos(15)=21.3 m/s and the vertical component is 22sin(15)=5.5 m/s.
