Answer:
Rs. 480.00
Explanation:
1kW = 1000W
therefore 500W = 0.5kW
20 × 24hrs = 480hrs in total.
0.5kW × 480hrs = 240kWh
if rs. 2 for 1kWh
then, 240kWh × 2 = Rs. 480.
Answer: The Earth's layer, which has the covering and layer, is made of a progression of things, or structural plates, that creep after some time. Along these lines, at intersecting limits, mainland outside is made and maritime covering is devastated. 2 plates slippy past each other structures a redesign plate limit.
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
Answer:
v = sqrt[2*(F*h*cot(theta)-mgh)/m]
Explanation:
Work = KE + Ug
F*r=1/2mv^2+mgh
1/2mv^2=F*r-mgh
v=sqrt[2(F*r-mgh)/m]
r=h/tan(theta)=h*cot(theta)