P= w/t and W= Work
In this case, W= 6,700j, and T= 45 seconds
Power is the ratio of work per unit time. When you perform a work in a given span of time, the ratio of work performed with respect to time is Called Power.
si unit for Power is Watt (W)
so, P= 6,700/45
P= 148
Final answer is P=148
Answer:
23. 4375 m
Explanation:
There are two parts of the rocket's motion
1 ) accelerating (assume it goes upto h1 height )
using motion equations upwards
Lets find the velocity after 2.5 seconds (V1)
V = U +at
V1 = 0 +5*2.5 = 12.5 m/s
2) motion under gravity (assume it goes upto h2 height )
now there no acceleration from the rocket. it is now subjected to the gravity
using motion equations upwards (assuming g= 10m/s² downwards)
V²= U² +2as
0 = 12.5²+2*(-10)*h2
h2 = 7.8125 m
maximum height = h1 + h2
= 15.625 + 7.8125
= 23. 4375 m
Answer:3.4 seconds
Explanation:
Initial velocity(u)=0
acceleration=34.5m/s^2
Height(h)=200m
Time =t
h=u x t - (gxt^2)/2
200=0xt+(34.5xt^2)/2
200=34.5t^2/2
Cross multiply
200x2=34.5t^2
400=34.5t^2
Divide both sides by 34.5
400/34.5=34.5t^2/34.5
11.59=t^2
t^2=11.59
Take them square root of both sides
t=√(11.59)
t=3.4 seconds
Answer:C.The tube should be held vertically, perpendicular to the ground.
Explanation: Power lines are mainly overhead power installation that transfer electric energy from one place to another. Electric power lines contains both Magnetic field and Electric field.
Potential difference is the change in the amount of energy carried by an electric circuit from one point to another. TO MAXIMIZE THE POTENTIAL DIFFERENCE BETWEEN ONE END OF THE TUBE AND ANOTHER? THE TUBE SHOULD BE HELD VERTICALLY,PERPENDICULAR TO THE GROUND.
