formula for gravitational P.E =mgh
Solution:-mass=3kg height=5metre and gravity=9.8 or 10m/sec² so P.E=mgh , 3×9.8×5=147kgm²/sec²
Answer:
the distance between the lawyer's home and her office is 124 miles
Explanation:
given information:
first lets assume that
x-axis (west = positive, east = negative)
y-axis (north = positive, south = negative)
thus,
distance of the house = (-88,18)
distance of the office = (13, -54)
thus, the distance between the lawyer's home and her office
R = √(x₂ - x₁)² + (y₂ - y₁)²
= √(13 - (-88))² + (-54 -18)²
= 124 miles
Answer:
Time period of the osculation will be 2.1371 sec
Explanation:
We have given mass m = (B+25)
And the spring is stretched by (8.5 A )
Here A = 13 and B = 427
So mass m = 427+25 = 452 gram = 0.452 kg
Spring stretched x= 8.5×13 = 110.5 cm
As there is additional streching of spring by 3 cm
So new x = 110.5+3 = 113.5 = 1.135 m
Now we know that force is given by F = mg
And we also know that F = Kx
So 

Now we know that 
So 

