Answer:
300 N/m
Explanation:
given,
Load attached to the spring, W = 54 N
length of stretch of the spring, x = 0.15 m
spring constant= ?
Force applied on the spring is calculated by the equation
F = k x
where k is the spring constant
x is the displacement of the spring due to applied load
now,
54 = k × 0.15
hence, the spring constant is equal to 300 N/m
Answer:
b) 7.00
Explanation:
N( t ) = -20( t - 5 )²
dN/ dt = -20 x 2 ( t - 5 )
For maximum N ( depth )
dN/dt = 0
- 40 ( t - 5 ) = 0
t = 5
So at 2 + 5 = 7 .00 am depth of water reaches its maximum.
Answer:
I guess the acceleration would be 8 meters a second
Explanation:
I can't think of any other fitting way to put the answer sorry if it's not right
Btu/(lb-°F) J/(g-°C i mean this is the correct answer
