Answer: 12Mg/h
Explanation:
Let the spring is compressed by a distance x,before the lift stops,then
Mg(h+x)= 1/2 kx^2 ............... 1
Kx - Mg = M ( 5g ) ............ 2
Make x the subject in equation 2
Kx = 5Mg + Mg
Kx = 6Mg
x = 6Mg/k ............ 3
Put equation 3 into 1
Mg ( h + x ) = 1/2 kx^2
Mgh + Mgx = 1/2kx^2
Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2
Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2
h =18Mg/k - 6Mg/h
K = 12Mg/h
∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)
About a mil sience 2014-2015
Answer:
B
Explanation:
nothing to do with black holes creating star or related
Answer:
The car stops in 7.78s and does not spare the child.
Explanation:
In order to know if the car stops before the distance to the child, you take into account the following equation:
(1)
vo: initial speed of the car = 45km/h
a: deceleration of the car = 2 m/s^2
t: time
xo: initial distance to the child = 25m
x: final distance to the child = 0m
It is necessary that the solution of the equation (1) for time t are real.
You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:


You take the first value t1 because it has physical meaning.
The solution for t is real, then, the car stops in 7.78s and does not spare the child.