Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:

where F is the force applied, and

is the stretch of the spring with respect to its equilibrium position. Using the data, we find

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
Answer:
For the first situation, we first need to find the mass of the second train car.
In order to do that, we apply the conservation of the amount of movement:

and we have as a result:
m2 = 289.6875
For the second situation, also we will apply the conservation of the amount of movement:

and we have as a result:
V = 2.64 (it is moving to the right)
Answer:
b
Explanation:
n = m(g +a)
n= normal force (N)
m=mass (kg)
g=acceleration of gravity
a= acceleration of elevator
rearrange:
a= n/m - g
a= (810 N/73 kg) - 9.8 m/s ^2
a= 1.3 m/s ^2 up
and the acceleration is upwards bc her weight is less than the scale reading
There is nothing here? What do you need help with?