Answer:
F = 25530 N
Explanation:
F(net) = ma
F(net) = 2553 kg•(10) m/s² = 25530 N
Suppose car A is moving with a velocity Va, and car b with a velocity Vb,
According the principle of conservation of momentum:
Va x Ma + Vb x Mb = (Ma + Mb) V
V = (Va x Ma + Vb x Mb)/(Ma +Mb)
V = speed of cars after coupling
V = (Va x 20 mg + Vb x 15 mg)/(20 mg + 15 mg)
Put in the values of Va and Vb, and get the V
Answer:
4 days
either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2
n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2
Explanation:
Answer:
Explanation:
Given that,
Mass of a box, m = 18 kg
It is pushed with a force of 250 N
The coefficient of friction is 0.8
We need to find the acceleration of the box.
Normal force acting on the box,
N = mg
M = 18 × 10
N = 180 N
The force acting on the force in terms of coefficient of friction is given by :
So, the acceleration of the box is 
Answer:
m = 82.1 kg
Explanation:
For this exercise we must use Hook's law that states that the force exerted by a spring is proportional to the displacement
Fe = -k x
Let's use Newton's second law to establish equilibrium, the elastic force up and the body weight down
Fe - W = 0
Fe = W = mg
k x = m g
m = k x / g
Let's reduce the distance to SI units
x = 25 cm (1 m / 100cm) = 0.250 m
m = 3220 0.250 /9.8
m = 82.1 kg
