More compressed. moving up = apparent weight (i.e., your norma force) is greater. this means you’ll weighr more and push those springs down even more than you would at rest.
M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore;
(1060× 16) + (1830 ×0) = (1060 +1830) V
16960 = 2890 V
V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s
temperature decreases at higher altitudes because as air rises the pressure decreases.
Answer: 39.2 m/s
Explanation:
You can use the kinematic equation:
We know the final velocity because it says it came to a stop. So now all we gotta do is plug in.
Answer:
(a) T= 38.4 N
(b) m= 26.67 kg
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Kinematics
d= v₀t+ (1/2)*a*t² (Formula 2)
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
v₀=0, d=18 m , t=5 s
We apply the formula 2 to calculate the accelerations of the blocks:
d= v₀t+ (1/2)*a*t²
18= 0+ (1/2)*a*(5)²
a= (2*18) / ( 25) = 1.44 m/s²
to the right
We apply Newton's second law to the block A
∑Fx = m*ax
60-T = 15*1.44
60 - 15*1.44 = T
T = 38.4 N
We apply Newton's second law to the block B
∑Fx = m*ax
T = m*ax
38.4 = m*1.44
m= (38.4) / (1.44)
m = 26.67 kg
