Acceleration=(change in speed)/(time for the change). 43/0.28 = 153.6 m/s^2.
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
Answer:
Height h= 1.7 m
Explanation:
Supposing we have to find height in meter.
1 feet = 0.3048 m
1 inch = 0.0254 m
Given that:
5 feet
= 5×0.3048
= 1.524 m
and 7 inch = 7×0.0254= 0.1778 m
Therefore total height of a man in meter
5 feet 7 inch = 1.5424+0.1778 =1.7 m
Height h= 1.7 m
Answer:
8.3m/s
Explanation:
Given parameters:
mass of clay ball = 5kg
Speed of clay ball = 25m/s
mass of clay ball at rest = 10kg
speed of clay ball at rest = 0m/s
Unknown:
Velocity after collision = ?
Solution:
Since the balls stick together, this is an inelastic collision:
m1v1 + m2v2 = v(m1 + m2)
5(25) + 10(0) = v (5 + 10)
125 = 15v
v = 8.3m/s
