Answer:
8.9m/s
Explanation:
Final velocity = initial velocity + acceleration *time
Vf=vi+at
Vf=1.4 m/s
Vi=?
A=-2.5m/s^2
T=3
1.4m/s=vi+(-2.5m/s^2)(3s)
Vi=8.9m/s
Answer:
a) 3.7 m/s^2
b) 231.8 N
Explanation:
Let m1 be mass of the first object (m1 = 38.0 kg) and let m2 be the mass of the second object (m2 = 17.0 kg ). Let a be the acceleration of the two objects. Let F1 be the force of gravity exerted on m1 and F2 be the force of gravity exerted on m2. Let M = m1 +m2
a)
F1 = m1g and F2 = m2g
So Fnet = F1 + F2
Since the pulleys will move in different directions when accelerating...
Fnet = F1 - F2
M×a = m1g - mg2
M×a = g×(m1 -m2)
a = g×(m1 - m2)/M
a = 9.8×(38 - 17)/(38 + 17)
a = 3.7 m/s^2
b)
Looking at the part for m2
Fnet = T - m2g
-m2×a = T - m2g
T = m2(g - a)
T = 231.8 N
Answer:
oop false have a great day
Explanation:
i hope i helped u
Answer:
1.19cm^3 of glycerine
Explanation:
Let Vo= 150cm^3 for both aluminum and glycerine, using expansion formula:
Volume of spill glycerine = change in volume of glycerine - change in volume of aluminum
Volume of glycerine = coefficient of volume expansion of glycerine * Vo* change in temperature - coefficient of volume expansion of Aluminum*Vo* change temperature
coefficient of volume expansion of aluminum = coefficient of linear expansion of aluminum*3 = 23*10^-6 * 3 = 0.69*10^-4 oC^-1
Change in temperature = 41-23 = 18oC
Volume of glycerine that spill = (5.1*10^-4) - (0.69*10^-4) (150*18) = 4.41*10^-4*2700 = 1.19cm3
