The initial velocity of the 3250 Kg mass is 2.1 m/s. The distance covered by the larger mass in 5s is 4.7 cm.
In this problem, we have to apply the law of conservation of linear momentum. Note that;
Momentum before collision = Momentum after collision
m1u1 + m2u2 = m1v1 + m2v2
(2150 × 10) + (3250u1) = (2150 + 3250)5.22
21500 + 3250u1 = 5400 × 5.22
3250u1 = 28188 - 21500
u1 = 28188 - 21500/3250
u1 = 2.1 m/s
2) Again from the principle of conservation of linear momentum;
(0.40 × 3.5) + (0.60 × 0) = (0.40 × 0.70) + (0.60 × v2)
1.4 = 0.28 + 0.60v2
1.4 - 0.28 = 0.60v2
v2 = 1.87 cm/s
Using;
s = 1/2 ( u + v)t
s = 1/2(0 + 1.87) × 5
s = 4.7 cm
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The force acting in the front direction is the 130N.
The frictional force is acting backwards 30N.
1) The net force is 130N - 30N = 100N
2) s = ut + (1/2)at^2 u = 0, Start from rest, s = 25m t =5.
25 = 0*5 + (1/2)* a * 5^2.
25 = 0 + 25/2 * a.
25 = (25/2)a. Divide 25 from both sides.
1 = (1/2)* a. Cross multiply.
2 = a.
a = 2 m/s^2.
3) Mass of the box
Net Force, F = ma
100 = m*2. Divide both sides by 2.
100/2 = m
50 = m.
m = 50 kg.
4) Final velocity , v = u + at.
v = 0 + 2*5 = 10 m/s.
Kinetic Energy, K = (1/2) * mv^2.
= 1/2 * 50 * 10 * 10.
= 2500 J.
Answer:
36 m³
Explanation:
From gas laws, at a constant temperature, the product of volume and pressure are directly proportional hence expressed as p1v1=p2v2 where p is pressure while v represent volume. Subscripts 1 and 2 represent the initial and respectively. Making v2 the subject then
Substituting 120 kpa for P1, 12m³ for V1 and 40 kpa for p2 then
Therefore, the volume will be 36 m³
Answer:
Final angular velocity is 35rpm
Explanation:
Angular velocity is given by the equation:
I1w1i + I2w2i = I1w1f -I2w2f
But the two disks are identical, so Ii =I2
wf can be calculated using
wf = w1i - w2i/2
Given: w1i =50rpm w2i= 30rpm
wf= (50 + 20) / 2
wf= 70/2 = 35rpm