Answer:
2.48 m/s
Explanation:
We can use the kinematic equation,
s = ut +½at²
Where
s = displacement
u = initial velocity
t = time taken
a = acceleration
Using the equation in vertical direction,
321 = 0×t +½×g×t², u = 0 because initial vertical velocity is 0
We get t = 8.01 s
Using the equation in the horizontal direction,
52 = u×8.01 +½×0×(8.01)²,. a = 0 because no unbalanced force act on object in that direction
So u = 2.48 m/s
315g/95gmol-1
3.315 moles of MgCl2
Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
Answer:
16.2 s
Explanation:
Given:
Δx = 525 m
v₀ = 0 m/s
a = 4.00 m/s²
Find: t
Δx = v₀ t + ½ at²
525 m = (0 m/s) t + ½ (4.00 m/s²) t²
t = 16.2 s
Answer:
D
Explanation:
Work is not a vector but it is a scalar
