Answer:
The distance traveled by the ball is 8.5 m
Explanation:
Initial height of the ball, h₁ = 1.5 m above the ground
final height of the ball, h₂ = 5m
Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m
Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m
Total distance traveled = upward distance + downward distance
Total distance traveled = 3.5 m + 5m = 8.5 m
Therefore, the distance traveled by the ball is 8.5 m
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
Answer:
b. 88, 222
Explanation:
235U₉₂ ----→ Alpha --------→ 231P₉₀ ----→- beta -----→ 231Q₉₁ ------→-beta -------→231R₉₂--------→-alpha ------→-227S₉₀ ------→ gamma -----→-227S₉₀ ----------→ neutron ------→-226T₉₀-----------→ alpha --------→222 X ₈₈
Atomic No is 88 , atomic weight = 222 .
