Answer:
The time after which the two stones meet is tₓ = 4 s
Explanation:
Given data,
The height of the building, h = 200 m
The velocity of the stone thrown from foot of the building, U = 50 m/s
Using the II equation of motion
S = ut + ½ gt²
Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building
The equation for the stone dropped from top of the building becomes
x = 0 + ½ gtₓ²
The equation for the stone thrown from the base becomes
S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)
Adding these two equations,
x + (S - x) = U tₓ
S = U tₓ
200 = 50 tₓ
∴ tₓ = 4 s
Hence, the time after which the two stones meet is tₓ = 4 s
Answer:
D is the answer
Explanation:
D is the most highest one so
the answer is D
Answer:
Densities increase down the group
MP and BP decrease down the group
Softness increased going down the group
Speed of reacting increases going down the group
Answer:
A. 3.4 m
Explanation:
Given the following data;
Force = 56.7N
Workdone = 195J
To find the distance
Workdone is given by the formula;
Making "distance" the subject of formula, we have;
Substituting into the equation, we have;
Distance = 3.4 meters.
