In nature, hot air tends to rise and cool air tends to sink. This is behind the concept of convection heating that is used in ov
2 answers:
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Answer:
14 m/s
Explanation:
The motion of the stone is a free fall motion, so an accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground. So, we can use the following SUVAT equation:
where
v is the final speed of the stone as it reaches the water
u = 0 is the initial speed
g = 9.8 m/s^2 is the acceleration
h = 10 m is the distance covered by the stone
Solving for v, we find
The force of gravity between two objects is given by:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the mass of the object is
, while the Earth's mass is 
. Their separation is 
, therefore the gravitational force exerted on the object is
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the mass of the object is
Answer:
The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s
Explanation:
The given parameters of the ball are;
The initial speed of the ball = 15 m/s
The direction in which the ball is launched = 50° above the horizontal
The location of the other tennis player when the ball is launched = 10 m from the ball
The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched
The height at which the ball is hit back = 2.1 m above the height from which the ball is launched
The vertical position, 'y', at time, 't', of a projectile motion is given as follows;
y = (u·sinθ)·t - 1/2·g·t²
When y = 2.1 m, we have;
2.1 = (15·sin(50°))·t - 1/2·9.8·t²
∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0
Solving with the aid of a graphing calculator function, we get;
t = 0.199776187257 s or t = 2.14525782198 s
Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds
The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;
x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m
The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m
Therefore, we have;
The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s
The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m
The minimum average speed the other player has to move with, = d/t₂
∴ = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s
The minimum average speed the opponent must move so that he is in position to hit the ball, ≈ 5.79 m/s.