Answer:
58.8 N
Explanation:
The normal force is calculated as equal to the perpendicular component of the gravitational force.
Thus; N = mg
We are given m = 6 kg
Thus;
N = 6 × 9.8
N = 58.8 N
Thus, magnitude of normal force on the rock = 58.8 N
Alot as far as i know unless you need it in formal terms.
Answer:
Explanation:
Speed given = 125 m /min
125 /60 m /s
In 450 second it will travel
= 450 x 125 / 60
=937.5 m.
As the distance is covered in less than 450 seconds , The distance must be less than 937.5 m
In 400 seconds , it will travel
= 400 x 125 / 60
833.33 m
Since the distance is covered in more than 400 seconds , the distance must be more than ie 833.33 .
Hence the distance covered is more than .833 m but less than 937.5
In either case these distance are more than .8 km .
Answer:
The tank is losing 
Explanation:
According to the Bernoulli’s equation:
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume 
≅ 0 ;
then 
can be determined as:![\sqrt{[2g (h_1- h_2)]](https://tex.z-dn.net/?f=%5Csqrt%7B%5B2g%20%28h_1-%20h_2%29%5D)
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
![v_2 = \sqrt{[2*9.81*(20 - 15)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%2820%20-%2015%29%5D)
![v_2 = \sqrt{[2*9.81*(5)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%285%29%5D)
as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J = 
J = πr²
J =
J =
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh
₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is : 
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