Answer:
quantity A is mass and quantity B is wright
Answer:
v₂ = 97.4 m / s
Explanation:
Let's write the Bernoulli equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Index 1 is for tank and index 2 for exit
We can calculate the pressure in the tank with the equation
P = F / A
Where the area of a circle is
A = π r²
E radius is half the diameter
r = d / 2
A = π d² / 4
We replace
P = F 4 / π d²2
P₁ = 397 4 /π 0.058²
P₁ = 1.50 10⁵ Pa
The water velocity in the tank is zero because it is at rest (v1 = 0)
The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa
Since the pipe is horizontal y₁ = y₂
We replace on the first occasion
P₁ = P₂ + ½ ρ v₂²
v₂ = √ (P1-P2) 2 / ρ
v₂ = √ [(1.50-1.013) 10⁵ 2/1000]
v₂ = 97.4 m / s
The force exerted on the car during this stop is 6975N
<u>Explanation:</u>
Given-
Mass, m = 930kg
Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s
Time, t = 2s
Force, F = ?
F = m X a
F = m X s/t
F = 930 X 15/2
F = 6975N
Therefore, the force exerted on the car during this stop is 6975N
When the object is at rest, there is a zero net force due the cancellation of the object's weight <em>w</em> with the normal force <em>n</em> of the table pushing up on the object, so that by Newton's second law,
∑ <em>F</em> = <em>n</em> - <em>w</em> = 0 → <em>n</em> = <em>w</em> = <em>mg</em> = 112.5 N ≈ 113 N
where <em>m</em> = 12.5 kg and <em>g</em> = 9.80 m/s².
The minimum force <em>F</em> needed to overcome <u>maximum</u> static friction <em>f</em> and get the object moving is
<em>F</em> > <em>f</em> = 0.50 <em>n</em> = 61.25 N ≈ 61.3 N
which means a push of <em>F</em> = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.
So:
(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.
(b) Friction has a magnitude of 15 N because it balances the pushing force.
(c) The object is in equilibrium and not moving, so the acceleration is zero.
The firearm is unloaded.
The ramrod establishes whether the firearm is loaded or unloaded. It thrusts the bullet down the barrel. It is marked by wrapping some tape around it ( at the muzzle end ). This is to effectively signal if the muzzleloader is empty or loaded.
