Answer:
a) F = 4.9 10⁴ N, b) F₁ = 122.5 N
Explanation:
To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height
1) pressure is defined by the relation
P = F / A
to lift the weight of the truck the force of the piston must be equal to the weight of the truck
∑F = 0
F-W = 0
F = W = mg
F = 5000 9.8
F = 4.9 10⁴ N
the area of the pisto is
A = pi r²
A = pi d² / 4
A = pi 1 ^ 2/4
A = 0.7854 m²
pressure is
P = 4.9 104 / 0.7854
P = 3.85 104 Pa
2) Let's find a point with the same height on the two pistons, the pressure is the same
where subscript 1 is for the small piston and subscript 2 is for the large piston
F₁ = 
the force applied must be equal to the weight of the truck
F₁ = 
F₁ = (0.05 / 1) ² 5000 9.8
F₁ = 122.5 N
Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
Answer:
The toy must calculate the person's speed/velocity
Explanation:
Since the school toy given to Henry can be used to tell how fast someone is moving, the toy must be able to calculate the person's speed/velocity using the <u>average distance</u> covered by the person divided by <u>time taken</u> to cover the distance; average distance ÷ time taken.
The toy must be able to determine the parameters (average distance and time taken) in order to be able to calculate the person's speed/velocity
Answer:
A) ω = 6v/19L
B) K2/K1 = 3/19
Explanation:
Mr = Mass of rod
Mb = Mass of bullet = Mr/4
Ir = (1/3)(Mr)L²
Ib = MbRb²
Radius of rotation of bullet Rb = L/2
A) From conservation of angular momentum,
L1 = L2
(Mb)v(L/2) = (Ir+ Ib)ω2
Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.
(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2
(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2
Divide each term by Mr;
vL/8 = (L²/3 + L²/16)ω2
vL/8 = (19L²/48)ω2
Divide both sides by L to obtain;
v/8 = (19L/48)ω2
Thus;
ω2 = 48v/(19x8L) = 6v/19L
B) K1 = K1b + K1r
K1 = (1/2)(Mb)v² + Ir(w1²)
= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)
= (1/8)(Mr)v²
K2 = (1/2)(Isys)(ω2²)
I(sys) is (Ir+ Ib). This gives us;
Isys = (19L²Mr/48)
K2 =(1/2)(19L²Mr/48)(6v/19L)²
= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152
Thus, the ratio, K2/K1 =
[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19
Answer:
I'm pretty sure it's 37.5 joules of energy
Explanation:
hope this helps!
