e) A escalator carries 60 people of average mass 70kg to a height of 5m in one minute. Find the power needed to do this? 4
1 answer:
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<u>Answer:</u> The weight of the object is 29.4 N
<u>Explanation:</u>
To calculate the weight of the object, we use the equation:
where,
m = mass of the object = 3 kg
g = acceleration due to gravity =
Putting values in above equation, we get:
Hence, the weight of the object is 29.4 N
Answer:
The acceleration of the electron is 1.457 x 10¹⁵ m/s².
Explanation:
Given;
initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s
distance traveled by the electron, d = 0.01 m
final velocity of the electron, v = 5.4 x 10⁶ m/s
The acceleration of the electron is calculated as;
v² = u² + 2ad
(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a
(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²
(2 x 0.01)a = 2.91375 x 10¹³
Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².
Answer:
τ (bc. max) =25.37 MPa
Explanation:
From the question, T = 1.3Kn.m;
(G = 77.2 GPa) and from the image of this solid shaft system i attached;
d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m
So ΣT = 0 → Ta + Tc = 1.3Kn.m
So the system is statically indeterminate.
Let's check at the equation that makes it compatible ;
ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0
[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =
ΣT = 0 and T(bc) = T(c)
ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m
Now,
[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0
So, T(c) = 273374 Nmm = 273.374Nm
T(a) = 1300Nm - 273.374Nm = 1026. 63Nm
From the beginning we saw that;
T(bc) = T(c) = 273.374Nm
Now let's find the maximum shear stress in shaft BC;
τ (bc. max) = {τ (bc) x r(bc)} / J(bc)
τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707
= 25.37 MPa
