Answer:
4
Explanation:
From the question given above, the following data were obtained:
Effort (E) = 80 lbs
Load (L) = 320 lbs
Mechanical advantage (MA) =?
Mechanical advantage is simply defined as the ratio of load to effort. Mathematically, it is expressed as:
Mechanical advantage = Load / Effort
MA = L / E
With the above formula, we can obtain the mechanical advantage as illustrated below:
Effort (E) = 80 lbs
Load (L) = 320 lbs
Mechanical advantage (MA) =?
MA = L / E
MA = 320 / 80
MA = 4
Thus, the mechanical advantage is 4
The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.
Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³
Given:
At ground level,
p₁ = 752 mm Hg
= (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
= 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
= 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
= 300.8 K
At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
= 9.7326 x 10³ Pa
T₂ = 235 K
If the volume at 36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
= (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
= 762.15 m³
Answer: 762.2 m³
Answer:
answer is 1000 N
formula used-
<em><u>F= m x (v-u/t)</u></em>
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Answer:
v_f = 3 m/s
Explanation:
From work energy theorem;
W = K_f - K_i
Where;
K_f is final kinetic energy
K_i is initial kinetic energy
W is work done
K_f = ½mv_f²
K_i = ½mv_i²
Where v_f and v_i are final and initial velocities respectively
Thus;
W = ½mv_f² - ½mv_i²
We are given;
W = 150 J
m = 60 kg
v_i = 2 m/s
Thus;
150 = ½×60(v_f² - 2²)
150 = 30(v_f² - 4)
(v_f² - 4) = 150/30
(v_f² - 4) = 5
v_f² = 5 + 4
v_f² = 9
v_f = √9
v_f = 3 m/s
