A motorcycle accelerates uniformly from rest at 7.9\,\dfrac{\text{m}}{\text{s}^2}7.9 s 2 m 7, point, 9, space, start fraction,
1 answer:
Answer:
t = 3.516 s
Explanation:
The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:
Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0
If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.
This can be done knowing that
1 km = 1000 m
1 h = 3600 s
Therefore
1 km/h = (1000/3600) m/s = 0.2777... m/s
100 km/h = 27.777... m/s
Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:
27.777 m/s = 7.9 m/s^2 t
Solving for t
t = (27.7777 / 7.9) s = 3.516 s
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Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
= 1 m³
= 10°C = 283 K
= 350 kPa
= 3 kg
= 35°C = 308 K
= 150 kPa
Now, lets apply the ideal gas equation;
=
R
=
R
/
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
= ( 3 × 0.287 × 308) / 150
= 265.188 / 150
= 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, =
/ R
= (350 × 1)/(0.287 × 283) = 350 / 81.221
= 4.309 kg
Total mass, =
+
= 4.309 + 3 = 7.309 kg
Total volume =
+
= 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
=
R
/
given that; final temperature = 20°C = 293 K
we substitute
= ( 7.309 × 0.287 × 293) / 2.77
= 614.6211119 / 2.77
= 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa