Answer:
The answer is below
Explanation:
A diver works in the sea on a day when the atmospheric pressure is 101 kPa. The diver uses compressed air to breathe under water. 1700 litres of air from the atmosphere is compressed into a 12-litre gas cylinder. The compressed air quickly cools to its original temperature. Calculate the pressure of the air in the cylinder.
Solution:
Boyles law states that the volume of a given gas is inversely proportional to the pressure exerted by the gas, provided that the temperature is constant.
That is:
P ∝ 1/V; PV = constant
P₁V₁ = P₂V₂
Given that P₁ = initial pressure = 101 kPa, V₁ = initial volume = 1700 L, P₂ = cylinder pressure, V₂ = cylinder volume = 12 L. Hence:
P₁V₁ = P₂V₂
100 kPa * 1700 L = P₂ * 12 L
P₂ = (100 kPa * 1700 L) / 12 L
P₂ = 14308 kPa
<span>Gravitational force is affected by: a. mass c. distance b. weight d. both a and c
Mass.
</span>When an object is above the Earth's surface it hasgravitational potential<span> energy (GPE). The amount of GPE an object has depends on its mass and its height above the Earth's surface</span><span>
</span>
Answer:

Explanation:
The initial velocity, u, of the car=15m/s
The final velocity, v, of the car =0m/s
Time, t, taken for the car to come to a stop=5s
Acceleration is calculated by,

By substitution,



The negative sign implies that the car has decelerated.
Answer:
The tension in string P is 25 N, while that of Q is 85 N.
Explanation:
Considering the conditions for equilibrium,
i. Total upward force = Total downward force
+
= 110 N
ii. Taking moment about P,
clockwise moment = anticlockwise moment
110 × (2.5 - 0.8) =
× (3 - 0.8)
110 × 1.7 =
× 2.2
187 = 2.2
= 
= 85 N
From the first condition,
+
= 110 N
+ 85 N = 110 N
= 110 - 85
25 N
Therefore, the tension in string P is 25 N while that of Q is 85 N.
Answer:
4.92°
Explanation:
The banking angle θ = tan⁻¹(v²/rg) where v = designated speed of ramp = 30 mph = 30 × 1609 m/3600 s = 13.41 m/s, r = radius of curve = 700 ft = 700 × 0.3048 m = 213.36 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the variables into the equation, we have
θ = tan⁻¹(v²/rg)
= tan⁻¹((13.41 m/s)²/[213.36 m × 9.8 m/s²])
= tan⁻¹((179.8281 m²/s)²/[2090.928 m²/s²])
= tan⁻¹(0.086)
= 4.92°