Answer:work is done, and temperature increases
Explanation:
In an adiabatic process, when gases are compressed, work is done on the liquid and the temperature increases
Answer:
f = 485.62 N
Explanation:
Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:
Unbalanced Force = Horizontal Component Applied Force - Frictional Force
Unbalanced Force = Fx - f
But, from Newtons Second Law of Motion:
Unbalanced Force = ma
comparing the equations:
ma = Fx - f
f = F Cos θ - ma
where,
f = frictional force = ?
F = Applied force = 593 N
m = mass of person = 49 kg
a = acceleration = 0.57 m/s²
θ = Angle with horizontal = 30°
Therefore,
f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)
f = 513.55 N - 27.93 N
<u>f = 485.62 N</u>
Answer:
Use a faster than normal approach and landing speed.
Explanation
For pilots, it is one of the critical moments of the flight that concentrates 12% of fatal accidents. The main difficulty lies in reaching enough speed to take flight within the space of the runway. At present, it ceased to be a challenge for the aircraft, since the engine power improved, so the takeoff ceased to be the most dangerous moment of the flight.
One of the risks that aircraft face today is that some of the engines fail while the plane accelerates. In that case, the pilot must decide in an instant whether it is better to take flight and solve the problem in the air or if it is preferable not to take off.
Although for many staying on the ground might seem the most sensible option, it is not as simple as it seems: to suddenly decelerate an aircraft, with the weight it has and the speed it reaches can cause accidents. However, today a special cement was designed that runs around the runways of the airports, which when coming into contact with the wheels of the aircraft the ground breaks and helps to slow down.
Answer:
3.135 kN/C
Explanation:
The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}
Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m
So, E = qz/{4πε₀[√(z² + R²)]³}
E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}
E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}
E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}
E = 6.839 × 10³ Cm²/[1.297 m]³F
E = 6.839 × 10³ Cm²/2.182 m³F
E = 3.135 × 10³ V/m
E = 3.135 × 10³ N/C
E = 3.135 kN/C
Answer:
the answer would be microwelds.
