Answer:
Torque = 0.47 Nm
Explanation:
Torque = BIAsin∅
Since the plane is parallel to the magnetic field, ∅ = 90⁰
sin 90 = 1
Magnetic field, B = 0.30 T
current, I = 2.0 A
Radius, R = 0.50 m
A = πR²
A = π(0.5)² = 0.785 m²
Torque = 0.3 * 2 * 0.785
Torque = 0.47 Nm
<h2>
Answer: Sensory adaptation</h2>
Explanation:
Sensory adaptation refers to the adjustment of the sensory capacity of a person following prolonged exposure to stimuli.
To better understand this, it is necessary to explain that environmental stimuli cause a change in the sensitivity of a person's sensory receptors. Then, depending on the type of environmental stimulus, the determined receptor will be stimulated (related to the five main senses of the human being: sight, smell, taste, touch and hearing).
However, when a person gets used to a stimulus, adaptation occurs. Therefore, <u>it will not respond to the stimulus in the same way as it did before.
</u>
A very common example is the relation with the smells (olfactory sense), because people get used quickly to the smells that surround them and then they stop "perceiving" the smell.
The punching bag applies the same amount of force to the boxer’s hand. (C)
Answer:
638.8kW
Explanation:
The flow rate of the steam m = 22kg/s
The Pressure of the steam at the inlet of the turbine P1 = 1.6MPa
The temperature of the steam at the inlet of the turbine T1 = 350*C
Steam quality at the exit of the turbine x2 = 1.0
The temperature of the steam at the exit of the turbine T2 = 30*C
Power produced = 12,350kW
Assuming the turbine is running on a steady state, hence we neglect the effect of kinetic and potential energy we get:
If you refer to the superheated steam table for the specific enthalpy at a pressure of 1.6MPa and temperature of 350*C, we get
h1 = 3,146kJ/kg
Refer to the steam table for saturated gas at temperature 30*C to get the specific enthalpy value h2 = 2,556.81kJ/kg
The heat that comes out from the turbine can be defined from the balance of energy in the system, and is represented as
Ein - Eout = change in system Energy = 0
Thus Ein = Eout
mh1 = mh2 + Wout + Qout
Qout = m(hi-h2) - Wout
Qout = 22 x (3146-255.6) - 12350
Qout = 638.8kW
Explanation:
Work done is given by the product of force and displacement.
Case 1,
1. A boy lifts a 2-newton box 0.8 meters.
W = 2 N × 0.8 m = 1.6 J
2. A boy lifts a 5-newton box 0.8 meters.
W = 5 N × 0.8 m = 4 J
3. A boy lifts a 8-newton box 0.2 meters.
W = 8 N × 0.2 m = 1.6 J
4. A boy lifts a 10-newton box 0.2 meters.
W = 10 N × 0.2 m = 2 J
Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.
