B. +Q, + W are the correct sign
The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E







Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
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Here As we can see the figure that the end of the rope is pulled by some force F
Now as we can see that Piano is connected by a pulley which is passing over the pulley so effectively net force on the piano upwards will be 2F as it is connected by 2 ropes by the pulley
Now for constant velocity of the piano we will say

since velocity is constant so acceleration must be ZERO
so here we have

as we know here that
mg = 1000 N
so we will have


so here force must be 500 N
So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
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