A window washer on a hanging platform cleans the outside of windows on a skyscraper. The washer hoists the platform up the side

Answer:

The required energy remains identical in both scenarios since the specific heat capacity (Cp) does not change with varying pressure.

Explanation:

Given;

initial temperature, t₁ = 50 °C

final temperature, t₂ = 80 °C

Temperature change, ΔT = 80 °C - 50 °C = 30 °C

Pressure for scenario one = 1 atm

Pressure for scenario two = 3 atm

The energy needed in both scenarios is expressed as;

Where;

Cp denotes specific heat capacity, which only varies with temperature and remains unaffected by pressure.

Hence, the energy required remains the same for both scenarios since specific heat capacity (Cp) is pressure-independent.

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at   ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The total time in the air equals twice the time to reach maximum height

2 t₁ = 3.896 s

The horizontal distance covered is

D = v × t

D = 3.896×19.09

   = 74.375 m

The ball was in the air for 3.896 seconds

U = 1794.005 × 10⁶ J. Explanation: Information provided indicates that the capacitance of the original capacitor is C = 1.27 F, and the potential difference applied to it is V = 59.9 kV, or 59.9 × 10³ V. The potential energy (U) for the capacitor is determined by the formula: U = (1/2) × C × V². Substituting the respective values, we find U = (1/2) × 1.27 × (59.9 × 10³)², resulting in U = 1794.005 × 10⁶ J.

Response:

1) An observer in B 'perceives the two events occurring at the same time

2) Observer B recognizes that the events happen at different times

3)  Δt = Δt₀ /√ (1 + v²/c²)

Clarification:

This scenario illustrates the concept of simultaneity in special relativity. It is important to keep in mind that light's speed remains constant across all inertial frames

1) Since the events are stationary within the frame S ', they propagate at the constant speed of light, resulting in them reaching observation point B'—located equidistantly between both events—simultaneously

Thus, an observer in B 'observes the two events occurring at the same time

2) For an observer B situated within frame S attached to the Earth, both events at A and B appear to take place at the same moment. However, the event at A covers a shorter distance, while the event at B travels a longer distance, since frame S 'is in motion at velocity + v. Hence, with a constant speed, the event covering the lesser distance is perceived first.

Consequently, observer B perceives that the events do not occur simultaneously

3) Let's determine the timing for each event

        Δt = Δt₀ /√ (1 + v²/c²)

where t₀ represents the time in the S' frame, which remains at rest for the events

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

The worker is positioned at 2.5 m