Answer:
163.8 ft
Explanation:
In triangle ABD
= 155 ft
Using Pythagorean theorem in triangle ADC
= distance between the anchor points
distance between the anchor points is given as
A body of mass 10 kg is moving at a constant speed of 20 m s over a frictionless surface the work done by the weight is
1 answer:
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Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
Answer:
Explanation:
The equation relating the the enthalphy, pressure and temperature is expressed as
Where P is the pressure, H is the enthalphy, and T is the temperature.
since the given values are
if we insert values, we arrive at