Answer:
Efficiency = 52%
Explanation:
Given:
First stage
heat absorbed, Q₁ at temperature T₁ = 500 K
Heat released, Q₂ at temperature T₂ = 430 K
and the work done is W₁
Second stage
Heat released, Q₂ at temperature T₂ = 430 K
Heat released, Q₃ at temperature T₃ = 240 K
and the work done is W₂
Total work done, W = W₁ + W₂
Now,
The efficiency is given as:
or
Work done = change in heat
thus,
W₁ = Q₁ - Q₂
W₂ = Q₂ - Q₃
Thus,
or
or
also,
or
thus,
thus,
or
or
Efficiency = 52%
Answer:
a) T = 608.22 N
b) T = 608.22 N
c) T = 682.62 N
d) T = 533.82 N
Explanation:
Given that the mass of gymnast is m = 62.0 kg
Acceleration due to gravity is g = 9.81 m/s²
Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.
So;
To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;
T = mg
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs the rope at a constant rate tension in the string is
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs up the rope with an upward acceleration of magnitude
a = 1.2 m/s²
the tension in the string is T - mg = ma (Since acceleration a is upwards)
T = ma + mg
= m (a + g )
= (62.0 kg)(9.81 m/s² + 1.2 m/s²)
= (62.0 kg) (11.01 m/s²)
= 682.62 N
When the gymnast climbs up the rope with an downward acceleration of magnitude
a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)
T = mg - ma
= m (g - a )
= (62.0 kg)(9.81 m/s² - 1.2 m/s²)
= (62.0 kg)(8.61 m/s²)
= 533.82 N
Answer:
Explanation:
given,
refractive index of lens, n = 1.70
Radius of curvature of front surface. R₁ = 20 cm
Radius of curvature of the back surface, R₂ = 30 cm
focal length= ?
R₁ = +20 cm
R₂ = -30 cm
n = 1.70
the focal length of the lens is equal to 17.15 cm
Answer:
c
Explanation:
though c is wider it has more water.
