The formula we can use in this case is:
v = v0 + a t
where v is final velocity, v0 is initial velocity, a is
acceleration and t is time
So finding for v0:
v0 = v – a t
v0 = 43.7 – (2.5) 2.7
v0 = 36.95 m/s
Answer:
The energy returns to the weightlifter's muscles, where it is dissipated as heat.
Explanation:
The energy returns to the weightlifter's muscles, where it is dissipated as heat. As long as the weightlifter controls the weight's descent, their muscles are acting as an overdamped shock absorber, as if the weight were sitting on a piston containing very thick fluid, slowly compressing it downward (and slightly heating up the fluid in the process). Since muscles are complicated biological systems and not simple pistons, they require metabolic energy to maintain tension throughout the controlled descent, so the weightlifter feels like they're putting energy into the weight, even though the weight's gravitational potential energy is being converted into heat within the lifter's muscles.
Constant velocity means the netto force = 0, therefore F(gravity) = F(astronaut).
175N divided by 87,5kg = 2.00kg/N
The correct answer is Metals.
Generally, the specific heat of metals is low. Very high specific heat exists in water.A physical feature of matter known as heat capacity or thermal capacity is the quantity of heat that must be applied to an object in order to cause a unit change in temperature. Heat capacity is measured in joules per kelvin (J/K), the SI unit. A broad property is heat capacity. Use the following equation to determine heat capacity: heat capacity = E / T, where E is the quantity of delivered heat energy and T is the change in temperature. The formula would be as follows, for instance, if it takes 2,000 Joules of energy to raise a block's temperature by 5 degrees Celsius: 2,000 Joules per °C is the heat capacity.
Learn more about heat capacity here :-
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Answer:
Explanation:
Given:
Steam Mass rate, ms = 1.5 kg/min
= 1.5 kg/min × 1 min/60 sec
= 0.025 kg/s
Air Mass rate, ma = 100 kg/min
= 100 kg/min × 1 min/60 sec
= 1.67 kg/s
A.
Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.
xf, quality = 0.9.
Tsat = 89.9°C
hf = 376.57 kJ/kg
hfg = 2283.38 kJ/kg
Using the equation for specific enthalpy,
hi = hf + (hfg × xf)
= 376.57 + (2283.38 × 0.9)
= 2431.552 kJ/kg
The specific enthalpy of the outlet, h2 = hf
= 376.57 kJ/kg
B.
Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy
= ms × (hi - h2)
= 0.025 × (2431.552 - 376.57)
= 0.025 × 2055.042
= 51.37455 kW
= 51.38 kW.
