A garden hose has a radius of 0.0120 m, and water initially comes out at a speed of 2.88m/s. Dasha puts her thumb over the end , which cuts its area to 1.05×10^-4 m^2.what is the new velocity of the water coming out ?

Aroll in a paper machine is powered by a separately excited dc machine. operating point 1 (under load): the roll typically operates at 764 rpm and must provide 301 nm of torque to oppose the load of the paper. the armature voltage is 230v for this operating point operating point 2 (no load): when the load is removed the speed of the roll increases to 900 rpm (the field characteristics are unchanged) la ea dev m (a) calculate the maximum torque (units of nm) that the motor can supply with this applied voltage (b) calculate the value of the armature resistance, ra (in ohms) (c) at operating point 1 (loaded condition), the field current, i, and the total input power (armature field), are measured to be to be 10 a and 28,600 w respectively. use this information to calculate the value of the field resistance, rf (in ohms). (d) (d) at operating point 1 (loaded condition), calculate the motor efficiency. express your answer as a percent to at least 2 significant digits. (d) if the field voltage is reduced by 1/2, calculate the new stall torque (in nm) of the motor (f). if the field voltage is reduced by 1/2, calculate the new no-load speed (in rad/sec)

Aheat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. the same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. heat pumps are widely used for home heating. you can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat qh, warming the indoors. perhaps this seems a little silly, but consider the following. electricity can be directly used to heat a home by passing an electric current through a heating coil. this is a direct, 100% conversion of work to heat. that is, 19.0 \rm kw of electric power (generated by doing work at the rate 19.0 kj/s at the power plant) produces heat energy inside the home at a rate of 19.0 kj/s. suppose that the neighbor's home has a heat pump with a coefficient of performance of 4.00, a realistic value. note: with a refrigerator, "what you get" is heat removed. but with a heat pump, "what you get" is heat delivered. so the coefficient of performance of a heat pump is k=qh/win. an average price for electricity is about 40 mj per dollar. a furnace or heat pump will run typically 200 hours per month during the winter. what does one month's heating cost in the home with a 16.0 kw electric heater? what does one month's heating cost in the home of a neighbor who uses a heat pump to provide the same amount of heating?

Acommercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at -35°c by rejecting waste heat to cooling water that enters the condenser at 18°c at a rate of 0.25 kg/s and leaves at 26°c. the refrigerant enters the condenser at 1.2 mpa and 50°c. if the compressor consumes 3.3 kw of power, determine (a) the mass flow rate of the refrigerant, (b) the refrigeration load, (c) the cop, and (d) the minimum power input to the compressor for the same refrigeration load.