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zhenek [66]
3 years ago
6

Two fishing boats depart a harbor at the same time, one traveling east, the other south. the eastbound boat travels at a speed 1

mi/h faster than the southbound boat. after 5 h the boats are 25 mi apart. find the speed of the southbound boat.
Physics
1 answer:
Novosadov [1.4K]3 years ago
6 0
<span>they are travelling at right angles to each other.
 At any given instant they form a right triangle with their starting point
 </span>South bound <span>= x  [mi/h]
</span> East bound <span> = x+1 [mi/h]
 after five hours they will be
 d=5x
 and
 d=5(x+1)
 miles away from the starting point 
 (5x)^2+(5(x+1))^2=625
 25x^2+(5x+5)^2=625
 25x^2+25x^2+50x+25=625
 50</span>x^2+50x-600=0
<span> x^2+ x - 12=0
 (x+4)(x-3)=0
 take the postive value
 x= 3 mph the speed of south bound
 4mph east bound </span>
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The answer is B. desert. Deserts don't get much rainfall to begin with and most of it evaporates.
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2. A 55 kg woman has a momentum of 200 kg m/s. What is her velocity?
NeTakaya

Answer:

\boxed {\tt 3.63636364 \ m/s}

Explanation:

Velocity can be found using the following formula:

v=\frac{p}{m}

where p is the momentum and m is the mass.

The woman has a mass of 55 kilograms and a momentum of 200 kilogram meters per second.

p= 200 \ kgm/s\\m=55 \ kg

Substitute the values into the formula.

v=\frac{200 \ kg m/s}{55 \ kg}

Divide. Note that the kilograms, or kg, will cancel each other out.

v=\frac{200 \ m/s}{55}

v= 3.63636364 \ m/s

The woman's velocity is 3.63636364 meters per second.

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3 years ago
What is the resistance of a 600 W kettle that draws a current of 5.0 A? please answer with steps
lukranit [14]
P=I^2 *R

600 =5.0^2 *R

R=24

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5 1
3 years ago
Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet
kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

h_i = h_0 + WW = h_i -h_0

Using the relation T-P we can find the final temperature:

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\

\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

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3 years ago
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Answer:

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4 0
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