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Tems11 [23]
3 years ago
12

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00

m above the ground and experience negligible air resistance. How far from where the child is standing does the ball hit the ground?
Physics
1 answer:
mr_godi [17]3 years ago
5 0

Answer:

Explanation:

initial velocity, u = 8 m/s

vertical height, h = 1 m

θ = 40°

Let the horizontal distance is d and the time taken is t.

Use second equation of motion in vertical direction

h = ut + 1/2 at²

1 = 8 Sin 40 x t + 0.5 x 9.8 t²

1 = 5.14 t - 4.9t²

4.9t² - 5.14 t + 1 = 0

so, t = 0.26 s (smaller value)

So, the horizontal distance is

d = u cos 40 x t

d = 8 cos 40 x 0.26

d = 1.6 m

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4 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
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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

4 0
3 years ago
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