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mr Goodwill [35]
4 years ago
10

A space shuttle is sitting stationary at the launch pad.

Physics
1 answer:
beks73 [17]4 years ago
3 0

Answer: D. potential energy: high

kinetic energy: zero

Explanation:

I'm not a certified expert so please correct me if i'm wrong

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A ball is thrown vertically up from the edge of a cliff with a speed of 8 m/s, how high is
anygoal [31]

Explanation:

s=d/t

d=s×t

d=8×16.4

d=131.2

distance is 131.2m.

7 0
3 years ago
A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant
Juliette [100K]
<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force F applied to it, <u>as long as the spring is not permanently deformed</u>:

F=k (x-x_{o})    (1)

Where:

k is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

x_{o} is the length of the spring without applying force.

x is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth W of the block:

W=m.g   (2)

Where m=250g=0.25kg is the mass of the block and g=9.8\frac{m}{s^{2}}  is the gravity acceleration.

W=(0.25kg)(9.8\frac{m}{s^{2}})   (3)

W=2.45N   (4)

Knowing the force applied W and x=18cm=0.18m and x_{o}=0, we can substitute the values in equation (1) and find k:

W=k (x-x_{o})    (5)

2.45N=k (0.18m-0m)    (6)

<u>Finally:</u>

k=13.61\frac{N}{m}  

4 0
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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