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Nikolay [14]
2 years ago
11

Can you hit the monkey at a fast speed with gravity on?

Physics
1 answer:
Ierofanga [76]2 years ago
7 0

Answer:

It depends how far the monkey is and how hard you throw the object because If the monkey is close, then you will most likely hit it

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Correctly round the following number to the ten thousandths place<br> 3.68274
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The answer to this question is 3.69
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What is the average mass of a neutron?
expeople1 [14]
A 1 amu is the answer
4 0
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Read 2 more answers
Compare and contrast instantaneous and average speed.
Harlamova29_29 [7]

Answer:

instantaneous velocity is a velocity covered at an instant while average velocity is the change in distance/ the change in time taken

6 0
3 years ago
A 500 kg sack of coal falls vertically onto a 2000 kg railroad flatcar which was initially moving horizontally at 3 m/s. no exte
Zinaida [17]

Since there are no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, we would assume that momentum is conserved. This means that

total momentum of car before collision = total momentum of car after collision.

Recall,

momentum = mass x velocity

From the information given,

mass of car before collision = 2000

velocity of car before collision = 3

Thus,

total momentum of car before collision = 2000 x 3 = 6000

Also,

mass of sack = 500

mass of car and sack after collision = 500 + 2000 = 2500

velocity after collision = v

momentum after collision = 2500 x v = 2500v

Since momentum is conserved, then

6000 = 2500v

v = 6000/2500

v = 2.4

the speed of the flatcar is 2.4 m/s

6 0
1 year 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

4 0
3 years ago
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