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RSB [31]
3 years ago
5

A cannon ball is fired with an arching trajectory such that at the highest point of the trajectory the cannon ball is traveling

at 98 m/s. If the acceleration of gravity is 9.81 m/s^2, what is the radius of curvature of the cannon balls path at this instant?
Engineering
1 answer:
ELEN [110]3 years ago
8 0

Answer:

The radius of curvature is 979 meter

Explanation:

We have given velocity of the canon ball v = 98 m/sec

Acceleration due to gravity g=9.81m/sec^2

We know that at highest point of trajectory angular acceleration is equal to acceleration due to gravity

Acceleration due to gravity is given by a_c=\frac{v^2}{r}, here v is velocity and r is radius of curvature

So \frac{98^2}{r}=9.81

r = 979 meter

So the radius of curvature is 979 meter

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Answer:

The essence including its problem is listed throughout the clarification section following.

Explanation:

Projects build deliverable that seem to be the products of the venture or indeed the implementation of the project. This ensures that perhaps the agile methodology may be as broad as either the goal of the study itself as well as the coverage that would be part of a much larger venture.

For every other production to have been marked as "deliverable" within the same project, this should satisfy a few eligibility requirements:

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So that the above seems to be the right answer.

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3 years ago
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A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a
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Answer:

Q_{cv}=-339.347kJ

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

P_iV_i=m_iRT_i (i could be 1 for initial and 2 for the end)

State1

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State2

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m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

u_1=210.49kJ/kg

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h_1=295.17kJ/kg

For the second state the Specific internal Energy (6bar, 350K)

u_2=250.02kJ/kg

At the end we make a Energy balance, so

U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e

No work done there is here, so clearing the equation for Q

Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)

Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)

Q_{cv}=-339.347kJ

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

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