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padilas [110]
3 years ago
10

What is the output of the following program fragment. Choose appropriate data-types of variables to match output.

Engineering
1 answer:
Alexxx [7]3 years ago
3 0

Answer:

1

Explanation:

Given :

i=7;j=8;k=9;

printf(“%d”,(i+10)%k/j);

The printf function only displays the result

%d - is used for the display format

However, the actual calculation is in the expression: (i+10)%k/j

Given the variables :

i = 7 ;

j = 8 ;

k = 9 ;

(i + 10) = (7 + 10) = 17

(i + 10) % k = 17 % 9

% = remainder value after division

17 % 9 = (17 / 9) = 1 remainder 8

17 % 9 = 8

Hence,

(i + 10) % k = 8

(i + 10) % k / j

j = 8

8 / 8

= 1

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The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
Answer my question I will mark brainliest
Iteru [2.4K]

Answer:

150

Explanation:

Mark me Brainliest

6 0
3 years ago
Read 2 more answers
What must engineers keep in mind so that their solutions will be appropriate?
vekshin1

Answer:

Context

Explanation:

It is of great value for an engineer to keep the context of his/her experiment in mind.

7 0
3 years ago
A body weighs 50 N and hangs from a spring with spring constant of 50 N/m. A dashpot is attached to the body. If the body is rai
lbvjy [14]

Answer:

a) 3.607 m

b) 1.5963 m

Explanation:

See that attached pictures for explanation.

3 0
3 years ago
The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24C, and a wet-bulb temperature of 17C. Using the psychrome
TEA [102]

Answer:

(a) Relative Humidity = 48%,

Specific humidity = 0.0095

(b) Enthalpy = 65 KJ/Kg of dry sir

Specific volume = 0.86 m^3/Kg of dry air

(c/d) 12.78 degree C

(e) Specific volume = 0.86 m^3/Kg of dry air

8 0
3 years ago
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