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skad [1K]
3 years ago
15

What is projectile motion​

Engineering
2 answers:
Alenkinab [10]3 years ago
5 0

\boxed{\large{\bold{\blue{ANSWER~:) }}}}

  • If an object is given an initial velocity in any direction and then allowed to travel freely under gravity only, it is called a projectile motion
OleMash [197]3 years ago
5 0

Explanation:

of course, it is called a projectile motion

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A program contains the following function definition: int cube(int number) { return number * number * number; } Write a stateme
Nonamiya [84]

Answer:

The statement can be written as

int result = cube(4);

Explanation:

A function is a block of reusable codes to perform some tasks. For example, the function in the question is to calculate the cube of a number.

A function can also operate on one or more input value (argument) and return a result. The <em>cube </em>function in the question accept one input value through its parameter <em>number </em>and the <em>number</em> will be multiplied by itself twice and return the result.  

To call a function, just simply write the function name followed with parenthesis (e.g. <em>cube()</em>). Within the parenthesis, we can include zero or one or more than one values as argument(s) (e.g. <em>cube(4)</em>).

We can then use the "=" operator to assign the return output of the function to a variable (e.g. <em>int result = cube(4)</em>)

8 0
3 years ago
Convert A'B'C'D' + A'B'C'D + A'B'CD' + A'BC'D + AB'C'D' + AB'C'D+ AB'CD' to SOP form
bazaltina [42]

Answer:

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6 0
3 years ago
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Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3
goldenfox [79]

Answer:

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.

Step-by-step solution:

Step 1 of 5

Given data:-

The velocity of water is .

The water flow rate is.

3 0
3 years ago
In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length
masya89 [10]

Answer:

Time period  = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration  9.4 \times 10^{-5} kg/s

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration = 9.4\times 10^{-5} \times 62 = 582.8\times 10{-5} kg/s

total volume of air  infiltration= \frac{582.8\times 10{-5}}{1.156} = 5.04\times 10^{-3} m^3/s

Time period = \frac{210}{5.04\times 10^{-3}} = 41654.08 s

3 0
3 years ago
Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 10mA. The zener has an incremental resistance of rZ = 30
hram777 [196]

Answer:

The answer is given in the explanation.

Explanation:

The circuit is as indicated in the attached figure.

From the analytical description the zener voltage is given as

V_z=V_z_o+I_zr_z

Here

Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.

The equivalent model is shown in the attached figure.

From the above equation, Vzo is calculated as

V_z_o=V_z-I_zr_z

Here Vz is given as 7.5 V

Iz is given as 10 mA

rz is given as 30 Ω

Thus the Vzo is given as

V_z_o=V_z-I_zr_z\\V_z_o=7.4-30*10*10^{-3}\\V_z_o=7.5-0.3\\V_z_o=7.2 V

The value of I_L is given as 5 mA

Now the expression of current is as

I=I_z+I_L\\I=10mA+5mA\\I=15 mA

Now the resistance is calculated as

R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66

So the value of resistance is 186.66 Ω.

Considering the supply voltage is increased by 10%

V is 10-10%*10=10+1=11 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V

Considering the supply voltage is decreased by 10%

V is 10-10%*10=10-1=9 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V

Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA

so

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V

Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus

V_z_o=V_z-I_zr_z\\V_z_o=7.5-30*0.5*10^{-3}\\V_z_o=7.5-0.015\\V_z_o=7.485 V

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA

The load voltage is 7.485 V

8 0
3 years ago
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