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Kisachek [45]
2 years ago
13

Raul doesn’t feel like he needs to write down events that will happen months from now. Explain to him why it is important to use

the different types of calendars.
Engineering
2 answers:
MAXImum [283]2 years ago
4 0
The primary reason for having a calendar is to organize the days, weeks, months and years. It keeps a track of which day of the week events fall and when special events are going to happen. Historically speaking calendars were often uses to preserve religious holidays and events.
larisa86 [58]2 years ago
4 0

Answer: Calendars are important so that; Students know what’s coming up. Adequate time is planned for important events and studying different levels of detail are accounted for students may be organized and prepared for anything.

Explanation:

You might be interested in
1. Add:<br>(i) 5xy, -2xy, -11xy, 8xy<br>(iv) 3a - 2b + c, 5a + 8b -70​
Cerrena [4.2K]

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

8 0
1 year ago
Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b
grandymaker [24]

Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

3 0
2 years ago
Technician A that shielding gas nozzles may have different shapes. Technician B says that gelding gas nozzles is attached to the
Lilit [14]
B is correct! in this senecio
5 0
2 years ago
The radial component of acceleration of a particle moving in a circular path is always:________ a. negative. b. directed towards
lesya [120]

Answer:

d. all of the above

Explanation:

There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.

The radial acceleration is given by;

a_r = \frac{V^2}{R}

Where;

V is the velocity of the particle

R is the radius of the circular path

This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.

Therefore, from the given options in the question, all the options are correct.

d. all of the above

7 0
2 years ago
A 20-centimeter wrench is used to loosen a bolt. The force is applied 0.20m from the bolt. It takes 50 newtons to loosen the bol
defon

Answer:

The force would be 57.8 N if it was applied at a 30-degree angle from perpendicular.

Explanation:

We know that torque and force are represented by the following equation

τ =  r x F sinθ ---- (1)

where τ is Torque, r is radius, F is force and θ is the angle between the force and lever arm.

Given data:

F = 50 N

r = 0.2 m

we have to find the force when θ = 30°

Putting these values in equation 1, we get

τ  = 0.2 x 50 x sin(90°)

τ  = 10 Nm

so Torque is 10 Nm

Now we will calculate the force was applied at a 30 degree angle from perpendicular, in this case the equation 1 becomes as

τ =  r x F cosθ

Putting the values of τ , r, θ  and F in above equation

10 = 0.2 x F x cos(30°)

Rearranging the above equation

F = 10 / (0.2 x cos (30°))

F = 10 / 0.173

F = 57.8 N

8 0
2 years ago
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