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2 years ago

Electrical protective devices are designed to automatically

Engineering

2 answers:

lutik1710 [3]2 years ago

5 0

Answer:

can i have your number please

BabaBlast [244]2 years ago

5 0

The answer is 1 I just did the test hope this helps!

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Utility company power lines carry what kind of current?

vagabundo [1.1K]

Answer:

Alternating

Explanation:

8 0

3 years ago

Write a program to calculate overtime pay of 10 employees. Overtime is paid at the rate of Rs. 12.00

fgiga [73]

Answer:

Here is the code.

Explanation:

#include<stdio.h>

int main()

{

int i, time_worked, over_time, overtime_pay = 0;

for (i = 1; i <= 10; i++)

{

printf("\nEnter the time employee worked in hr ");

scanf("%d", &time_worked);

if (time_worked>40)

{

over_time = time_worked - 40;

overtime_pay = overtime_pay + (12 * over_time);

}

printf("\nTotal Overtime Pay Of 10 Employees Is %d", overtime_pay);

return 0;

}

Output :

Enter the time employee worked in hr 42

Enter the time employee worked in hr 45

Enter the time employee worked in hr 42

Enter the time employee worked in hr 41

Enter the time employee worked in hr 50

Enter the time employee worked in hr 51

Enter the time employee worked in hr 52

Enter the time employee worked in hr 53

Enter the time employee worked in hr 54

Enter the time employee worked in hr 55

Total Overtime Pay Of 10 Employees Is 1020.

6 0

3 years ago

What is an entrepreneur’s primary responsibility?

inessss [21]

Answer:

Entrepreneur's primary duty is to make money for himself and investors.

<Jayla>

6 0

3 years ago

Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch

Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 $\frac{ft^{3}}{sec}$

$D_{1}$ = 6 inch=0.5 ft

$D_{2}$ =2 inch=0.1667 ft

As we know that Q=AV

$A_{1}\times V_{1}=A_{2}\times V_{2}$

So $V_{2}=\frac{Q}{A_2}$

$V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}$

$V_{2$ =0.687 ft/sec

We know that Head loss due to sudden contraction

$h_{l}=K\frac{V_{2}^2}{2g}$

If nothing is given then take K=0.5

So head loss $h_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}$

=0.00366 ft

So head loss=0.00366 ft

4 0

3 years ago

The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume mus = 0.18.

barxatty [35]

The omitted part of the question shown in bold format

The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume μs = 0.18 and μk = 0.16. What couple must be applied to the shaft to exert a 30-lb force on the clamped object?

Answer:

C = 0.967 in. lb

Explanation:

Given that:

The lead of the threaded shaft of the C-clamp is 0.05 in.

∴ the pitch of the screw = 0.05 in

the mean radius of the thread is r = 0.15 in.

Assuming:

(μs)= 0.18 which implies the coefficient of the static friction

(μk) = 0.16 (coefficient of kinetic friction)

Force = 30-lb

What couple must be applied to the shaft to exert a 30-lb force on the clamped object?

To determine the Couple (C) that must be applied; we use the expression:

C = Fr tan ( $\theta_k$ + $\alpha$ )

where; F = force

r = mean radius

$\theta_k$ = angle of kinetic friction

$\alpha$ = pitch angle

NOW, let's take then one after the other.

From the coefficient of the static friction and the kinetic friction; we can solve for their respective angles, so we have:

Angle of static friction ( $\theta_s$ ) $= tan^{-1}(U_s)$

( $\theta_s$ ) = $tan^{-1}(0.18)$

( $\theta_s$ ) = 10.204°

Angle of kinetic friction ( $\theta_k$ ) $= tan^{-1}(U_k)$

$\theta_k$ $= tan^{-1}(0.16)$

$\theta_k$ = 9.0903°

To determine the pitch angle( $\alpha$ ); we apply the expression:

( $\alpha$ ) = $tan^{-1}(\frac{p}{2 \pi r} )$

( $\alpha$ ) = $tan^{-1}(\frac{0.05}{2 \pi 0.15} )$

( $\alpha$ ) = $tan^{-1}(0.0530516 )$

( $\alpha$ ) = 3.0368°

Have gotten our parameters to solve for Couple (C); then we have:

C = Fr tan ( $\theta_k$ + $\alpha$ )

substituting our values; we have:

C = (30 × 0.15) tan ( 9.0903 + 3.0368)

C = 4.5 × tan ( 12.1271)

C = 4.5 × 0.2148761968

C = 0.96669428854 in.lb

C = 0.967 in. lb

Therefore, 0.967 in. lb couple must be applied to the shaft to exert a 30-lb force on the clamped object.

6 0

4 years ago