What is NOT an example of professional networking?

Best use for a suspension bridge

babunello [35]

Answer:

over a rive or fast moving water or canyon

Explanation: you would use a suspension bridge in an area where you can't put supports down.

4 0

3 years ago

Who play 1v1 lol unblocked games 76

nikklg [1K]

Answer:

I did not what is it about?

8 0

3 years ago

Read 2 more answers

A 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?

seraphim [82]

Answer:

No.

Explanation:

$P_r$ = Power rating = 1 W

R = Resistance = $350\ \Omega$

V = Voltage = $24\ \text{V}$

Power is given by

$P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}$

$1.65\ \text{W}>1\ \text{W}$

So

$P>P_r$

Hence, the resistor is not operating within its power rating.

8 0

3 years ago

A hypothetical accumulator processor uses one of the following 16-bit instruction formats, depending on the instruction.(a) (10

Jlenok [28]

Answer:

Every part has been answered with complete detail. See the pictures.

Explanation:

See the pictures for detailed answer.

7 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

3 years ago