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3 years ago

A 14-foot boat operating under oars after dark must show what color light to prevent a collision?

1 answer:

7 0

<span>A "White" colored light must be shown by a 14-foot boat that is operating under oars after it is dark, in order to prevent a collision.

Boats which are less than 23 feet cannot exceed a maximum speed of 7 knots. They need to exhibit an all-around white light when it is sailing in the dark.

The other practicable sidelights can also be used instead of the white lights, when they are sailing in international waters.
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Does friction always oppose relative motion?

Ilya [14]

Answer:

Yes

Explanation:

Friction is a force that opposes relative motion between systems in contact. One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion or attempted motion of the systems relative to each other.

3 0

2 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

2 years ago

The angle of incidence and the angle of reflection are measured from the_____, the perpendicular to the surface.

kogti [31]

I believe it is angles

6 0

2 years ago

Suppose Joe, who can type 40 words per minute, takes 2.5 hours to type his report. To speed things up, his friend Mark offers to

Vika [28.1K]

Explanation:

When Joe works alone, the total number of words he typed can be given by:

Total words = (40 words per minute) x (60 minutes per hour) x (2.5 hours)

Total words = 6000 words

Now, when Joe and Mark work together, let 'y' be the number of hours for which they both work simultaneously:

Total words = Words Typed by Joe + Words Typed by Mark

6000 = {(40 words per minute) x (60 minutes per hours) x (y hours)} + {(20 words per minute) x (60 minutes per hours) x (y hours)}

6000 = 2400y + 1200y = 3600y

y = 1.67 hours = 1 hour and 40 minutes

Thus, working together simultaneously, Joe and Mark will take 1 hour and 40 minutes to complete the report.

5 0

2 years ago

Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to

Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

L = W

Area

A = W W = W²

R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

Length L = D= 2W

Area A = W L

R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

R₂ / R₁ = 2W²/L

6 0

2 years ago