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liraira [26]
3 years ago
11

Eq-45 what type of boat requires navigation lights?

Physics
1 answer:
irina [24]3 years ago
8 0

All the boats operating at night requires Navigation light.

Navigation light helps prevent collisions between boats and see if visibility is poor. The types of boat are : Rowboats, Tug boats, Vessels, Sailboat etc.

Various boats have different lightning color to show its use and side of the boat.

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Given f(x)=3x-11 which of the following is f(7)<br><br> A. 10<br> B. 32<br> C. -6<br> D. 6
Crazy boy [7]

Answer:

The answer is A. 10

Explanation:

<em>Given </em>

<em>f(</em><em>x</em><em>)</em><em> </em><em>=</em><em> </em><em>3x </em><em>-</em><em> </em><em>1</em><em>1</em>

<em>So, </em><em> </em>

<em>f(</em><em>7</em><em>)</em><em> </em><em>=</em><em> </em><em>3</em><em> </em><em>*</em><em> </em><em>7</em><em> </em><em>-</em><em> </em><em>1</em><em>1</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>2</em><em>1</em><em> </em><em>-</em><em> </em><em>1</em><em>1</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>1</em><em>0</em>

5 0
3 years ago
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When evaluating data, why is it better to make a graph instead of just looking at the raw data in the table?(8th grade Science s
ziro4ka [17]

Answer:All professors planned and thought about data before graph construction. When reflecting on their graphs, professors and graduate students focused on the ... feel more inclusive in the learning process and gain better science process skills, ... We organized data in a table instead of a paragraph with numbers

Explanation:

hope that helps

3 0
3 years ago
A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl
Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

Q = C_w BH^{3/2}

H = (\dfrac{Q}{C_wB})^{2/3}

H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}

H = 0.371\ m

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0
3 years ago
A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo
Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

E_{1}=E_{pot}+E_{kin}+E_{elas}\\

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

E_{2}=E_{kin}+E_{pot}

Now we can use the first statement to get the first equation:

E_{1}+W_{1-2}=E_{2}

where:

W₁₋₂ = work from the state 1 to 2.

E_{k}=\frac{1}{2} *m*v^{2} \\

E_{pot}=m*g*h

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5

58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]

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