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3 years ago

What is this thing on my boat?

Engineering

1 answer:

MaRussiya [10]3 years ago

5 0

It may be an engine cooler, or it may be for the boats speed

But I think it is for cooling the engine down

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Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charg

babunello [35]

Answer:

Explanation:

To find the electrical force per unit area on each sheet we start defining our variables,

$\sigma_A = -4.10^{-5}C/m^2$

$\sigma_B= -7.10^{-5}C/m^2$

$\sigma_C = -3.1^{-5}C/m^2$

We find the electric field for each one, this formula is given by,

$E= \frac{\sigma_i}{2\epsilon_0}$

Substituting each value from the three charged sheets, we have

$E_A= \frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_B= \frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})$

$E_C= \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

The electric field is

$E_{NET}= E_A+E_B+E_C$

$E_{NET}=\frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})+\frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})+ \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_{NET} = 0$

Force on each sheet is,

$F=E_{NET}\sigma ds$

$F=0$

The total force is 0

5 0

3 years ago

What are the laws that apply to one vehicle towing another?

Sergeu [11.5K]

The drawbar or other connections must be strong enough to pull all the weight of the vehicle being towed. The drawbar or other connection may not exceed 15 feet from one vehicle to the other.

5 0

3 years ago

Select the correct answer. Ruby wants to create a cube puzzle game. For that she has to create a cube. Which drafting tool do yo

Alexxx [7]

Answer:

D. a triangle and a T-Square

Explanation:

A T-Square is the best drawing tool to create squares. You would need a squares to create cubes.

3 0

3 years ago

Is a street the same as a avenue

-BARSIC- [3]

they're essentially the same thing so i'd say yes

5 0

3 years ago

Read 2 more answers

Steam enters a counterflow heat exchanger operating at steady state at 0.06 MPa with a quality of 0.9 and exits at the same pres

Pie

Explanation:

$\begin{array}{ll}\1 & \2 \\P_{1}=0-\_{\text {mpa }} \text { Steam } & P_{2}=0.06 mpq \\x=0.9 & x=0 \text { (satunated liquid) }\end{array}$

a) Term of entering steam will be Tsaturated of corresponding pressure

$so, At P =0.06 mpa , T_{s a t}=85.92^{\circ} C$

$\begin{aligned}Q &=m c_{p}\left(T_{0}-T_{i}\right) \\&=\frac{100}{60} \times 1.005(60-30)=50.25 W\end{aligned}$

5 0

4 years ago