All categories

3 years ago

At what angle North of East must the ship travel to reach its destination? Let East be 0◦ and North 90◦.

1 answer:

3 0

That kind of depends on where the ship is now, and where it wants to go.
We learn these kinds of things in navigation school, and I'll just bet that
it's the kind of information that's right there in the part of the question
that you didn't bother copying.

You might be interested in

What is the difference between Drift current & Diffusion current ?

dsp73

The main difference between the drift current and diffusion current is the corresponding reaction of the semiconductor materials in the electric fields. In electric fields, when semiconductor material's movement go along with the direction of the electric field it is called drift current, but when it tends to go from high concentrated area to lower concentrated area then it is diffusion current.

8 0

3 years ago

air passing over an airplanes wing travels ,and therefore exerts pressure.than air traveling beneath the wing.

pickupchik [31]

Bernoulli's principle of laminar/lamellar air flow, I think. High flow speed = low pressure, low flow speed = high pressure I think. So, the wings/aerofoils are designed to induce a low pressure on the top side of the wing and a high pressure on the underside of the wing, thus producing an "aerodynamic upthrust" (a static upthrust comes from an object in water via Archimedes) and LIFT.

Two "particles" of air one going topside and the other underside meet again at the end of their motion across the wing. So, top side has to travel faster than bottom side. So top side has a lower "dynamic pressure" than underside.

And all that for 5 points ????????? (If I'm right, of course ... )

5 0

3 years ago

A car starts from 0 m along a road and accelerates at 0.5 m/s^2 to the right. A second car starts from 1000 m along the road and

Brrunno [24]

Answer:

e) 31.6 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow s_1=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_1=\frac{1}{2}0.5t^2\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s_2=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_2=\frac{1}{2}1.5t^2\ m$

$s_1+s_2=1000$

$\\\Rightarrow 1000=\frac{1}{2}0.5t^2+\frac{1}{2}1.5t^2\\\Rightarrow 1000=\frac{0.5t^2+1.5t^2}{2}\\\Rightarrow 1000=\frac{2t^2}{2}\\\Rightarrow 1000=t^2\\\Rightarrow t=\sqrt{1000}\\\Rightarrow t=31.6\ s$

Time taken by the cars to meet 31.6 seconds.

5 0

3 years ago

In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push agains

Hoochie [10]

Answer:

$\frac{m_B}{m_E}=1.18$

Explanation:

$m_B$ = Mass of Bonzo

$m_E$ = Mass of Ender

$u_B$ = Initial Velocity of Bonzo = 2.2 m/s

$u_E$ = Initial Velocity of Ender = -2.6 m/s

From conservation of linear momentum

$m_Bu_B=m_Eu_E\\\Rightarrow \frac{m_B}{m_E}=\frac{u_E}{u_B}\\\Rightarrow \frac{m_B}{m_E}=\frac{2.6}{2.2}\\\Rightarrow \frac{m_B}{m_E}=1.18$

$\therefore \frac{m_B}{m_E}=1.18$

8 0

3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0

3 years ago