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Wewaii [24]
3 years ago
9

Please help!!! A river has a constant current of 3 km per hour. If a motorboat, capable of maintaining a constant speed of 20km

per hour, tries to reach a point directly opposite the dock, what will his actual speed and direction be? If the river is 2 km wide, how long will it take to cross?
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
4 0
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have: 

<span>theta = arcsin(3/20) = approx. 8.63° </span>

<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>

<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>

<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
<span>= approx. 91 seconds</span>
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The number of crates is 84580.

Explanation:

mass, m = 30 kg

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Power, P = 0.5 hp = 0.5 x 746 W = 373 W

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P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580

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a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde
Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

  • The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal
  • The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

Volume \ of \ a \ cone = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot h

  • Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

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  • \dfrac{dV}{dr} = \dfrac{d}{dr}  \left(\dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot  s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0

3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

\underline{\left  \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}

\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}

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An increase in the strength of intermolecular forces increases will lead to an increase in force applied to break the barriers posed by the strength of the molecules.

This increased intermolecular strength will cause a rise in boiling point,melting point, viscousity and surface tension.

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