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Tcecarenko [31]

2 years ago

13

A boat can travel in still water at 56 m/s. If the boats sails directly across a river that flows at 126 m/s. What is the boats

speed relative to the ground

1 answer:

MAVERICK [17]2 years ago

7 0

Answer:

The answer is below

Explanation:

The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:

V² = 56² + 126²

V² = 19012

V = 137.9 m/s

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Please help answer question

nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

4 0

2 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

6 0

2 years ago

Determine la fuerza aplicada sobre un émbolo en una prensa hidráulica si se quiere cargar una lanza de 2970kg; toma en cuenta qu

Artist 52 [7]

I’m not sure sorry I really wish I could help

4 0

2 years ago

a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres

Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

$\text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}$

Similarly,

$\text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}$

We know the below formulas,

$\text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}$

$\text {volume of wire}=\pi \times r^{2} \times l$

When equating both the equations, we can find length of wire as below, where $\pi=\frac{22}{7}$

$\frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l$

$\frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l$

The $\pi$ value gets cancelled as common on both sides, we get

$\frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l$

The $10^{-6}$ value gets cancelled as common on both sides, we get

$l=4 \times 9=36 m$

7 0

2 years ago

What is the speed of a wave in (m/s) with a 5 meter wavelength and a period of 20 seconds?

arlik [135]

Answer: 0.25 m/s

Explanation: Speed = wavelengt · frequency

v = λf and frequency is 1/period f = 1/T

Then v = λ/T = 5 m / 20 s = 0.25 m/s

6 0

2 years ago