Ask question

Ask question

All categories

3 years ago

8

A boat heads north in still water at 4.5 m/s directly across a river that is running east at 3.0 m/s. What is the velocity of th

e boat with respect to Earth?

2 answers:

Fynjy0 [20]3 years ago

4 0

<h2>Velocity of the boat with respect to Earth is 5 m/s 56.31° north of east</h2>

Explanation:

A boat heads north in still water at 4.5 m/s directly across a river that is running east at 3.0 m/s.

Let north be positive y axis and east be positive x axis.

We have

Velocity of boat = 4.5 j m/s

Velocity of river = 3 i m/s

Velocity of the boat with respect to Earth = 3 i + 4.5 j

$\texttt{Magnitude of velocity = }\sqrt{3^2+4.5^2}\\\\\texttt{Magnitude of velocity = }5.41m/s$

$\texttt{Angle = }tan^{-1}\left ( \frac{4.5}{3}\right )\\\\\texttt{Angle = }56.31^0$

That is 56.31° north of east.

Velocity of the boat with respect to Earth is 5 m/s 56.31° north of east

Ilya [14]3 years ago

3 0

Answer:

5.41 m/s

Explanation:

velocity of boat with respect to water , Vb,w = 4.5 j

velocity of water with respect to ground, Vw,g = 3 i

velocity of boat with respect to water = velocity of boat - velocity of water

Vb,w = Vb, g - Vw, g

4.5 i = Vb,g - 3i

Vb,g = 3 i + 4.5 j

The magnitude of velocity = $\sqrt{3^{2}+4.5^{2}}$ = 5.41 m/s

You might be interested in

A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release

iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= $2pie\frac sqrt {m}{k}$

$\frac {{f2}/times {fo}}$ =1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒ $\frac{f2} {f1}$ = $\frac sqrt{m1}[m2}$

⇒ $\frac{f2}{fo}$ = 1:3

⇒f2= $\frac{fo} {3}$

6 0

3 years ago

a boy takes 15min to reach his school on bicycle.if the bicycle have speed of 2m/s the what is the distance between the house an

Rom4ik [11]

<h2>Answer:</h2>

The distance will be 1800 m

<h2>Explanation</h2>

As in question

Time = 15 min

Time = 15 x 60 sec = 900 sec

Speed = 2 m/s

We know that

$Distance = Speed x Time$

$Distance = 2 x 900$

$Distance = 1800 m$

So, the answer is 1800 m

4 0

3 years ago

Which country or region is coal-rich and oil-poor?

zimovet [89]

THE ANSWER FOR UR QUESTION IS

WHICH COUNTRY OR REGION IS COAL RICH AND OIL POOR
ANS:- UNITED STATES IS THE COUNTRY WHICH IS COAL RICH AND OIL POOR

HOPE THIS HELPS!!!

^_^ HAPPY TO HELP U ^_^

5 0

3 years ago

What happens to the density of a given substance if you increase the amount of the substance that you have?

Amanda [17]

Have you ever looked up the density of a substance ? You ought to try it. Go ahead. Pick a substance, then go online or open up an actual book and find its density. You will never see any particular volume mentioned along with the density . . . because it doesn't matter. The whole idea of density is that it describes the substance, no matter how much or how little you have of it. The density of a tiny drop of water under a microscope is the same as the density of a supertanker-ful of water.

7 0

4 years ago

Read 2 more answers

In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar

katrin2010 [14]

Time period of any moon of Jupiter is given by

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

from above formula we can say that mass of Jupiter is given by

$M = \frac{4 \pi^2 r^3}{GT^2}$

now for part a)

$r = 4.22 * 10^8 m$

T = 1.77 day = 152928 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}$

$M = 1.9* 10^{27} kg$

Part B)

$r = 6.71 * 10^8 m$

T = 3.55 day = 306720 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}$

$M = 1.9* 10^{27} kg$

Part c)

$r = 10.7 * 10^8 m$

T = 7.16 day = 618624 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}$

$M = 1.89* 10^{27} kg$

PART D)

$r = 18.8 * 10^8 m$

T = 16.7 day = 1442880 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}$

$M = 1.889* 10^{27} kg$

6 0

3 years ago