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3 years ago

A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled?

Physics

2 answers:

zhannawk [14.2K]3 years ago

5 0

Answer:

A boat travels for three hours with a... A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in still water?

Explanation:

Ganezh [65]3 years ago

4 0

Answer:

20 km/hr x 3hr = 60 km

Explanation:

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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

A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit

JulsSmile [24]

I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ = $\frac{y*d}{D} = \frac{3.55*10^{-3}*1.25*10^{-4} }{2.25} = 1.97*10^{-7} m$
λ ~ 1.97 ×10⁻⁷m

7 0

3 years ago

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo

LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0

3 years ago

A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration

babymother [125]

acceleration = change in velocity/change in time

so...

a = 20 m/s / 2 seconds

a = 10

hope that helps :)

P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.

7 0

3 years ago

Say you want to make a sling by swinging a mass M of 1.7 kg in a horizontal circle of radius 0.048 m, using a string of length 0

DENIUS [597]

Answer:

Tension in the string is equal to 58.33 N ( this will be the strength of the string )

Explanation:

We have given mass m = 1.7 kg

radius of the circle r = 0.48 m $F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N$

Kinetic energy is given 14 J

Kinetic energy is equal to $KE=\frac{1}{2}mv^2$

So $\frac{1}{2}\times 1.7\times v^2=14$

$v^2=16.47$

v = 4.05 m/sec

Centripetal force is equal to $F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N$

So tension in the string will be equal to 58.33 N ( this will be the strength of the string )

3 0

3 years ago