Timmy drove 2/5 of a journey at an average speed of 20 mph.

A flea walking along a ruler moves from the 45 cm mark to the 27 cm mark. It does this in 3 seconds. What is the speed? What is

ad-work [718]

Answer:

Explanation:

In the calculation of distance we do not take into account the direction of movement .So only magnitude of movement is taken care of.

On the other hand , in respect of displacement , we take into account even direction of movement along with magnitude.

Distance covered by flea = 45 - 27 = 18 cm

Displacement by flea = 27 - 45 = - 18 cm

Speed = Distance covered / time

= 18 / 3 = 6 cm / s

Velocity = - 18 / 3 = - 6 cm /s.

3 0

3 years ago

A man at point A directs his rowboat due north toward point B, straight across a river of width 100 m. The river current is due

prohojiy [21]

Answer:

1.35208 m/s

Explanation:

Speed of the boat = 0.75 m/s

Distance between the shores = 100 m

Time = Distance / Speed

$Time=\frac{100}{0.75}=133.33\ s$

Time taken by the boat to get across is 133.33 seconds

Point C is 150 m from B

Speed = Distance / Time

$Speed=\frac{150}{\frac{100}{0.75}}=1.125\ m/s$

Velocity of the water is 1.125 m/s

From Pythagoras theorem

$c=\sqrt{0.75^2+1.125^2}\\\Rightarrow c=1.35208\ m/s$

So, the man's velocity relative to the shore is 1.35208 m/s

3 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 curve in a stretch of highway has radius R. The road is not banked in any way. The coefficient of static friction between the

adelina 88 [10]

Answer:

maximum possible velocity = $\sqrt{ugR}$

Explanation:

centripetal acceleration when the car is going in the circle must be less than the maximum friction for the car to not slip.

centripetal acceleration $\frac{mv^{2}}{r}$

where v is the velocity of car and r is the radius of circle

maximum friction = umg

where u is the coefficient of static friction.

therefore $umg\geq \frac{mv^{2}}{R}$

therefore maximum possible velocity = $\sqrt{ugR}$

6 0

3 years ago

In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b

lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

Diameter of ride=5 m

radius(r)=2.5 m

Static friction coefficient range=0.60-1

Here Frictional force will balance weight

And limiting frictional force is provided by Centripetal force

$f=\mu N=\mu m\omega ^2\cdot r$

weight of object=mg

Equating two

f=mg

$\mu m\omega ^2\cdot r=mg$

$\omega ^2=\frac{g}{\mu r}$

$\omega =\sqrt{\frac{g}{\mu r}}$

$\omega =2.55 rad/sec$

8 0

3 years ago