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3 years ago

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If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you

r speed is decreased by the water's speed. The current in a river flows at 0.52 m/s. In still water you can swim at 1.73 m/s. If you swim downstream a certain distance, then back again upstream, how much longer, in percent, does it take compared to the same trip in still water?

1 answer:

Blababa [14]3 years ago

5 0

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is $t_1$ and when he swims in downward direction is $t_2$

Lets distance is d and it will be remain constant in both the case

$d=(u+v)t_1$

$d=(u-v)t_2$

$(1.73+0.52)t_1=(1.73-0.52)t_2$

$t_2=1.85t_1$

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

$t_1=0.44d\ sec$

$t_2=0.82d\ sec$

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

$percentage\ time =\dfrac{1.28-1.15}{1.15}$

Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

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Answer:

Explanation:

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4 years ago

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.

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Answer:

Mass of the planet = 1.48 × 10²⁵ Kg

Mass of the star = 5.09 × 10³⁰ kg

Explanation:

Given;

Diameter = 1.8 × 10⁷ m

Therefore,

Radius = $\frac{\textup{Diameter}}{\textup{2}}$ = $\frac{\textup{1.8}\times10^7}{\textup{2}}$

or

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Radius of star = 2.2 × 10¹¹ m

Orbit period = 407 earth days = 407 × 24 × 60 × 60 seconds = 35164800 s

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Now,

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g = $\frac{\textup{GM}}{\textup{R}^2}$

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on substituting the respective values, we get

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or

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T² = $\frac{\textup{4}\pi^2}{\textup{G}M_{star}}(R_{star})^3$

on substituting the respective values, we get

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3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

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Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

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Explanation:

a) given:

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4J + 18J = 22J

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Kinetic energy =

$\frac{1}{2} * 4Vf^2$

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Total energy =

2Vf² - 0.024

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4 years ago

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Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

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To solve the problem it is necessary to take into account the concepts related to energy efficiency in the engines, the work done, the heat input in the systems, the exchange and loss of heat in the soupy the radius between the work done the lost heat ( efficiency).

By definition the efficiency of the heat engine is

$\epsilon = 1- \frac{T_c}{T_h}$

Where,

$T_c =$ Temperature at the room

$T_h =$ Temperature of the soup

The work done is defined as,

$dW = \epsilon(dQ_h)$

Where $Q_h$ represents the input heat and at the same time is defined as

$dQ_h =c_v (dT_h)$

Where,

$c_V =$ Specific Heat

The change at the work would be defined then as

$dW = \epsilon(dQ_h)$

$dW = \epsilon c_v (dT_h)$

$dW = (1-\frac{T_c}{T_h})c_v (dT_h)$

$W = \int dW = \int (1-\frac{T_c}{T_h})c_v (dT_h)$

$W = c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})$

On the other hand we have that the heat lost by the soup is equal to

$dQ_h =c_v (dT_h)$

$Q_h =c_v (T_h-T_c)$

The ratio between both would be,

$\frac{W}{Q_h} = \frac{c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})}{c_v (T_h-T_c)}$

$\frac{W}{Q_h} = \frac{1+ln(\frac{T_h}{T_c})}{1-\frac{T_h}{T_c}}$

Replacing with our values we have,

$\frac{W}{Q_h} = 1+\frac{ln(\frac{340K}{300K})}{1-\frac{340K}{300K}}$

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Therefore the fraction of heat lost by the soup that can be turned into useable work by the engine is 0.0613.

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3 years ago