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1 year ago

6

One particular descent goes from 2100m to 1600m. Assuming work done against friction is 90% of the potential energy change of th

e cyclist and the cycle, what increase in speed in Km/h can a rider attain by the end of the
descent?

1 answer:

alisha [4.7K]1 year ago

7 0

Answer:

1/2 M V^2 = .1 M g H where 10% of PE goes into KE

V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2

V = 31.1 m/s increase in speed during descent

1 km / hr = 1000 m / 3600 sec = .278 m/s

V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr

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Name the nutrients required for the<br>body.

Anettt [7]

Answer:

1- water

2- fat

3- carbohydrates

4- vitamins

5- minerals

6 0

2 years ago

during a lunar mission, it is necessary to increase the speed of a spacecraft by 2.2 m/s when it is moving at 400 m/s relative t

wel

The initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

<h3>What fraction of the initial mass of the spacecraft?</h3>

Increase the speed: Vf-Vi = 2.2 m/s

Speed of aircraft: Vr = 400 m/s

Speed of ejected products: Vrel = 1000 m/s

The answer is:

$V_f - V_i = V_{rel} log_{e} (\frac{mi}{mf})\\\\2.2 = 1000 log_{e} (\frac{mi}{mf})\\\\ log_{e} (\frac{mi}{mf} ) = \frac{2.2}{1000} \\\\ log_{e} (\frac{mi}{mf} ) = 0.0022$

$\frac{mi}{mf} = e^{0,0022} \\\\\frac{mf}{mi} = e^{-0,0022} \\\\\frac{mi-mf}{mi} = 1 - e^{-0,0022} = 0,00219$

So, the initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

Learn more about spaceship speed fraction brainly.com/question/28256735

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3 0

8 months ago

A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?

Advocard [28]

Answer:

$v=697.2km/h$

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

$V=\frac{d}{t}$

The distance is clearly 1743 km and the time is:

$t=2h+30min*\frac{1h}{60min} =2.5h$

Thus, the velocity turns out:

$v=\frac{1743km}{2.5h}\\ \\v=697.2km/h$

Which is a typical velocity for a plane to allow it be stable when flying.

Best regards.

5 0

2 years ago

The LR5 is the specialist submarine for underwater rescue. The average density of seawater is 1028 kg/ m3.

sladkih [1.3K]

Answer:

P = 7196 [kPa]

Explanation:

We can solve this problem using the expression that defines the pressure depending on the height of water column.

P = dens*g*h

where:

dens = 1028 [kg/m^3]

g = 10 [m/s^2]

h = 700 [m]

Therefore:

P = 1028*10*700

P = 7196000 [Pa]

P = 7196 [kPa]

5 0

3 years ago

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0

2 years ago