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Yuki888 [10]
2 years ago
5

The Slingshot is a ride for two people. It consists of a single passenger cage, two towers, and two elastic bands. Potential ene

rgy is stored in the elastic bands and the passenger cage is released. On the way up, this potential energy in the elastic bands is converted into the kinetic energy of the cage. At the maximum height of the ride, the energy has been converted into gravitational potential energy of the cage. The slingshot has two towers of height h = 76 m. The towers are a distance d = 31 m apart. Each elastic band has an unstretched length of L0 = 41 m and a spring constant of k = 310 N/m. The total mass of the passengers and cage is m = 410 kg. The car is pulled down to the ground in the middle of the two towers.
Calculate the maximum height, in meters, of the ride.
Physics
1 answer:
Vitek1552 [10]2 years ago
8 0

my bad I was in a herie last time can you please answer my question , I am going to give you the 5 points back for this question and extra 30 points ,

Explanation:

first its going to say 10 points for my question but after that I well make answer a small question and give you 30points. like whats your favorite color . stay tuned .

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A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng
frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

Q_{1} =\frac{W}{n}

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

Q_{1} =\frac{750}{0.35} =2142.86kJ

Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ

c) The efficiency of Carnot engine is:

n=1-\frac{300K}{550K} =0.455

The heat transfer to the heat engine is:

Q_{1c} =\frac{750}{0.455} =1648.35kJ

The heat transfer from the heat engine is:

Q_{2c} =1648.35-750=898.35kJ

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Which trailer has more downward pressure where it attaches to the car?
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The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.

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Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.
Aleonysh [2.5K]

De broglie wavelength, \lambda = \frac{h}{mv}, where h is the Planck's constant,  m is the mass and v is the velocity.

h = 6.63*10^{-34}

Mass of hydrogen atom,  m = 1.67*10^{-27}kg

v = 440 m/s

Substituting

   Wavelength \lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m

1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm

So  the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0
2 years ago
A 4kg block is sliding on a horizontal frictionless floor at a speed of 2.5ms and runs into a horizontal spring. The spring has
alexgriva [62]

Answer:

Explanation:

Given

mas of block m=4\ kg

speed of block v=2.5\ m/s

spring constant k=30\ N-m

As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring

\frac{1}{2}mv^2=\frac{1}{2}kx^2

x=v\sqrt{\frac{m}{k}}

x=2.5\times \sqrt{\frac{4}{30}}

x=0.912\ m

8 0
2 years ago
Read 2 more answers
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
rosijanka [135]

Answer:

v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s

Explanation:

For this case we know that the initial velocity is given by:

v_i = 7 \frac{m}{s}

The final velocity on this case is given by:

v_f = 13 \frac{m}{s}

And we know that it takes 8 seconds to go from 7m/s to 13m/s. We can use the following kinematic formula in order to find the acceleration during the first interval:

v_f = v_i +at

If we solve for the acceleration we got:

a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}

So for the other traject we assume that the acceleration is constant and the train travels for 16 s. The initial velocity on this case would be 13m/s from the first interval and we can find the final velocity with the following formula:

v_f = v_ i +a t

And if we replace we got:

v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s

5 0
2 years ago
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