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Margarita [4]
1 year ago
8

Suppose that air resistance cannot be ignored. For the position at which the person has jumped from the platform and the cord re

aches its maximum stretch length, how does the stretch length of the cord in the case with air resistance compare with the stretch length in the case in which air resistance is ignored? Briefly state your reasoning.
Physics
1 answer:
Kamila [148]1 year ago
6 0

Answer:

The stretch cord stores potential energy as a result of stretching but due to kinetic energy, it will move back to its original state. Since air resistance is not being ignored in this case, it will experience a slight delay in stretching at first.

Explanation:

In case, where air resistance is being ignored the stretch cord will stretch as it normally does.              

  • Air resistance is a force that any object experiences as a result of its motion through the air.  

There are various factors that affect air resistance like speed, the density of air, area, the shape of an object etc. Meanwhile, the density of air changes with temperature or altitude. <em>Hence this force is not constant but is thought to be constant during short time frames.  </em>

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Taya2010 [7]
In a circular motion scenario, the force that pulls the revolving object towards the centre is the force that produces the centripetal acceleration. So, in this case, the tension on the string is the force that pulls the puck towards the centre.

Therefore, it is the tension in the string that causes the centripetal acceleration of the puck

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8 0
1 year ago
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Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he sw
Natasha_Volkova [10]

Answer:

Explanation:

Given

average speed of Phelps v_{avg}=2\ m/s

total distance d=200\ m

he swims first 100 m at an average speed if 1.8 m/s

so time taken is t_1=\frac{100}{1.8}=55.55\ s

suppose t_2 is the time taken to swim remaining half

average velocity is v_{avg}=\frac{displacement}{total\ time}

v_{avg}=\frac{100+100}{55.55+t_2}

t_2+55.55=\frac{200}{2}=100

t_2=44.45\ s

so velocity in the second half is

v_2=\frac{100}{45.45}

v_2=2.19\approx 2.2\ m/s                                      

3 0
1 year ago
Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller
jeka94

Answer:

Explanation:

Torque on smaller wheel

= F x r

50 x .30

= 15 Nm

Torque on larger wheel

= F x .5

For equilibrium

F x .5 = 15

F = 15 / .5

= 30 N

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2 years ago
In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to
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1 year ago
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Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m
OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

E_m = \frac{1}{2} m\omega_1^2A_1^2

The energy of the system having mass 2m is,

E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2

For the two expressions mentioned above remember that the variables mean

m = mass

\omega =Angular velocity

A = Amplitude

The energies of the two system are same then,

E_m = E_{2m}

\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2

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Remember that

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Replacing this value we have then

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\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}

\frac{A_1}{A_2} = 1

Therefore the ratio of the oscillation amplitudes it is the same.

5 0
2 years ago
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