Physics report Essay Example

  • Category:
    Physics
  • Document type:
    Assignment
  • Level:
    High School
  • Page:
    2
  • Words:
    997

PHYSICS REPORT

Problem one:

  1. speed of the rider

The centre of mass for the system travels through an arc

Radius of the arc = 14.0 m

Friction is negligible

The riders are close to the centre of mass

Using the formula; v2 = g x r

v is the speed of the riders

g is the acceleration due to gravity (9.81 m/s2)

r is the radius of the arc

We have;
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  1. Centripetal acceleration at the bottom of the arc

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  1. Free body diagram indicating the forces experienced by a rider at the bottom of the arc

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Free body diagram

  1. Force exerted on a 60.0 Kg rider by the ride and its comparison with the weight of the rider

Weight of the rider = Mass of the rider x gravitational acceleration

But mass of the rider = 60.0 kg and;

Gravitational acceleration = 9.81 m/s

Therefore;

Weight of the rider = 60.0 kg x 9.81 m/s

Weight of the rider = 588.6 N

Force exerted on a 60.0 Kg rider by the ride = M x (v2/r)

Force exerted on a 60.0 Kg rider by the ride = 60.0 kg x (9.812/14) = 412.4 N

  1. Discussion of the reasonability of the answer in (4) above

The answer obtain in (4) above in the two cases is no reasonable because it is ideally expected that the force exerted by the rider should be equal to her weight. However, in this case the weight is more than the force exerted by the rider.

Problem Two:

  1. Force in the cables

Considering moments on the right hand side and those on the left hand side at the hinge point and then equating them for the system to remain in equilibrium;

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For equilibrium; Anticlockwise moments = Clockwise moments

(T sin 40° x (7.5 + 1.5)) = (1.5 x 2500 x 9.81)

(T sin 40° x 9.0) = (1.5 x 2500 x 9.81) + (4.5 x 900 x 9.81)

T = [(1.5 x 2500 x 9.81) + (4.5 x 900 x 9.81)] / (sin 40° x 9.0) = 13227 N

Therefore force in the cables (T) = 13227 N

  1. Magnitude and direction of the force that the hinges exert on the bridge

Upward forces = downward forces

(F + 13227 Sin 40°) = [(2500 x 9.81) + (900 x 9.81)]

F = [(2500 x 9.81) + (900 x 9.81)] – [13227 Sin 40°]

Magnitude, F = 24.85 KN

Direction = Tan-1 (24859 /1.5) = 89°

Problem Three:

  1. Showing that a vertical spring is able to exert a force of 2.00 mg in the upward direction on object at the lowest point

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Upward force excerted by the spring

= Mass x gravitational acceleration x ?x

Upward force excerted by the spring = m x g x ?x

Upward force excerted by the spring = m x g x 2 = 2.00mg

  1. Finding the amplitudes of the oscillations

Amplitude = Maximum displacement by the spring

Force constant for the spring, k = 10.0 N/m

Mass of objects attached to spring, m = 0.25 Kg

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Amplitude = 10.0 x 0.25 x 2.0 =20 mm

  1. Maximum velocity

Maximum velocity = ω x A = 2π x f x A

But f = 1/ ω = 1/40 = 0.025

Therefore; Maximum velocity = ω x A = 2π x 0.025 x 40 = 6.28 m/s

Problem Four:

  1. Area of array for photovoltaic array of solar cells if the average intensity of sunlight on one day is 700 W/m2

Efficiency of photovoltaic array = 10.0%

Average intensity of sunlight = 700W/m2

Rate of energy gathering = 100W

Area of array = (Rate of energy gathering / Average intensity of sunlight) x Efficiency of photovoltaic array

Area of array = (100 W / 700 W/m2) x 10.0% = 0.014 m2

  1. Maximum cost of array

Period of operation = 2 years

Duration of operation per day = 10.0 hours/day

Rate of earning money = 9.00 ¢ per kilowatt-hour

Maximum cost of array = 10.0 hours/day x 100 W x 9.00 per kilowatt-hour = 9000 ¢ in a day

Problem Five:

  1. Echo times

For temperature = 5 oC

V = 331 √1 + (T/273)

Where V = Speed of sound in air at a temperature of 5 oC in m/s and T is the temperature of in degree Celsius

V = 331 √1 + (5/273) = 334 m/s

Therefore Time = Distance / (2 x Speed of sound in air at a temperature of 5 oC

Time = 3 / (2 x 334 m/s) = 0.0040 seconds

For temperature = 35 oC

V = 331 √1 + (T/273)

Where V = Speed of sound in air at a temperature of 5 oC in m/s and T is the temperature of in degree Celsius

V = 331 √1 + (35/273) = 352 m/s

Therefore Time = Distance / (2 x Speed of sound in air at a temperature of 35 oC

Time = 3 / (2 x 352 m/s) = 0.0043 seconds

  1. percentage uncertainty

These temperature and times cause a very small percentage uncertainty that could range between 5 to 10 per cent for the bats in in locating their prey.

  1. Significance of percentage uncertainty

The percentage of uncertainty caused to the bats in locating the insects or their prey is significance in the sense that it demonstrates the effectiveness of this mode of feeding by the bat. The percentage of uncertainty in the use of sound echoes by bats plays a significant role in their lives. In this case, the bats find themselves in a position to approach the insect in a strategic manner to lower the chances of uncertainties. The uncertainties experienced by the bats also cause them to become highly specialized when it comes to locating obstacles. The percentage of uncertainty also indicates the manner in which a bat traces an insect (Pickover, 124).

Work Cited

Pickover, Clifford A. The Physics Book: From the Big Bang to Quantum Resurrection, 250

Milestones in the History of Physics. New York: Sterling Pub, 2011. Print.