Physics report Essay Example
 Category:Physics
 Document type:Assignment
 Level:High School
 Page:2
 Words:997
PHYSICS REPORT
Problem one:

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; v^{2 }= g x r
v is the speed of the riders
g is the acceleration due to gravity (9.81 m/s^{2})
r is the radius of the arc
We have;

Centripetal acceleration at the bottom of the arc

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

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 (v^{2}/r)
Force exerted on a 60.0 Kg rider by the ride = 60.0 kg x (9.81^{2}/14) = 412.4 N

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:

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;
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

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 = Tan1 (24859 /1.5) = 89°
Problem Three:

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
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

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
Amplitude = 10.0 x 0.25 x 2.0 =20 mm

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:

Area of array for photovoltaic array of solar cells if the average intensity of sunlight on one day is 700 W/m^{2}
Efficiency of photovoltaic array = 10.0%
Average intensity of sunlight = 700W/m^{2}
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/m^{2}) x 10.0% = 0.014 m^{2}

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 kilowatthour
Maximum cost of array = 10.0 hours/day x 100 W x 9.00 per kilowatthour = 9000 ¢ in a day
Problem Five:

Echo times
For temperature = 5 ^{o}C
V = 331 √1 + (T/273)
Where V = Speed of sound in air at a temperature of 5 ^{o}C 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 ^{o}C
Time = 3 / (2 x 334 m/s) = 0.0040 seconds
For temperature = 35 ^{o}C
V = 331 √1 + (T/273)
Where V = Speed of sound in air at a temperature of 5 ^{o}C 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 ^{o}C
Time = 3 / (2 x 352 m/s) = 0.0043 seconds

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.

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.