I have an exam coming soon and trying to revise previous papers. Can you solve the questions in the attachment and show me how to reach the correct answer? Please remember to keep track of significant figures and units.
Some of the questions from attachment are listed below to give you some idea about the kind of questions. Please see the attachment for complete and all the questions.
3. Two otherwise identical wires are made of different materials, the second having twice the resistivity of the first. The first wire has a resistance of 2 Ohms, what is the resistance of the second wire?
6. If the effective spring constant of a 2000 kg older car (with poor shocks) is 6.0 x 10^4 N/m, what is the period of its vibration after hitting a bump?
8. What is the frequency of the second harmonic of a 2.00 m long stretched wire if the wave speed is 171.5 m/s?
15. In the figure below (in the attachment), a light ray traveling from air enters an unknown substance. What is the index of refraction for the unknown substance?
19. An object is placed 2.0 cm in front of diverging lens, which has a power of -25 diopters. Where is the image located at?© BrainMass Inc. brainmass.com October 25, 2018, 2:44 am ad1c9bdddf
Please refer to the attachment.
1. Speed (v), frequency (ν) and wave length (λ) are related as: v = νλ
or ν = v/λ = 350/0.1 = 3500 Hz
2. Unit of current is ampere.
3. Resistance of a wire R = ρl/A where ρ is the resistivity, l is the length and A is the cross sectional area. Hence, l and A being equal, resistance is directly proportional to resistivity. The second wire will have double the resistance of the first i.e. 4 ohm.
4. As per Coulomb's law, the force between the charges is inversely proportional to square of the distance between them. Hence, on doubling the distance, the force reduces by a factor of 4.
5. Fundamental frequency on a stretched string is given by: ν1 = (1/2L)√T/m
Substituting ν1 = 144 Hz, length of the string L = 0.9 m, mass per unit length m = 2x10-3/0.9 = 2.22x10-3 kg/m
144 = (1/2x0.9)√T/(2.22x10-3)
√T/(2.22x10-3) = 144x2x0.9 = 259.2
Squaring both the sides: T/(2.22x10-3) = 6.72x104
T = 149 N
6. Time period of oscillations of a mass m resting on a spring of constant k is given by:
Tension in cord, Period of vibration, Resistance and Force between charges are examined.
Environmental stressors are briefly discussed.
Use these THREE environmental stressors,
Evaluate how these stressors affect individuals.
ALSO discuss STRATEGIES to manage the above THREE STRESSORS (including utilizing technology).View Full Posting Details