1 ) The Sun emits electromagnetic waves (including light) equally in all directi

Question

1 ) The Sun emits electromagnetic waves (including light) equally in all directions. The intensity of the waves at the Earth's upper atmosphere is 1.5 kW/m2. At what rate does the Sun emit electromagnetic waves? unit in W

2 ) When the tension in a cord is 79 N, the wave speed is 130 m/s. What is the linear mass density of the cord? unit in g/m

3 ) Two strings, each 18.3 m long, are stretched side by side. One string has a mass of 78.0 g and a tension of 180.0 N. The second string has a mass of 58.0 g and a tension of 151.2 N. A pulse is generated at one end of each string simultaneously. On which string will the pulse move faster?

B ) Once the faster pulse reaches the far end of its string, how much additional time will the slower pulse require to reach the end of its string? unit in ms

Explanation / Answer

1) The intensity at a distance d from the source is given by I = P/A A is the surface area of the sphere of radius d P is the power emitted per unit time unit area from the source P = IA P = (1500 W/m2)(4*(1.5*1011 m)2)            (surface area of the sphere A = 4R2) P = 4.239*1026 W 2) The tension in the string is T = 79 N The speed of the wave is v = 130 m/s The wave speed in a stretched string is given by v = (T/) is the linear density of the string 130 m/s = (79 N/) = 4.675*10-3 kg/m 3) The lengths of the strings are L = 18.3 m The mass of string 1 is m1 = 78 g The tension applied is T1 = 180 N The mass of the string 2 is m2 = 58 g The tension applied is T2 = 151.2 N The wave speed on string 1 is v1 = (T1L/m1) = 205.5 m/s The wave speed on string 2 is v2 = (T2L/m2) = 218.4 m/s The wave speed in string 2 is greater than that of string 1 The time taken by the wave on string 1 to reach the other end is t1 = L/v1 = 89.05 ms The time taken by the wacv on string 2 to reach the other end is t2 = L/v2 = 83.79 ms The additional time taken by the slower pulse to reach the other end is t = 5.26 ms 130 m/s = (79 N/) = 4.675*10-3 kg/m 3) The lengths of the strings are L = 18.3 m The mass of string 1 is m1 = 78 g The tension applied is T1 = 180 N The mass of the string 2 is m2 = 58 g The tension applied is T2 = 151.2 N The wave speed on string 1 is v1 = (T1L/m1) = 205.5 m/s The wave speed on string 2 is v2 = (T2L/m2) = 218.4 m/s The wave speed in string 2 is greater than that of string 1 The time taken by the wave on string 1 to reach the other end is t1 = L/v1 = 89.05 ms The time taken by the wacv on string 2 to reach the other end is t2 = L/v2 = 83.79 ms The additional time taken by the slower pulse to reach the other end is t = 5.26 ms The wave speed on string 2 is v2 = (T2L/m2) = 218.4 m/s The wave speed in string 2 is greater than that of string 1 The time taken by the wave on string 1 to reach the other end is t1 = L/v1 = 89.05 ms The time taken by the wacv on string 2 to reach the other end is t2 = L/v2 = 83.79 ms The additional time taken by the slower pulse to reach the other end is t = 5.26 ms The time taken by the wacv on string 2 to reach the other end is t2 = L/v2 = 83.79 ms The additional time taken by the slower pulse to reach the other end is t = 5.26 ms

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.