Question

the speed of sound in air at zero degree celsius is 332 metre per second if it increases the rate of 0.6 metre per second per degree what will be the temperature when the velocity has increased to 344 metre per second.??

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Answer 1

ANSWERThe speed of sound in air at 00 ℃ = 332m/s=332m/s

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃Thus, 344 = 332 + 0.6x344=332+0.6x

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃Thus, 344 = 332 + 0.6x344=332+0.6x ⇒ 344 - 332 = 0.6x344−332=0.6x

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃Thus, 344 = 332 + 0.6x344=332+0.6x ⇒ 344 - 332 = 0.6x344−332=0.6x ⇒ 12 = 0.6x12=0.6x

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃Thus, 344 = 332 + 0.6x344=332+0.6x ⇒ 344 - 332 = 0.6x344−332=0.6x ⇒ 12 = 0.6x12=0.6x ∴ x = \frac{12}{0.6}x=

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃Thus, 344 = 332 + 0.6x344=332+0.6x ⇒ 344 - 332 = 0.6x344−332=0.6x ⇒ 12 = 0.6x12=0.6x ∴ x = \frac{12}{0.6}x= 0.6

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃Thus, 344 = 332 + 0.6x344=332+0.6x ⇒ 344 - 332 = 0.6x344−332=0.6x ⇒ 12 = 0.6x12=0.6x ∴ x = \frac{12}{0.6}x= 0.612

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃Thus, 344 = 332 + 0.6x344=332+0.6x ⇒ 344 - 332 = 0.6x344−332=0.6x ⇒ 12 = 0.6x12=0.6x ∴ x = \frac{12}{0.6}x= 0.612

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃Thus, 344 = 332 + 0.6x344=332+0.6x ⇒ 344 - 332 = 0.6x344−332=0.6x ⇒ 12 = 0.6x12=0.6x ∴ x = \frac{12}{0.6}x= 0.612

The speed of sound in air at 00 ℃ = 332m/s=332m/sNow, If it increases at the rate of 0.6m/s0.6m/s per degree,Then, Speed at xx ℃ = (332 + 0.6x)=(332+0.6x) ℃Thus, 344 = 332 + 0.6x344=332+0.6x ⇒ 344 - 332 = 0.6x344−332=0.6x ⇒ 12 = 0.6x12=0.6x ∴ x = \frac{12}{0.6}x= 0.612 = 20=20 ℃