Answer:
Answer is 257°C
Explanation:
Given
frequency = 628 Hz
wavelength = 0.49 m
speed of sound at 0°C = 325 m/s
Find
1.speed of sound at t °C=?
2.temperature = ?
Formula
speed of sound at t °C = speed of sound at 0°C [1+t/273]
Solution
1. speed of sound at t °C = f*∧
= 628*0.49
=307.7m/s
2.speed of sound at t °C = speed of sound at 0°C [1+t/273]
307.7 m/s = 325 m/s [1+ t/ 273]
307.7/325 = 1+t/273
0.94 = 1+t/273
multiplying 273 on both sides
(0.94)(273) = 1+t/273 (273)
258 = 1+ t
258 - 1 = t
257°C = t
I would go with the third one
Answer:
5.00
Explanation:
a sinθ=mλ
You are missing (a) so you move sinθ to the other side than solve from there.
Explanation:
1) For a positive point charge, the lines radiate outwards
for a negative point charge, the lines converge inwards
2) F = 2.3 X 10^-26 N
k = 9 X 10^9 N.m²/C
q1 = q2 = e = 1.6 X 10^-19 C
r = ?
F = kq1q2/r²
r² = kq1q2/F
r = √[kq1q2/F ]
r = √0.0100
r = 0.10m
The two protons are 0.10 m apart
3) The unit if electric field intensity is Newton-per-coulomb N/C