A body accelerates uniformly from rest at \(2ms^{-2}\). Calculate the velocity after traveling 9m
36\(ms^{-1}\)
18\(ms^{-1}\)
6\(ms^{-1}\)
4.5\(ms^{-1}\)
4.24\(ms^{-1}\)
Correct answer is C
\(V^2= U^2+2as\) =\(0+ 2\times 2ms^{-2}\times 9 =36ms^{-1}\) thus
V=\(\sqrt{36m^2s^{-2}}\) = \(6ms^{-1}\)
21.2\(ms^{-1}\)
30\(ms^{-1}\)
300\(ms^{-1}\)
450\(ms^{-1}\)
900\(ms^{-1}\)
Correct answer is B
\(v^2 = u^2+2as=0^2 + 2\times 10\times 45 \Rightarrow
v^2 = 900m^2s^{-2} \Rightarrow \\
v = \sqrt{900m^2s^{-2}} \Rightarrow
v = 30ms^{-1}\)
0.15A
0.20A
2.10A
3A
6.6A
Correct answer is A
I =\(\frac{E}{R+r} = \frac{1.5}{2.5+0.5+7}\) = 0.15A
A note of frequency of 2000Hz has a velocity of \(400ms^{-1}\). What is the wavelength of the note?
800m
200m
5m
2m
0.2m
Correct answer is E
\(\lambda = \frac{v}{f} = \frac{400}{2000} = 0.2\)m
0.8 x \(10^{-7}\)m
1 x \(10^{-7}\)m
2.3 x \(10^{-7}\)m
3.8 x \(10^{-7}\)m
12.4 x \(10^{-7}\)m
Correct answer is C
\(\theta = E - E_f = h_f = \frac{v}{\lambda} \Rightarrow = \lambda = \frac{h_v}{\theta}
\frac{6.6\times 10^{-34} \times 3 \times 10^{8}}{8.6 \times 10^{-19}} = 0.8 \times 10^{-7}\)m