Answer:
the first one is D
Explanation:
so if the others u put are right the the second would be c
Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
Answer:
The focal lenth (F) =+10.0cm
Explanation:
The formular for combined focal length (F) is given as;
In this question,
F1 = 20cm
F2 = -30cm
Plugging the values into the formuar above,
![1/f = 0.05 - 0.033[tex]1/f = -0.017f = [tex]1/ -0.017](https://tex.z-dn.net/?f=1%2Ff%20%3D%200.05%20-%200.033%3C%2Fp%3E%3Cp%3E%5Btex%5D1%2Ff%20%3D%20-0.017%3C%2Fp%3E%3Cp%3Ef%20%3D%20%5Btex%5D1%2F%20-0.017)
f = 58.82cm
i.e. the combination behaves as a converging lens (because of the postive sign) of focal length 58.82cm .
F = 750 N (Force)
d = 10 m (displacement
)
t = 25 s (time)
L = ? (Mechanical work
) = (Energy)
P = ? (Power)
Solve:
L = F × d = 750 × 10 = 7500 Joules
P = L / t = 7500 / 25 = 300 Watts
