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strojnjashka [21]
3 years ago
6

Please help on this one?

Physics
1 answer:
olganol [36]3 years ago
6 0

the object distance of both lenses are positive.

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It pushes the currents to opposite sides
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What is the difference between direct current and alternating current?
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In a direct current, the electric charge, or current, only flows in one direction. In an alternating current, the electric charge changes periodically.  
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A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m
belka [17]

Answer:

Energy stored in the capacitor is U=2.7\times 10^{-5}\ J        

Explanation:

It is given that,

Charge, q=1.5\ \mu C=1.5\times 10^{-6}\ C

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

U=\dfrac{1}{2}q\times V

U=\dfrac{1}{2}\times 1.5\times 10^{-6}\times 36  

U = 0.000027 J

U=2.7\times 10^{-5}\ J

So, the potential energy is stored in the capacitor is U=2.7\times 10^{-5}\ J. Hence, this is the required solution.

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Maxwell shoots a rubber band at his friend Jimmy. Which type of energy is converted into kinetic energy?
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A stretched rubber band is storing <em>elastic potential energy. (A)</em>

3 0
3 years ago
A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

The tension in the wire is 67.9064 N

7 0
2 years ago
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