A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its
2 answers:
Answer:
The speed is maximum and the acceleration is zero
Explanation:
- The speed of the mass in simple harmonic motion can be found by using the law of conservation of energy. In fact, the total mechanical energy of the mass-spring system is sum of kinetic energy and elastic potential energy:
where
m is the mass
v is the speed
k is the spring constant
x is the displacement
As we can see, when the displacement is zero (x=0), the term representing the kinetic energy is maximum, so v (the speed) is also maximum.
- The acceleration of the mass in simple harmonic motion is proportional to the restoring force acting on the mass, which is given by Hook's law
where
k is the spring constant
x is the displacement
When x = 0, F = 0, so the net force acting on the mass is zero. Therefore, this also means that the acceleration of the mass is also zero: a = 0.
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Answer
given,
current (I) = 16 mA
circumference of the circular loop (S)= 1.90 m
Magnetic field (B)= 0.790 T
S = 2 π r
1.9 = 2 π r
r = 0.3024 m
a) magnetic moment of loop
M= I A
M=
M=
M=
b) torque exerted in the loop
The answers to your questions are as written below:
- The objects that represents a negatively charged particle is : Image B
- The object that represents a positively charged molecule is : Image A
- The object that represents an uncharged molecule is : Image C
- The object the will not move when in an electric fied is : Image C
A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.
A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.
Hence we can concude that the answers to your questions are as listed above.
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attached below is the missing image
Answer:
Explanation:
It is given that,
The number of lines per unit length, N = 900 slits per cm
Distance between the formed pattern and the grating, l = 2.3 m
n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm,
Let d is the slit width of the grating,
For the first wavelength, the position of maxima is given by :
For the other wavelength, the position of maxima is given by :
So,
or
So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.