Answer:
Explanation:
mass m = 3 kg
spring constant be k
k x .8 = 40 N
k = 40 / .8 = 50 N /m
angular frequency ω = √ ( k / m )
= √ ( 50 / 3 )
= 4.08 rad /s
Let amplitude of oscillation be A .
1/2 k A² = 1/2 m v²
50 A² = 3 x 1²
A = .245 m = 24.5 cm
For displacement , the equation of SHM is
x = A sinωt
= 24.5 sin4.08 t
x = 24.5 sin4.08 t
Here, angle 4.08 t is in radians .
Answer: -33.3 * 10^9 C/m^2( nC/m^2)
Explanation: In order to solve this problem we have to use the gaussian law, the we have:
Eoutside =0 so teh Q inside==
the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.
Then we simplify and get
σ= -4.6/(2*π*b)= -33.3 nC/m^2
Explanation:
To find the answer use the equation speed of light=wavelength multiplied by frequency (c=lambda*f) by substituting the value for the frequency the the speed of light
The answer is
Pitch of the buzzer increased (higher tone) as it moves towards the observer
