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Nookie1986 [14]
3 years ago
6

A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?

Physics
2 answers:
rjkz [21]3 years ago
8 0

Answer:

Maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2              

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec

Maximum acceleration is given by a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2

So maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2

Mnenie [13.5K]3 years ago
5 0

Answer:

1.11 m/s²

Explanation:

Amplitude, A = 0.43 m

Time period, T = 3.9 second

Let a be the maximum acceleration and ω be the angular frequency.

ω = 2π/T

ω = ( 2 x 31.4) / 3.9

ω = 1.61 rad/s

Maximum acceleration

a = ω²A

a = 1.61 x 1.61 x 0.43

a = 1.11 m/s²

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If the force is 4 newtons between two charged spheres separated by 3 centimeters, calculate the force between the same spheres s
il63 [147K]

Answer:

1 Newton

Explanation:

F=9*10^9*q0q1/r^2]]

F=9*10^9*(q0q1)/ r^2

r=3cm

F=4N

F=9*10^9*(q0q1)/3^2

4=9*10^9*(q0q1)/9

4=10^9 q0q1

q0q1=4/10^9

q0q1=4*10^-9

To calculate the force between the forces at a distance of 6 cm

F=9*10^9*(q0q1)/ r^2

=9*10^9*(4*10^-9)/6^2

=9*10^9*(4*10^-9)/36

=10^9*4*10^-9/4

=10^9*10^-9

=1 Newton

7 0
2 years ago
A boy throws a ball with an initial velocity of 19.6 m/s. What maximum height does the ball reach?
Wewaii [24]
<h2>Hello!</h2>

The answer is: 19.59 m

<h2>Why?</h2>

Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.

When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:

We will use the following formula:

Vf^2=Vi^2+2*g*s

Where:

Vf= Final velocity = 0

Vi= Initial velocity = \frac{19.6m}{s}

g = Gravity Acceleration = \frac{9.81m}{s^{2} }

s = Traveled distance

0=19.6^2+2*-9.81*s\\s=\frac{19.6^2}{2*9.81}=\frac{384.16}{19.62}=19.59m

Have a nice day!

8 0
3 years ago
A parallel beam of light of wavelength 4.5 x 10^-7 m is incident on a pair of slits that are 5.0 x 10^-4 m apart. The interferen
RSB [31]

Answer:

1.8x10⁻³m

Explanation:

From the question above, the following information was used to solve the problem.

wavelength λ = 4.5x10⁻⁷m

Length L = 2.0 meters

distance d = 5 x 10₋⁴m

ΔY = λL/d

= 4.5x10⁻⁷m (2) / 5 x 10₋⁴m

= 0.00000045 / 0.0005

= 0.0000009/0.0005

= 0.0018

= 1.8x10⁻³m

from the solution above The separation between two adjacent bright fringes is most nearly 1.8x10⁻³m

thank you!

3 0
3 years ago
You sit at the middle of a large turntable at an amusement park as it is set spinning on nearly frictionless bearings, and then
jolli1 [7]

Answer:

Decrease

Explanation:

If you crawl to the rim the rotational speed will decrease. The law of conservation of angular momentum supports this answer. And it states that :

"When the net external torque acting on a system about a given axis is. zero , the total angular momentum of the system about that axis remains constant."

3 0
3 years ago
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A
cluponka [151]

Answer:

D) -Q

Explanation:

The charge inserted will induce -Q charge on the inner surface and + Q on the outer surface of the shell . This charge is called bound charge because it remained attached with opposite charge inserted inside.

5 0
3 years ago
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