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dezoksy [38]
2 years ago
14

3. A ball of mass 2.0 kg is attached to a vertical spring. The spring constant is 196 N/m. What is the period if it were to be i

n SHM?
-0.063
-0.63 5
-6.3 s
-63 s
PLEASEHELPME
Physics
1 answer:
Reil [10]2 years ago
4 0

The period of the ball in simple harmonic motion is 0.634 s.

<h3>What is period?</h3>

Period can be defined as the time ( in seconds) taken for a body to complete one oscillation.

To calculate the period of the ball in SHM, we use the formula below.

Formula:

  • T = 2π√(m/k)............... Equation 1

Where:

  • T = Period of the ball
  • m = mass of the ball
  • k = spring constant
  • π  = pie

From the question,

Given:

  • m = 2 kg
  • k = 196 N/m
  • π = 3.14

Substitute these values into equation 1

  • T = 2(3.14)√(2/196)
  • T = 6.28√(0.010204)
  • T = 0.634 s

Hence, The period of the ball in simple harmonic motion is 0.634 s.  

Learn more about period here: brainly.com/question/21924087

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The primary purpose of a switch in a circuit is to
Vanyuwa [196]

Answer:

See below

Explanation:

A switch regulates the transparency and closeness of an electrical circuit. This enables the current flow of the circuit to be regulated (not even possible to reach it and manually sever the wires). Switches are important components of any circuit that involve an input system or control

Hope this helps.

4 0
3 years ago
the illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the saurce?
Lelechka [254]
L = illuminance
A = surface
i = intensity

L = i / A ==: i = L * A

i = 6 lux * 4 m^2 = 24 lumen
8 0
3 years ago
A 3kg book falls from a 2m tall bookshelf what is the speed <br><br> PICTURE included
mafiozo [28]

Answer:

39.2m/s

Explanation:

The potential energy the book has right before it falls is equal to the kinetic energy in falling.

PE = KE

mgh = (1/2)mv

2gh=v

v=(2)(9.81)(2)

v=39.24m/s

8 0
3 years ago
Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object
ANEK [815]

Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem.  Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them.  So let's call them F1 and F2, with F1 arger than F2.  Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.  

Can you solve this system of equations seeing them like this, or do you need more help?

6 0
2 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
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