Ask question

Ask question

All categories

3 years ago

14

When a mass of 29 g is attached to a certain spring, it makes 20 complete vibrations in 3.1 s. What is the spring constant of th

e spring? Answer in units of N/m.

1 answer:

never [62]3 years ago

4 0

Answer:

The spring constant of the spring is 47.62 N/m

Explanation:

Given that,

Mass that is attached with the spring, m = 29 g = 0.029 kg

The spring makes 20 complete vibrations in 3.1 s. We need to find the spring constant of the spring. We know that the number of oscillations per unit time is called frequency of an object. So,

$f=\dfrac{20}{3.1}$

f = 6.45 Hz

The frequency of oscillator is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

k is the spring constant

$k=4\pi^2f^2m$

$k=4\pi^2\times (6.45)^2\times 0.029$

k = 47.62 N/m

So, the spring constant of the spring is 47.62 N/m. Hence, this is the required solution.

You might be interested in

3 PHYSICAL SCIENCE: In order to figure out what is in a sealed box, you perform some tests. The box seems heavy for its size and

rjkz [21]

Some sort of magnetic metal

Metals are heavier per cubic unit than other materials such as air or water, and also are much more magnetic than other materials

7 0

3 years ago

a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

aleksandrvk [35]

Hope this helps!

-Lilly

3 0

3 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$

$F_{net} = 300 - 150$

$F_[net} = 150 N$

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

Tabitha is getting ready for a family trip by getting her clothes and toys together. She suddenly gets concerned and tells her m

Kisachek [45]

The answer is Hypothesis because she can predict that she needs a bigger bag

7 0

3 years ago

Read 2 more answers