Ask question

Ask question

All categories

Juli2301 [7.4K]

3 years ago

15

A clock is designed that uses a mass on the end of a spring as a timing mechanism. If the oscillation time needed is exactly one

second, what is the spring constant required if the mass is 1.3 kg

1 answer:

user100 [1]3 years ago

5 0

Answer:

The value is $k = 51.34 \ N/ m$

Explanation:

From the question we are told that

The mass is $m = 1.3 \ kg$

The needed oscillation time is $T = 1 \ s$

Generally the spring constant is mathematically represented as

$k = \frac{4 \pi^2 * m }{ T^2}$

=> $k = \frac{4* 3.142^2 * 1.3 }{ 1^2}$

=> $k = 51.34 \ N/ m$

You might be interested in

Suppose the gravitational force between two spheres is 30 N. If the magnitude of both masses doubles, and the distance between t

mixer [17]

Answer:

60 N

Explanation:

because when we double the 30N, we will get 60N as a force

8 0

3 years ago

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

7 0

4 years ago

A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

m_a_m_a [10]

Answer:

a The kinetic energy is $KE = 0.0543 J$

b The height of the center of mass above that position is $h = 1.372 \ m$

Explanation:

From the question we are told that

The length of the rod is $L = 1.4m$

The mass of the rod $m = 140 = \frac{140}{1000} = 0.140 \ kg$

The angular speed at the lowest point is $w = 1.09 \ rad/s$

Generally moment of inertia of the rod about an axis that passes through its one end is

$I = \frac{mL^2}{3}$

Substituting values

$I = \frac{(0.140) (1.4)^2}{3}$

$I = 0.0915 \ kg \cdot m^2$

Generally the kinetic energy rod is mathematically represented as

$KE = \frac{1}{2} Iw^2$

$KE = \frac{1}{2} (0.0915) (1.09)^2$

$KE = 0.0543 J$

From the law of conservation of energy

The kinetic energy of the rod during motion = The potential energy of the rod at the highest point

Therefore

$KE = PE = mgh$

$0.0543 = mgh$

$h = \frac{0.0543}{9.8 * 0.140}$

$h = 1.372 \ m$

5 0

3 years ago

Read 2 more answers

Two identical stones are thrown from the top of a tall building. Stone 1 is thrown vertically downward with an initial speed v,

Molodets [167]

If the resistance of the Air is ignored, we can use the theory given by Galileo in which he warned that the thermal velocity of a body in free fall was given by

$v= \frac{1}{2}gt$

Where

g = Gravitational acceleration

t = time

As we can see the speed of objects in free fall is indifferent to the position that is launched (as long as the resistance of the air is ignored) or its mass.

Both bodies will end with the same thermal speed.

5 0

3 years ago

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener

defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s$

(a) the force constant of the spring

$\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m$

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

$A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m$

7 0

3 years ago