All categories

3 years ago

How can the time period of a pendulum be decreased or increased

Physics

1 answer:

erastovalidia [21]3 years ago

5 0

Explanation:

The period of a simple pendulum is:

T = 2π √(L/g)

where L is the length of the pendulum and g is the acceleration due to gravity.

The period of a pendulum can be decreased by decreasing the length of the pendulum or by moving the pendulum to a place with a higher gravity.

The period of a pendulum can be increased by increasing the length of the pendulum or by moving the pendulum to a place with a lower gravity.

You might be interested in

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

3 0

2 years ago

What is the difference between rutherford's model of the atom and bohr's model of the atom

Amanda [17]

Answer:

Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits. Bohr built upon Rutherford's model of the atom. ... So it was not possible for electrons to occupy just any energy level.

Explanation:

7 0

2 years ago

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

2 years ago

Ground reaction force acting on carter

mezya [45]

We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.

I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.

8 0

2 years ago

A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density

Amiraneli [1.4K]

Answer:

a) K = 2/3 π G m ρ R₁³ / R₂ , b) U = - G m M / r

Explanation:

The law of universal gravitation is

F = G m M / r²

Part A

Let's use Newton's second law

F = m a

The acceleration is centripetal

a = v² / R₂

G m M / R₂² = m v² / R₂

v² = G M / R₂

They give us the density of the planet

ρ = M / V

V = 4/3 π R₁³

M = ρ V

M = ρ 4/3 π R₁³

v² = 4/3 π G ρ R₁³ / R₂

K = ½ m v²

K = ½ m (4/3 π G ρ R₁³ / R₂)

K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

F = - dU / dr

∫ dU = - ∫ F. dr

The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1

U- U₀ = G m M ∫ dr / r²

U - U₀ = G m M (- r⁻¹)

We evaluate for

U - U₀ = -G m M (1 / $r_{f}$ - 1 / $r_{i}$ )

They indicate that for ri = ∞ U₀ = 0

U = - G m M / r

6 0

2 years ago