Ask question

Ask question

All categories

3 years ago

14

A 45.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping

force Fx= −bvx acts on the egg, and the amplitude of the motion decreases to 0.100 m in a time of 5.00 s. Calculate the magnitude of the damping constant b. Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures.

1 answer:

leva [86]3 years ago

6 0

Answer:

0.02896 kg/s

Explanation:

$A_1$ = Initial displacement = 0.5 m

$A21$ = Final displacement = 0.1 m

t = Time taken = 0.5 s

m = Mass of object = 45 g

Displacement is given by

$x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)$

At maximum displacement

$cos(\omega t+\phi)=1$

$\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s$

The magnitude of the damping coefficient is 0.02896 kg/s

You might be interested in

What is the charge of an atom that has lost one electron?

Luden [163]

Answer:

positive

An atom has a even amount of protons, electrons, and a little less neutrons. If the atom lost one electron, then it now has a positive charge.

Hope it helps!

5 0

3 years ago

In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, fla

lara31 [8.8K]

Answer:

Explanation:

Given

Speed of ball $u=3.2\ m/s$

Plane is inclined at an angle $20^{\circ}$

To win the Game we need to hit the target at $x=2.4\ m$ away

Launch angle of ball $\theta$

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is $g\sin 20$

Range of Projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

for $R=2.4\ m$

$2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}$

$\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}$

$\sin 2\theta =0.7855$

$2\theta =51.77$

$\theta =25.88^{\circ}$

so ball must be launched at an angle of $25.88^{\circ}$

4 0

2 years ago

Electrical current is defined as

Dmitry_Shevchenko [17]

Explanation:

The flow of electrical charge per unit time

8 0

2 years ago

Which list the terrestrial planets in order from smallest to largest

andrew-mc [135]

Answer:

The terrestrial planets in order from smallest to largest are :

Pluto

Mercury

Mars

Venus

Earth

Neptune

Uranus
Saturn
Jupiter

Explanation:

6 0

2 years ago

Read 2 more answers

URGENT. Physics quiz on force, distance, etc. will reward brainliest.

goblinko [34]

13a) 9 J

The work done is equal to the area under the curve between x=0 cm and x=30 cm. However, first we should find the magnitude of the force for x=30 cm. If we notice that the force is proportional to the stretching x, we can set the following proportion to find the value of F for x=30 cm:

$10 N : 5 cm = x : 30 cm$

$x=\frac{30 cm \cdot 10 N}{5 cm}=60 N$

And so, the work done is

$W=Area=\frac{1}{2}(base)(height)=\frac{1}{2}(0.30 m)(60 N)=9 J$

13b) 24.5 m/s

The kinetic energy gained by the arrow is equal to the work done in stretching the bow:

$K=W=9 J$

Given the formula for the kinetic energy:

$K=\frac{1}{2}mv^2$

we can find the speed v of the arrow:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\cdot 9J}{0.030 kg}}=24.5 m/s$

13c) 30.6 m

If shot vertically upward, at the point of maximum height all the initial kinetic energy of the arrow is converted into gravitational potential energy:

$\frac{1}{2}mv^2 = mgh$

Re-arranging the formula and using the initial speed of the arrow, we can find its maximum height h:

$h=\frac{v^2}{2g}=\frac{(24.5 m/s)^2}{2(9.81 m/s^2)}=30.6 m$

14) 20 m/s

We can solve the problem by using the work-energy theorem. In fact, the work done by the frictional force of the brake is equal to the change in kinetic energy of the car:

$W=\Delta K=K_f -K_i$

$Fd=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$F=-2500 N$ is the force applied by the brakes (with a negative sign, since it is opposite to the displacement of the car)

$d=100 m$ is the displacement of the car

$m=1000 kg$ is the car's mass

$v$ is the final speed of the car

$u=30 m/s$ is the initial speed of the car

By re-arranging the equation, we can find v:

$v=\sqrt{\frac{2(Fd+\frac{1}{2}mu^2)}{m}}=20 m/s$

15) 5.0 m/s

We can solve the problem by using the law of conservation of energy:

$U_i + K_i = U_f + K_f\\mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where

m is the mass of the pendulum

$h_i=1.2 m$ is the initial height of the pendulum

$u=3 m/s$ is the initial speed of the pendulum

$h_f=0.4 m$ is the final height of the pendulum

$v$ is the final speed of the pendulum

Re-arranging the equation, we can find v:

$v=\sqrt{2gh_i + u^2 - 2gh_f}=5.0 m/s$

16) Point B (at the top of the loop)

Gravitational potential energy is defined as:

$U=mgh$

where m is the mass, g is the gravitational acceleration and h is the height above the ground. Therefore, we see that the potential energy is proportional to h: the higher the ball above the ground, the greater its potential energy. In this example, the point of maximum height is point B, therefore it is the point where the ball has the largest potential energy.

17) Law of conservation of energy: the total mechanical energy of an isolated object is conserved (if no frictional force act on it)

Example: A stone left falling from rest from a cliff. Let's call h the height of the cliff, m the mass of the stone. The mechanical energy of the stone is constant, and it is sum of the potential energy and kinetic energy:

$E=U+K$

At the top of the cliff, the kinetic energy is zero (the stone is at rest), so all its energy is potential energy:

$E_i = U_i = mgh$

When the stone falls, its energy is converted into kinetic energy. Just before hitting the ground, the height has become zero, h=0, so the potential energy is zero and all the mechanical energy is now kinetic energy:

$E_f=K_f=\frac{1}{2}mv^2$

since the mechanical energy must be conserved, we can write

$E_i=E_f\\mgh = \frac{1}{2}mv^2\\2gh=v^2$

6 0

3 years ago