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2 years ago

When is the magnitude of the acceleration of a mass on a spring at its maximum value?

Physics

2 answers:

Alex73 [517]2 years ago

6 0

Answer:

A. when the mass has a speed of zero

Andreas93 [3]2 years ago

5 0

Answer:

A. when the mass has a speed of zero

Explanation:

In mass-spring system, the velocity and the acceleration are in anti-phase, which means that when one of the two quantities is maximum, the other one is zero, and vice-versa.

In fact:

- When the displacement of the spring is zero (x=0), the velocity is maximum, due to conservation of energy. In fact, as the displacement is zero, the elastic potential energy of the system (given by $\frac{1}{2}kx^2$ ) is zero, therefore the kinetic energy (given by $\frac{1}{2}mv^2$ ) must be maximum, and so the velocity (v) is also maximum. On the cotnrary, acceleration (a) is directly proportional to the restoring force of the spring, given by

$F=-kx$

so we see that when x=0, then the force is zero: F=0, and so the acceleration is zero as well.

- When the displacement of the spring is maximum, the velocity is zero, due to conservation of energy. In fact, as the displacement is maximum, the elastic potential energy of the system (given by $\frac{1}{2}kx^2$ ) is maximum, therefore the kinetic energy (given by $\frac{1}{2}mv^2$ ) must be zero, and so the velocity (v) is also zero. On the cotnrary, since acceleration (a) is directly proportional to the restoring force of the spring, given by

$F=-kx$

so we see that when x=maximum, then the force is maximum, and so the acceleration is maximum as well.

Based on this, the correct answer is

A. when the mass has a speed of zero

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Which of the following statements is accurate?

Volgvan

Answer: B. If an object's velocity is changing,it's either experiencing acceleration or deceleration.

Acceleration is defined as the rate at which an object changes its velocity. This implies that if an object is changing it's velocity it is experiencing acceleration/ deceleration.

Acceleration is a vector quantity that has both a magnitude and time.

It is represented as

Acceleration= change in velocity/time.

The SI unit for acceleration is m/s^2

8 0

3 years ago

Read 2 more answers

A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain

Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

For maximum speed

$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16$

At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$

The speed of the block is 5.71314 m/s

4 0

2 years ago

Read 2 more answers

3. The smallest bird is the Cuban bee hummingbird, which has a mass of only

Firlakuza [10]

Answer:

2.6 m

Explanation:

The work done by the bird is given by

$W=Fd$

where

F is the force exerted

d is the distance covered

In this problem, we know:

$W=8.8\cdot 10^{-4} J$ is the work

$F=3.4\cdot 10^{-4} N$ is the force

Solving the equation for d, we find the distance covered by the bird:

$d=\frac{W}{F}=\frac{8.8\cdot 10^{-4}}{3.4\cdot 10^{-4}}=2.6 m$

8 0

3 years ago

Inside a television picture tube there is a build-up of electrons (charge of 1. 602 × 10–19 C) with an average spacing of 38. 0

Olin [163]

The magnitude of the electric field inside the picture tube is 998.476 kN/C.

<h3>What is an electric field?</h3>

It is an area in the space where a charged particle experiences a force. It can be calculated by the formula,

$E =\dfrac {kQ}{r^2 }$

Where,

$E$ = electric field produced

$Q$ = charge = $1.602 \times 10^{-19}\rm \ C$

$r$ = distance = $38 \times 10^{-9}\rm \ m$

$k$ = Coulomb's constant = $9 \times 10^9 \rm \ N. m2 / C2$

Put the values in the formula,

$E = \dfrac {9 \times 10^9\times 1.602 \times 10^{-19}}{(38 \times 10^{-9})^2}\\\\E = 998.476 \rm \ kN/C$

Therefore, the magnitude of the electric field inside the picture tube is 998.476 kN/C.

Learn more about the electric field:

brainly.com/question/3783640

8 0

2 years ago

A person with mass mp = 75 kg stands on a spinning platform disk with a radius of R = 1.59 m and mass md = 186 kg. The disk is i

Ad libitum [116K]

Answer:

1. $I_0=424.7208\,kg.m^2$

2. $I_f=256.1808\,kg.m^2$

3. $\omega_f=3.3158\,rad.s^{-1}$

4. $\Delta KE=558.8432\,J$

5. $a_c=5.8271\,m.s^{-2}$

6. $\omega_0=2\,rad.s^{-1}$

Explanation:

Given:

mass of the person, $m_p=75\,kg$

radius of the disk, $r=1.59\,m$

mass of the disk, $m_d=186\,kg$

initial angular speed of the disk, $\omega_0=2\,rad.s^{-1}$

distance of the person from the center of the disk, $d=0.53\,m$

Initial moment of inertia of the system when the man stands at the rim of disk:

Moment of inertia of the disc:

$I_D=\frac{1}{2} .m_d.R^2$

$I_D=\frac{1}{2} \times 186\times 1.59^2$

$I_D=235.1133\,kg.m^2$

Now for the person, we treat the mass to be a point revolving around R:

$I_P=m_p.R^2$

$I_P=75\times 1.59^2$

$I_P=189.6075\,kg.m^2$

∴We have the moment of inertia of the system in this case as:

$I_0=I_D+I_P$

$I_0=235.1133+189.6075$

$I_0=424.7208\,kg.m^2$

Moment of inertia when the person stands at 0.53 m from the center of the disk:

Moment of inertia of the disk will be constant:

$I_D=235.1133\,kg.m^2$

For the person, we treat the mass to be a point revolving around radius 0.53 m:

$I_P=75\times 0.53^2$

$I_P=21.0675\,kg.m^2$

∴We have the moment of inertia of the system

$I_f=I_D+I_P$

$I_f=235.1133+21.0675$

$I_f=256.1808\,kg.m^2$

The final angular velocity of the disk:

Using the conservation of angular momentum:

$I_0.\omega_0=I_f.\omega_f$

$424.7208\times 2=256.1808\times \omega_f$

$\omega_f=3.3158\,rad.s^{-1}$

Change in Kinetic Energy:

∵ $KE=\frac{1}{2} I.\omega^2$

∴ $\Delta KE= \frac{1}{2} (I_f.\omega_f^2-I_0.\omega_0^2)$

$\Delta KE=\frac{1}{2} (256.1808\times 3.3158^2-424.7208\times 2^2)$

$\Delta KE=558.8432\,J$

Centripetal acceleration of the person when she is at R/3:

Centripetal acceleration is given as:

$a_c=r'.\omega^2$

we have ω=3.3158 radian per second at R=0.53 m

$a_c=\frac{R}{3} .\omega^2$

$a_c=0.53\times 3.3158^2$

$a_c=5.8271\,m.s^{-2}$

If the person now walks back to the rim of the disk:

Then by the law of conservation of angular momentum the initial angular speed of $\omega_0=2\,rad.s^{-1}$ will be restored.

8 0

3 years ago