Ask question

Ask question

All categories

2 years ago

8

A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top

of the first at the moment the spring is extended to the end (x = A). Which of the following is true?
A. The maximum velocity does not change.
B. The amplitude of the oscillation A decreases.
C. The frequency of oscillations f does not change.
D. The frequency of oscillations f increases.
E. The period of oscillation increases.

1 answer:

elixir [45]2 years ago

8 0

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

You might be interested in

Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water

tangare [24]

Answer:

$T_b=107.3784\ ^{\circ}C$

Explanation:

Given:

thickness of the base of the kettle, $dx=0.52\ cm=5.2\times 10^{-3}\ m$

radius of the base of the kettle, $r=0.12\ m$

temperature of the top surface of the kettle base, $T_t=100^{\circ}C$

rate of heat transfer through the kettle to boil water, $\dot Q=0.409\ kg.min^{-1}$

We have the latent heat vaporization of water, $L=2260\times 10^3\ J.kg^{-1}$

and thermal conductivity of aluminium, $k=240\ W.m^{-1}.K^{-1}$

<u>So, the heat rate:</u>

$\dot Q=\frac{0.409\times 2260000}{60}$

$\dot Q=15405.67\ W$

<u>From the Fourier's law of conduction we have:</u>

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}$

where:

$A=$ area of the surface through which conduction occurs

$T_b=$ temperature of the bottom surface

$15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}$

$T_b=107.3784\ ^{\circ}C$ is the temperature of the bottom of the base surface of the kettle.

6 0

3 years ago

Please help me!!!!<br> Describe the roles these organisms play in their ecosystem.

creativ13 [48]

Answer:

Explanation:

Deer = primary consumer

Green plants = producer

Fungi = decomposers

3 0

3 years ago

An apple falls from a tree and hits the ground 9.98 m below. with what speed will it hit the ground

cupoosta [38]

The answer to this question is <span>13,537</span>

8 0

3 years ago

Mirage is due to...

Irina18 [472]

<span>Mirage is due to (A). unequal heating of different parts of the atmosphere,
because that produces unequal density of different parts of the atmosphere,
which bends the light.
</span>

3 0

3 years ago

base your answer to this question on the information below and on your knowledge of physics. A toy launcher that is used to laun

Nadya [2.5K]

Answer: v = 2.24 m/s

Explanation: The <u>Law</u> <u>of</u> <u>Conservation</u> <u>of</u> <u>Energy</u> states that total energy is constant in any process and, it cannot be created nor destroyed, only transformed.

So, in the toy launcher, the energy of the compressed spring, called <u>Elastic</u> <u>Potential</u> <u>Energy (PE)</u>, transforms into the movement of the plastic sphere, called <u>Kinetic</u> <u>Energy (KE)</u>. Since total energy must be constant:

$KE_{i}+PE_{i}=KE_{f}+PE_{f}$

where the terms with subscript i are related to the initial of the process and the terms with subscript f relates to the final process.

The equation is calculated as:

$\frac{1}{2}kx^{2}+0=0+\frac{1}{2}mv^{2}$

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

$\frac{1}{2}50(0.1)^{2}=\frac{1}{2}(0.1)v^{2}$

$v^{2}=\frac{50(0.1)^{2}}{0.1}$

$v=\sqrt{50(0.1)}$

$v=\sqrt{5}$

v = 2.24

The maximum speed the plastic sphere will be launched is 2.24 m/s.

3 0

3 years ago