Which is a “big idea” for matter and change?

A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10

Charra [1.4K]

Answer:

$u(t)=1.15 \sin (8.68t)cm$

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

$\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s$

Where $g=980 cm/s^2$

$u(t)=Acos8.68 t+Bsin 8.68t$

u(0)=0

Substitute the value

$A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t$

Substitute u'(0)=10

$8.68B=10$

$B=\frac{10}{8.68}=1.15$

Substitute the values

$u(t)=1.15 \sin (8.68t)cm$

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

6 0

3 years ago

What does addition of two vectors give you?

mash [69]

Answer:

Explanation:

To add or subtract two vectors, add or subtract the corresponding components. Let →u=⟨u1,u2⟩ and →v=⟨v1,v2⟩ be two vectors. The sum of two or more vectors is called the resultant. The resultant of two vectors can be found using either the parallelogram method or the triangle method .

4 0

4 years ago

How would the terminal velocity of a piece of tissue paper compare to the terminal velocity of a rock?

matrenka [14]

Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.

Case 1: Terminal velocity of a piece of tissue paper.

The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.

Downward gravitational force, F = mg

Upward air resistance or friction or drag force will be $f_{1}$

So, paper will attain terminal velocity when mg = $f_{1}$

Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.

Downward force on rock, F = Mg

Drag force = $f_{2}$

Rock will attain terminal velocity when Mg = $f_{2}$

Mg > mg

so, $f_{2}$ > $f_{1}$

And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

5 0

3 years ago

A parallel RLC circuit contains an inductor with a value of 400 microhenries and a capacitor with a value of 0.3 microfarads wha

ipn [44]

Answer:

The resonant frequency of this circuit is 14.5 kHz.

Explanation:

Given that,

Inductance of a parallel LCR circuit, $L=400\ \mu H=400\times 10^{-6}\ H$

Capacitance of parallel LCR circuit, $C=0.3\ \mu F=0.3\times 10^{-6}\ F$

At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

$f=\dfrac{1}{2\pi \sqrt{400\times 10^{-6}\times 0.3\times 10^{-6}} }$

$f=14528.79\ Hz$

or

f = 14.5 kHz

So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.

4 0

3 years ago

As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th

Fofino [41]

Answer:

$5.04\cdot 10^8 A$

Explanation:

The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:

$\phi = 3.950\cdot 10^{-19}J$

So, the energy of the incoming photon hitting on the metal must be at least equal to this value.

The energy of a photon is given by

$E=\frac{hc}{\lambda}$

where

h is the Planck's constant

c is the speed of light

$\lambda$ is the wavelength of the photon

Using $E=\phi$ and solving for $\lambda$ , we find the maximum wavelength of the radiation that will eject electrons from the metal:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.950\cdot 10^{-19} J}=5.04\cdot 10^{-7}m$

And since

1 angstrom = $10^{-15}m$

The wavelength in angstroms is

$\lambda=\frac{5.04\cdot 10^{-7} m}{10^{-15} m/A}=5.04\cdot 10^8 A$

3 0

3 years ago