Ask question

Ask question

All categories

3 years ago

6

A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was

released from rest at
t = 0
from the position
x = 0.0480 m.
Determine the location of the mass at
t = 5.85 s?

1 answer:

Rudiy273 years ago

7 0

Answer:

$X=0.0389m$

Explanation:

From the question we are told that:

Period of spring $T_s=2.25s$

Initial Position of Mass $x=0.0480m$

Final Mass period $T_f=5.85s$

Generally the equation for the Mass location is mathematically given by

$X=xcos*\frac{2\pi T_s}{T_f}$

$X=0.048*cos*\frac{2\pi 5.85}{2.25}$

$X=0.0389m$

You might be interested in

A 3.20 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C. a. What is the solubility (in g sal

muminat

Answer:

The solubility of the salt is 35.16 (g/100 g of water).

It would take 71.09 grams of water to dissolve 25 grams of salt.

The percentage of salt that dissolves is 52.7 %

Explanation:

<h3>a.</h3>

We know that 3.20 grams of salt in 9.10 grams of water gives us a saturated solution at 25°C. To find how many grams of salt will gives us a saturated solution in 100 grams of water at the same temperature, we can use the rule of three.

$\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{100 \ g \ water}$

Working it a little this gives us :

$x = 100 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water}$

$x = 35.16 \ g \ salt$

So, the solubility of the salt is 35.16 (g/100 g of water).

<h3>b.</h3>

Using the rule of three, we got:

$\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{25 \ g \ salt}{x \ g \ water}$

Working it a little this gives us :

$x = \frac{25 \ g \ salt}{ \frac{3.20 \g \ salt}{9.10 \ g \ water}}$

$x = 71.09 g \ water$

So, it would take 71.09 grams of water to dissolve 25 grams of salt.

<h3>C.</h3>

Using the rule of three, we got that for 15.0 grams of water the salt dissolved will be:

$\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{15.0 \ g \ water}$

Working it a little this gives us :

$x = 15.0 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water}$

$x = 5.27\ g \ salt$

This is the salt dissolved

The percentage of salt dissolved is:

$percentage \ salt \ dissolved = 100 \% * \frac{g \ salt \ dissolved}{g \ salt}$

$percentage \ salt \ dissolved = 100 \% * \frac{ 5.27\ g \ salt }{ 10.0 \ g \ salt}$

$percentage \ salt \ dissolved = 52.7 \%$

3 0

4 years ago

Which scientist first proposed physical laws to mathematically describe the effect of forces on the motions of bodies?

Whitepunk [10]

Answer:

Sir Issac Newton

Explanation:

These are the effects of gravity.

4 0

3 years ago

What is the frequency of blue light that has a wavelength of 473nm?

ankoles [38]

Answer:

6.34 x 19^14 Hz

Explanation:

Wavelength, λ = 473 nm = 473 x 10^-9 m

the relation between the frequency and the wavelength is given by

v = f x λ

where, v is the velocity of light in vacuum which is equal to 3 x 10^8 m/s.

So, f = v / λ

f = ( 3 x 10^8) / ( 473 x 10^-9)

f = 6.34 x 19^14 Hz

6 0

4 years ago

1pt The process by which rock minerals are changed by natural processes into new substances is known as:

vagabundo [1.1K]

The answer is A because you have to have erosion

8 0

3 years ago

Light with a wavelength range of 141–295 nm shines on a silicon surface in a photoelectric effect apparatus, and a reversing pot

Ksju [112]

Answer:

a) the longest wavelength of the light that will eject electrons from the silicon surface is 258.7891 nm

b) maximum kinetic energy will electrons reach the anode is 0.5098 eV

Explanation:

Given:

Wavelength range = 141-295 nm

Potential of 3.5 V

For the silicon, the work function is Φ = 4.8 eV = 7.68x10⁻¹⁹J

Questions:

a) What is the longest wavelength of the light that will eject electrons from the silicon surface, λ = ?

b) With what maximum kinetic energy will electrons reach the anode,

a) The longest wavelength that will eject electrons:

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

Here

h = Planck's constant = 6.625x10⁻³⁴J s

c = speed of light = 3x10⁸m/s

Substituting values:

$\lambda =\frac{6.625x10^{-34}*3x10^{8} }{7.68x10^{-19} } =2.588x10^{-7} m=258.7891nm$

b) The maximum kinetic energy (one electron):

$K=\frac{hc}{\lambda } -\phi =\frac{6.625x10^{-34}*3x10^{8} }{141x10^{-9} } -7.68x10^{-19} =6.416x10^{-19} J=4.0098eV$

Now, you need to calculate the potential difference:

$K'=eV$

Here

e = charge of electron = 1.6x10⁻¹⁹C

Substituting:

$K'=1.6x10^{-19} *3.5=5.6x10^{-19} J=3.5eV$

Now, the maximum kinetic energy of the electrons:

Kmax = 4.0098 - 3.5 = 0.5098 eV

6 0

4 years ago