A 12 N net force is applied to an object as it moves a distance of 3.0 m: Use the

An object with a mass M = 250 g is on a plane inclined at 30º above the horizontal and is attached by a string to a mass m = 150

malfutka [58]

Answer:

0.495 m/s

Explanation:

$T$ = tension force in the string connecting the two objects

$M$ = Mass of the object on inclined plane = 250 g = 0.250 kg

$m$ = Mass of the hanging object = 150 g = 0.150 kg

$a$ = acceleration of each object

From the force diagram, force equation for the motion of the object on the inclined plane is given as

$T - Mg Sin30 = Ma\\T = Mg Sin30 + Ma$

From the force diagram, force equation for the motion of the hanging object on the inclined plane is given as

$mg - T = ma\\T = mg - ma$

Using the above two equations

$Mg Sin30 + Ma = mg - ma$

$(0.250)(9.8) Sin30 + (0.250) a = (0.150) (9.8) - (0.150)a$

$a = 0.6125 ms^{-2}$

$h$ = height dropped by the hanging object = 10 cm = 0.10 m

$v$ = Speed gained by the object

Speed gained by the object can be given as

$v = sqrt(2ah)\\v = sqrt(2(0.6125)(0.20))\\v = 0.495 ms^{-1}$

5 0

3 years ago

How do you know the speed of an electromagnetic wave in vacuum?

larisa [96]

Answer:

when huburxvkruxv hfgy

8 0

3 years ago

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Chapter 38, Problem 001

Norma-Jean [14]

Answer:

a) $\lambda=2.95x10^{-6}m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=\frac{c}{\lambda}$

If we replace the last equation into the E formula we got:

$E=h\frac{c}{\lambda}$

And if we solve for $\lambda$ we got:

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

Using the value of the constant $h=4.136x10^{-15} eVs$ we have this:

$\lambda=\frac{4.136x10^{15}eVs (3x10^8 \frac{m}{s})}{0.42eV}=2.95x10^{-6}m$

$\lambda=2.95x10^{-6}m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^{-6}m$

4 0

3 years ago

What is the magnification of a virtual image if the image is 60.0 cm from a

Hatshy [7]

Answer:

4cm

Explanation:

Magnification of the virtual image

= image distance / object distance

Given

Image distance = 60.0cm

Object distance = 15.0cm

Therefore,

Magnification = 60.0/15.0

= 4 cm

6 0

3 years ago

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What is the longest wavelength that can be observed in the third order for a transmission grating having 5200 slits/cm?

Anarel [89]

Answer:

641 nm.

Explanation:

Given that,

A transmission grating has 5200 slits/cm.

We need to find the longest wavelength that can be observed in the third order. Using grating equation as follows :

$d\sin\theta=m\lambda$ ...(1)

d is slit spacing

No fo slit per unit length :

$N={5200}\ slit/cm\\\\=520000\ slits/m$

We know that, N = 1/d

For longest wavelength, θ = 90°

From equation (1)

$\dfrac{\sin\theta}{m\lambda}=\dfrac{1}{d}\\\\520000=\dfrac{\sin(90)}{3\lambda}\\\\\lambda=\dfrac{1}{520000\times 3}\\\\=6.41\times 10^{-7}\ m\\\\=641\ nm$

Hence, the longest wavelength in third order for a transmission grating is 641 nm.

3 0

3 years ago