Solved, can someone check it over!

Calculate the total energy of 2.0 kg object moving horizontally at 10 m/s 50 meters above the surface

mina [271]

We have:

Total Energy: KE + GPE
KE (Kinetic Energy) = $\frac{1}{2} m*v^2$
GPE (Gravitational Potential Energy) = $m*g*h$

Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²

Formula:

Total Energy: KE + GPE
$TE = \frac{1}{2} m*v^2 + m*g*h$

Solving:
$TE = \frac{1}{2} m*v^2 + m*g*h$
$TE = \frac{1}{2} *2.0*10^2 + 2.0*10*50$
$TE = \frac{2.0*100}{2} + 1000$
$TE = \frac{200}{2} + 1000$
$TE = 100 + 1000$
$\boxed{\boxed{TE = 1100\:Joule}}\end{array}}\qquad\quad\checkmark$

5 0

3 years ago

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Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

3 years ago

Which source is considered a primary source of scientific information? an original research article in a scientific journal a re

Nitella [24]

The primary source would be the original article published in a scientific journal. All other choices would be based on information from the original article.

Newspapers would only pick up the information from the journal itself, or from the authors. Books follow after the original article, after it has gained momentum among the research community. The public lecture at a museum would be based on work from the journal article.

6 0

3 years ago

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Rank these electromagnetic waves on the basisof their speed (in vacuum). Rank from fastest to slowest. To rank items as equivale

adelina 88 [10]

Answer: On the basis of speed they are all equivalent.

Yellow light = Fm radio wave = Green light = X-ray = AM radio wave = Infrared wave

Explanation:

Yellow light, Fm radio wave, Green light ,X-ray, AM radio wave and Infrared wave are all electromagnetic waves, and all electromagnetic waves move at the same vacuum speed which is the speed of light and is approximately 3.0x10^8 m/s.

They only differ in wavelength and frequency

c = λf

c (speed of light) = λ (wavelength) x f (frequency)

Therefore; on the basis of speed they are all equivalent.

Yellow light = Fm radio wave = Green light = X-ray = AM radio wave = Infrared wave

6 0

3 years ago

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

Alexus [3.1K]

Answer:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field is given as

E = $\sqrt{2I/ce_{0} }$

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = 1.55 x 10^-8 v/m

6 0

4 years ago

Solved, can someone check it over! ​

Solved, can someone check it over!