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3 years ago

Which equation correctly represents the gravitational potential energy of a

Physics

2 answers:

Lerok [7]3 years ago

7 0

Answer:

GPE = ME − 1/2 mv^2

Explanation:

Just did it

arsen [322]3 years ago

5 0

Answer:

$PE=ME-\dfrac{1}{2}mv^2$

Explanation:

The mechanical energy of an object is equal to the sum of kinetic energy and the potential energy. Let ME is mechanical energy, KE is kinetic energy then its potential energy PE is given by :

$ME=PE+KE$

Since, KE = (1/2) mv^2

$PE=ME-KE\\\\PE=ME-\dfrac{1}{2}mv^2$

So, $PE=ME-\dfrac{1}{2}mv^2$ correctly represents the gravitational potential energy of a .system

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Suppose that you measure the light of a star between 640 and 680 nm and you get a very strong peak around 650 nm. what can you s

Llana [10]

Answer: The star is moving away from Earth

Explanation:

Let's begin by explaining what the Doppler shift is:

This is related to the Doppler effect and refers to the change in a wave perceived frequency (or wavelength=color) when the emitter of the waves, and the observer move relative to each other.

From there, it is deduced that the farther the object is, the more redshifted it is in its spectrum. For example, as a star moves away from the Earth, its espectrum turns towards the <u>red</u> and as the star moves toward the Earth, its espectrum turns towards the <u>blue</u>.

In this sense, since the peak in wavelength is around 650 nm, which corresponds to red, it can be said the star is moving away from Earth.

3 0

4 years ago

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to p

katrin [286]

Answer:

Required charge $q=2.6\times 10^{9}C$ .

$n=1.622\times 10^{10}\ electrons$

Explanation:

Given:

Diameter of the isolated plastic sphere = 25.0 cm

Magnitude of the Electric field = 1500 N/C

now

Electric field (E) is given as:

$E =\frac{kq}{r^2}$

where,

k = coulomb's constant = 9 × 10⁹ N

q = required charge

r = distance of the point from the charge where electric field is being measured

The value of r at the just outside of the sphere = $\frac{25.0}{2}=12.5cm=0.125m$

thus, according to the given data

$1500N/C=\frac{9\times 10^{9}N\times q}{(0.125m)^2}$

$q=\frac{0.125^2\times 1500}{9\times 10^{9}}$

Required charge $q=2.6\times 10^{9}C$ .

Now,

the number of electrons (n) required will be

$n=\frac{required\ charge}{charge\ of\ electron}$

$n=\frac{2.6\times 10^{-9}}{1.602\times 10^{-19}}$

$n=1.622\times 10^{10}\ electrons$

6 0

4 years ago

A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic fr

GrogVix [38]

Answer:

The speed of the block when is fallen 30cm

v=0.726 $\frac{m}{s}$

Explanation:

∑F= (m2)g - ƒ - (m1)g*sin(θ) = (m1)a

g = 9.81 m/s²

ƒ = μN = μ(m1)g

$(m2)*g - u*(m1)*g - (m1)*g*sin(\alpha ) = (m1)*a$

(0.200)(9.81) - (0.1)(0.290)(9.81) - (0.290)(9.81)sin(30°) = (0.290)a

$(0.200kg)(9.81\frac{m}{s^{2}}) - (0.1)(0.290kg)(9.81\frac{m}{s^{2}}) - (0.290kg)(9.81\frac{m}{s^{2}})sin(30°) = (0.290kg)*a$

$0.511101=0.29*a\\a=0.879\frac{m}{s^{2} }$

$v_{f}^{2}=v_{o}^{2}+2*a(x_{f}^{2}-v_{o}^{2})\\v_{o}=0\\v_{o}=0\\v_{f}^{2}=2*a(x_{f})\\v_{f}=\sqrt{2*a(x_{f})}\\v_{f}=\sqrt{2*0.879\frac{m}{s^{2}}*0.30m} \\v_{f}=0.726 \frac{m}{s}$