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2 years ago

Increasing which two things would increase gravitational potential energy?

Physics

1 answer:

sukhopar [10]2 years ago

7 0

Answer:

Mass and height

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about $9.8 m/s2.$

Which is represented as;

$PE_g=mgh$

$PE_g$ stands for gravitational potantial energy,

m stands for mass of object,

g is the gravitational constant and

h is the height.

Here we see that mass of object and height is directly proportional to the gravitational potential energy.

That means increasing in mass and height will result in increasing gravitational potential energy.

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A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull

sdas [7]

In spring mass system we know that angular frequency is given as

$\omega = 2\pi f$

f = 8.38 Hz

$\omega = 2\pi(8.38)$

$\omega = 52.65 rad/s$

now we know that speed of SHM at its extreme position is given by

$v = A\omega$

here we know that

A = 17.5 cm

$v = 0.175 (52.65)$

$v = 9.21 m/s$

so maximum speed is 9.21 m/s

7 0

3 years ago

Suppose a person pushes thumbtack that is 1/5 centimeter long into a bulletin board, and the force (in dynes) exerted when the d

mario62 [17]

Answer:

W = 290.7 dynes*cm

Explanation:

d = 1/5 cm = 0.2 cm

The force is in function of the depth x:

F(x) = 1000 * (1 + 2*x)^2

We can expand that as:

F(x) = 1000 * (1 + 4*x + 4x^2)

F(x) = 1000 + 4000*x + 4000*x^2

Work is defined as

W = F * d

Since we have non constant force we integrate

$W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx$

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2

W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3

W = 200 + 80 + 10.7 = 290.7 dynes*cm

3 0

3 years ago

Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a wa

creativ13 [48]

Answer:

in oil film λ = 303.57 10⁻⁹ m

in the water film λ = 319.55 10⁻⁹ m

Explanation:

When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,

when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship

v = λ f

in the void we have

c = λ₀ f

we divide the two expression

c / v = λ₀ / λ

the refractive index is

n = c / v

n = λ₀ /λ

λ = λ₀ / n

let's calculate

in oil film

λ = 425 10⁻⁹ / 1.40

λ = 303.57 10⁻⁹ m

in the water film

λ = 425 10⁻⁹ / 1.33

λ = 319.55 10⁻⁹

those wavelengths are in the ultraviolet

3 0

2 years ago

A 25.0 kg pickle is accelerated from rest through a distance of 6.0m in 4.0s across a level floor . If the friction force betwee

SIZIF [17.4K]

Add the KE increase and the work done against friction.

The final velocity is twice the average, or 3.0 m/s
The final KE is (1/2)*25*3^2 = 112.5 J

The friction work done is 6*3.8 = 22.8 J
hope this is correct

8 0

3 years ago

If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 20° with a set of planes in a crystal causing first order constru

Flura [38]

Answer:

43.16°

Explanation:

λ = Wavelength = 1.4×10⁻¹⁰ m

θ₁ = 20°

n can be any integer

d = distance between the two slits

Since for the first bright fringe, n₁ = 1

n₂ = 2 for second order line

The relation between the distance of the slits and the angle through which it is passed is:

dsinθ=nλ

As d and λ are constant

$\frac{n_1\lambda}{sin \theta_1}=\frac{n_2\lambda}{sin \theta_2}\\\Rightarrow \frac{1}{sin20}=\frac{2}{sin\theta_2}\\\Rightarrow sin\theta_2=\frac{2}{\frac{1}{sin20}}\\\Rightarrow \theta_2=sin^{-1}{\frac{2}{\frac{1}{sin20}}}\\\Rightarrow \theta_2=43.16^{\circ}$

∴ Angle by which the second order line appear is 43.16°

5 0

2 years ago