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tester [92]
3 years ago
12

an object of mass 20kg is lifted to a 25m building. how much potential energy is stored on a mass?(take g=10m/s²)​

Physics
1 answer:
Goshia [24]3 years ago
4 0

Answer:

The answer is 5000 Joules

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A diffraction grating is placed 1.00 m from a viewing screen. Light from a hydrogen lamp goes through the grating. A hydrogen sp
Colt1911 [192]

Answer:

λ = 396.7 nm

Explanation:

For this exercise we use the diffraction ratio of a grating

           d sin θ = m λ

in general the networks works in the first order m = 1

we can use trigonometry, remembering that in diffraction experiments the angles are small

           tan θ = y / L

           tan θ = \frac{sin \theta}{cos \theta} = sin θ

           sin θ = y / L

we substitute

          d \  \frac{y}{L} = m λ

with the initial data we look for the distance between the lines

           d = \frac{m \lambda \ L}{y}

           d = 1 656 10⁻⁹ 1.00 / 0.600

            d = 1.09 10⁻⁶ m

for the unknown lamp we look for the wavelength

           λ = d y / L m

           λ = 1.09 10⁻⁶ 0.364 / 1.00 1

           λ = 3.9676 10⁻⁷ m

           λ = 3.967 10⁻⁷ m

         

we reduce nm

           λ = 396.7 nm

8 0
3 years ago
Why does a comet's tail point away from the Sun? 1Choice 1 The solar wind blows the tail away from the Sun. 2 It is being pulled
AnnyKZ [126]

Choice 1

The Sun's radiation and solar wind cause the dust and gas around the comet (coma) to stretch the coma. The solar wind electromagnetically blows the ions in the coma away.

4 0
3 years ago
The planet Krypton has a mass of 7.6 × 1023 kg and radius of 1.7 × 106 m. What is the acceleration of an object in free fall nea
olga55 [171]

Answer:

17.55 m/s²

Explanation:

Parameters given:

Mass of Krypton, M = 7.6 * 10^23 kg

Radius, R = 1.7 * 10^6 m

Gravitational constant, G = 6.6726 * 10^(-11) Nm²/kg²

Acceleration due to gravity of planet of mass M is given as:

g = GM/R²

Since the object is close to the surface of Krypton, we can say that the distance from the Centre of Krypton is the radius of the planet Krypton.

Therefore,

g = (6.6726 * 10^(-11) * 7.6 * 10^23)/(1.7 * 10^6)²

g = 17.55 m/s²

5 0
3 years ago
Given the equation p2 = a3, what is the orbital period, in days, for the planet venus? (venus is located 0.72 au from the sun?)
melisa1 [442]

The correct answer is 223 days.

The relationship between the duration of revolution and the separation between the sun is shown by Kepler's third law. Using the notions of circular motion and the gravitational and centripetal forces, we may obtain this equation.

According to Kepler's third rule, the semi-major axis of an orbit is linked to the orbital period of a planet around the sun as follows:

p² = a³

where an is the semi-major axis/distance to the star and p is the orbital period in years.

It is said that a = 0.72 AU for Venus.

P= √(0.72 AU)^3 = 0.61 years.

365 days in a year = 222.9 ≈ 223 days.

To learn more about Kepler's third rule refer the link:
brainly.com/question/1608361

#SPJ4

5 0
1 year ago
An object has a mass of 2,000kg. what is its weight on earth? show your work​
artcher [175]

Answer:

F=m*g is the formula and the answer is 19,620 kg

Explanation:

Since the formula is F=m*g and Earth's gravity is 9.81 m/s^2 all you need to do is multiply 2,000 by 9.81

6 0
3 years ago
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