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pav-90 [236]
1 year ago
15

A well is being dug. A 4.5-kg bucket is filled with 28.0 kg of dirt and pulled vertically upward at a constant speed through a d

istance of 9.0 m. Determine the work done on the filled bucket in raising out of the hole.
Physics
1 answer:
andreyandreev [35.5K]1 year ago
4 0

The work done on the filled bucket in raising out of the hole is 2, 925 Joules

<h3>How to determine the work done</h3>

Using the formula:

Work done = force * distance

Note that force = mass * acceleration

F = mg + ma

F = 4. 5 * 10 + 28 * 10

F = 45 + 280

F = 325 Newton

Distance = 9m

Substitute into formula

Work done = 325 * 9

Work done = 2, 925 Joules

Therefore, the work done is 2, 925 Joules

Learn more about work done here:

brainly.com/question/25573309

#SPJ1

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The bending of light rays is the result of refraction of light passing through the lens. A converging lens is curved on both sides such that the rays coming out of it come together at a point (converge). The point at which the right rays meet after refraction is called the focal point which is a real in the convex lens. 
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In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single
FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is 2.27\times 10^{39}

Explanation:

It is given that,

Distance between electron and proton, r=4.53\ A=4.53\times 10^{-10}\ m

Electric force is given by :

F_e=k\dfrac{q_1q_2}{r^2}

Gravitational force is given by :

F_g=G\dfrac{m_1m_2}{r^2}

Where

m_1 is mass of electron, m_1=9.1\times 10^{-31}\ kg

m_2 is mass of proton, m_2=1.67\times 10^{-27}\ kg

q_1 is charge on electron, q_1=-1.6\times 10^{-19}\ kg

q_2 is charge on proton, q_2=1.6\times 10^{-19}\ kg

\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}

\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}

\dfrac{F_e}{F_g}=2.27\times 10^{39}

So, the ratio of electric force to the gravitational force is 2.27\times 10^{39}. Hence, this is the required solution.

3 0
3 years ago
You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If  θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for  90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function  

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

4 0
3 years ago
In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having
Orlov [11]

Answer:

R = 5.28  103 km

Explanation:

The definition of density is

              ρ = m / V

              V = m /ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

            V = 4/3 π r³

Let's replace

             4/3 π r³ = m / ρ

             R =∛ ¾ m / ρ π

The mass of the planet is

              M = 5.5 Me

              R = ∛ ¾ 5.5 Me /ρ π

Let's reduce the density to SI units

             ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³

             ρ = 1.76 10³ kg / m³

Let's calculate

               R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

               R = ∛ 0.14723 10²¹

               R = 0.528 10⁷ m

               R = 0.528 104 km

               R = 5.28  103 km

8 0
3 years ago
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