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Sergio039 [100]
2 years ago
13

A 2 kg box of taffy candy has 40j of potential energy relative to the ground. Its height above the ground is​

Physics
2 answers:
lord [1]2 years ago
8 0

Answer:

2.04m

Explanation:

PE=mgh

h= PE/mg

h= 40/2(9.82)

h=40/19.64

h=2.04

Alona [7]2 years ago
7 0

Answer:

20m

Explanation:

Formula to find the potential energy is,

Potential Energy = mgh

Here,

m = mass

g = gravitational field

h = height

According to the question, we have to find the height.

Usually, we get g is 10ms^{-2}

Let us solve for the answer now.

P.E. =mgh\\40J =2kg*10ms^{-2} *h\\40 = 20h\\\frac{40}{20}= \frac{20h}{20} \\20m=h

Hope this helps you.

Let me know if you have any other questions :-)

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A hollow metal sphere has 7 cm and 11 cm inner and outer radii, respectively. The surface charge density on the inside surface i
Alexus [3.1K]

Answer:

1)  E = 0 , 2) zero, 3)    E = - 2,162 10⁴ N/C , 4) directed towards the center of the sphere , 5) E = 1.412 10⁴ N / C , 6)direction coming out of the sphere

Explanation:

The electric field is a vector quantity, therefore we can calculate the field due to each charge distribution and add vector  

          E = E₁ + E₂

To calculate each field we can use Gauss's law, which states that the flow is equal to the charge  by the Gaussian surface divided by ε₀

For this case, let's take a sphere as a Gaussian surface

          Ф = ∫E dA = q_{int} /ε₀

 The area of ​​a sphere is

        A = 4π r²

         E = 1 / 4πε₀   q_{int} / r²

1) r = 4 cm

This radius is smaller than the radius of the sphere, therefore the charge inside is zero and therefore the field is zero

            E = 0

2) there is no field

3) r = 8 cm

Let's calculate each field, for the inner surface

This radius is larger than the internal radius, so the field is

           σ = q_{int} / A

The area of ​​the sphere is

          V = 4 π R_in²

         Rho = q_{int} / 4π R_in²

          q_{int} = ρ 4π R_in²

         E₁ = 1 /ε₀ ρ r_in² / r²

For the outer surface

This radius is smaller so there is no load inside the Gaussian surface and therefore the field is zero

          E₂ = 0

Total E

         E = E₁ + 0

          E = 1 /ε₀  ρ₁ R_in² / r²

Let's calculate

           E = 1 /8,854 10⁻¹²   250 10⁻⁹ (7/8)²

           E = - 2,162 10⁴ N / C

4) as the electric field is negative, it is directed towards the center of the sphere

5) r = 12 cm

In this case the two surfaces contribute to the electric field,

Inner surface

        Q₁ = ρ₁ 4π R_in²

        E₁ = 1 /ε₀  ₁rho1 R_in² / r²

Outer surface

         Q₂ = ρ₂ 4π R_out²

          E₂ = 1 /ε₀  ρ₂ R_out² / r²

The total field is

          E = E₁ + E₂

          E = 1 /ε₀  | ρ | [- R_in² + R_out²] / r²

Let's calculate

          E = 1 /8,854 10⁻¹² 250 10⁻⁹ [- 7² + 11²] / 12²

          E = 1.412 10⁴ N / C

6) As the field is positive, it is directed radially with direction coming out of the sphere

6 0
3 years ago
Winston stands on the edge of a building's flat roof, 12 m above the ground, and throws a 147.0-g baseball straight down. the ba
Volgvan
Assume theres no air reaistance. (or else the ball will oppose the motion of the ball, hence decrease the velocity increase)
see if there are any problems!

6 0
3 years ago
Which adaptation is likely to increase the chances of survival of an animal in a rainforest?
34kurt
Black-spotted skin coat as camouflage while stalking prey.
Survival = avoiding predators or capturing prey successfully
6 0
3 years ago
Read 2 more answers
How much kinetic energy does a 1500 kg car have if it’s moving at 15 m/s
Tresset [83]

Answer:

Mass of car m = 1500 kg

velocity v = 15 m/s

kinetic energy = ½ mv2

= ½ x 1500 x (15)2    

= 1687501   kg m2 /s2

  = 168750J

Explanation:

6 0
3 years ago
Two paths lead to the top of a big hill. One is steep and direct, while the other is twice as long but less steep. How much more
umka2103 [35]

Answer:

None.

Explanation:

  • Gravitational potential energy, depends only of the mass on which the gravity is doing work, and the displacement produced by this force.
  • As displacement depends only on the final and initial positions (in this case, the height of the hill), if we choose as our zero reference level the bottom of the hill, the change in gravitational potential energy will be as follows:

       \Delta U = U_{f} -U_{0} = m*g*h - 0  = m*g*h

  • As we can see, the only value of distance involved is the height of the hill , so it is independent of the distance travelled.
6 0
3 years ago
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