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Alborosie
3 years ago
11

In example 10 the kingda ka roller coaster starts with a speed of 6.0 m/s at the top of the drop and attains a speed of 50.3 m/s

when it reaches the bottom. if the roller coaster were to then start up an identical hill, what speed would it attain when it reached the top? assume that friction and air resistance are absent.
Physics
1 answer:
vovikov84 [41]3 years ago
4 0
<span>The speed of the roller coaster when it has to start a similar identical hill the speed would be exactly the same that is 6.0 m/s. And the fact that there is an absence of both friction and air resistance helps us get to this conclusion that it is same as in the previous case.</span>
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The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d
yulyashka [42]

Answer:

a)   F = 3.2 10⁻¹⁰ N , b)       v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

       F = q E

The electric field is related to the potential reference

     V = E d

     E = V / d

Let's replace

    F = e V / d

Let's calculate

    F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

    F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

          v² = v₀ + 2 a d

          v = √ 2 ad

Acceleration can be found with Newton's second law

        e V / d = m a

        a = e / m V / d

        a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

        a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

       v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

       v = √ (98,448 10¹⁴)

       v = 9.9 10⁷ m / s

3 0
2 years ago
Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa
lyudmila [28]

Answer:

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Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

give an electric field of zero at point (0,0,0)

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