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Tamiku [17]
3 years ago
13

10. At point A, a runner is jogging 5 m/s. Forty seconds later, at point B

Physics
1 answer:
kondaur [170]3 years ago
5 0

Answer:

Deceleration of 0.075m/s²

Explanation:

Given parameters:

Initial velocity  = 5m/s

Time  = 40s

Final velocity  = 2m/s

Unknown:

The jogger's acceleration  = ?

Solution:

To solve this problem, we use;

         a  = \frac{v - u}{t}  

a is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

      a  = \frac{2 - 5}{40}   = - 0.075m/s²

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I think it's B hope it helps
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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:

  • No, this doesn't mean the electric potential equals zero.

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)

8 0
3 years ago
The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8
Rasek [7]

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) \tau = \dfrac{\Delta L}{\Delta t}

   \tau = \dfrac{0.8-3}{1.5}

        τ = -1.467 N m

b) angle at which fly wheel will turn

   \theta= \omega t +\dfrac{1}{2}\alpha t^2

   \theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2

   \theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2

        θ = 20.35 rad

c) work done on the wheel

     W = τ x θ

     W = -1.467 x 20.35 rad

    W = -29.86 J

d) average power of wheel

    P_{av} =-\dfrac{W}{t}

    P_{av} =-\dfrac{(-29.86)}{1.5}

     P_{av} =19.91\ W          

7 0
2 years ago
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1) Which of the following is the best definition of the scientific method?
zysi [14]
D is the correct answer
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Starting at 1.3 m/s, a runner accelerates at a constant 0.22 m/s2 for 6.0 s. What is the runner’s displacement during this time
Anettt [7]

Answer:

answer is 11.76 meter

Explanation:

use 2nd equation of motion

S=ut+1/2at^2

7 0
3 years ago
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