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Nat2105 [25]
4 years ago
5

A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The d

istance it moves in the third second.
Physics
1 answer:
ValentinkaMS [17]4 years ago
8 0

Answer:

Explanation:

Initial velocity , u = 30 m/s

final velocity , v = 10 m/s

time , t = 5 seconds

1. Acceleration = v - u / t

= 10 - 30 / 5

= -20 / 5

= <u><em>- 4 m/s</em></u>

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Positive charge Q is distributed uniformly along the x-axis from x = 0 to x = a. A positive point charge q is located on the pos
ANTONII [103]

Answer:

(a): k\dfrac{Q}{r(a+r)}..

(b): 0

(c):k\dfrac{qQ}{r(a+r)}..

(d): Along positive x axis.

Explanation:

Given that the charge is distributed uniformly on the rod, extended from x=0 to x=a and having total charge Q.

Let \lambda be the linear charge density of the rod, such that,\lambda = \dfrac Qa.

The electric field due to a charge q at a point r distance away is given by

E=\dfrac{kq}{r^2}

where k is the Coulomb's constant whose value is 9\times 10^9] Nm^2/C^2.

Now consider a small line element of the rod of length dx at distance r from x =a+r.

The electric field at point x = a + r due to this element is given by

dE = \dfrac{k\ dq}{x^2}

dq is the charge on the line charge, then,

\lambda = \dfrac{dq}{dx}\\\Rightarrow dq=\lambda dx

Using this value,

dE = \dfrac{k\lambda\ dx}{x^2}.

The electric field due to the whole rod is given by

E=\int dE\\=\int\limits_{r}^{a+r} \dfrac{k\lambda\ dx}{x^2}\\=k\lambda\int\limits^{a+r}_{r} \dfrac{ dx}{x^2}\\=k\lambda\int\limits^{a+r}_{r} x^{-2}\ dx\\=k\lambda \left (-x^{-1} \right )\limits^r_{a+r}\ dt\\=-k\lambda \left (  \dfrac1{a+r}-\dfrac 1r\right ) \\=-k\lambda \left ( \dfrac{r-(a+r)}{r(a+r)}\right ) \\=k\lambda \left ( \dfrac{a}{r(a+r)}\right )\\=k\dfrac{Q}{a} \left ( \dfrac{a}{r(a+r)}\right )\\=k\dfrac{Q}{r(a+r)}.

This electric is along the x axis only.

(a):

The x component of the electric field at this point = k\dfrac{Q}{r(a+r)}..

(b):

The y component of the electric field at this point = 0.

(c):

The electric field and the electric force are related as

F=qE.

The magnitude of the force that this rod exerts on the charge q = k\dfrac{qQ}{r(a+r)}..

(d):

The direction of this force is along the positive x axis.

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