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Roman55 [17]
3 years ago
5

A car starts from rest and accelerates uniformly over a time of 18 seconds for a distance of 390 m. Determine the acceleration o

f the car.

Physics
1 answer:
Sergeeva-Olga [200]3 years ago
5 0

Answer:

a=2.4\ m/s^2

Explanation:

Given that,

The initial speed of a car, u = 0

Time, t = 18 s

Distance, d = 390 m

We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.

d=ut+\dfrac{1}{2}at^2

or

d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2

So, the acceleration of the car is 2.4\ m/s^2.

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Read 2 more answers
A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R. The centra
Dmitriy789 [7]

Answer:

(a). If z = 0, The electric field due to the rod is zero.

(b). If z =  ∞, The electric field due to the rod is E\propto\dfrac{1}{z^2}.

(c). The positive distance is \dfrac{R}{\sqrt{2}}

(d). The maximum magnitude of electric field is 1.54\times10^{4}\ N/C

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

We need to calculate the electric field due to the rod

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

E=0

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}

E\propto\dfrac{1}{z^2}

(c). In terms of R,

We need to calculate the positive distance

If E\rightarrow E_{max}

Then, \dfrac{dE}{dz}=0

\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0

Taking only positive distance

z=\dfrac{R}{\sqrt{2}}

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}

E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}

E_{max}=15418.7\ N/C

E_{max}=1.54\times10^{4}\ N/C

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z =  ∞, The electric field due to the rod is E\propto\dfrac{1}{z^2}.

(c). The positive distance is \dfrac{R}{\sqrt{2}}

(d). The maximum magnitude of electric field is 1.54\times10^{4}\ N/C

6 0
3 years ago
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