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bixtya [17]
3 years ago
11

A car moving at constant speed in a straight line with the engine providing a driving force equal to the resistive force F.

Physics
2 answers:
Sergeu [11.5K]3 years ago
7 0

Answer:

Option (b)

Explanation:

Let the initial speed of the car is u and the final speed is zero. Let mass of the car is m.

Case I

Acceleartion of the car, a = - F/m

Use III equation of motion,

v^{2} = u^{2} + 2 a s

0^{2} = u^{2} - 2 \frac{F}{m} \times 100

u^{2} = \frac{200 F}{m}     ...... (1)

Case II

Acceleration of the car, a = - 0.8 F/m

Use III equation of motion,

v^{2} = u^{2} + 2 a s

0^{2} = u^{2} - 2 \frac{0.8F}{m} \times s

0^{2} = u^{2} - 2 \frac{0.8F}{m} \times s

u^{2} = \frac{1.6F}{m} \times s      ..... (2)

Dividing equation (2) by equation (1), we get

s = 125 m

Lunna [17]3 years ago
6 0

125 b

simultaneous kinematic equations two variables are F and stopping distance

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Answer:

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Explanation:

Given:

Two long, parallel wires separated by a distance,

d = 3.50 cm = 0.035 meter

Currents,

I_{1}=1.55\ A\\I_{2}=3.15\ A

To Find:

Magnitude of the force per unit length that one wire exerts on the other,

\dfrac{F}{l}=?

Solution:

Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

\dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}

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Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

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