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3 years ago

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6. A car, 1110 kg, is traveling down a horizontal road at 20.0 m/s when it locks up its brakes. The coefficient of friction betw

een the tires and road is 0.901. How much distance will it take to bring the car to a stop?

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

$x=22.65m$

Explanation:

We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:

$v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}$

The acceleration can be calculated using Newton's second law:

$\sum F_x:F_f=ma\\\sum F_y:N=mg$

Recall that the maximum force of friction is defined as $F_f=\mu N$ . So, replacing this:

$\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}$

Now, we calculate the distance:

$x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m$

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Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

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E⃗ 2

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$E_4 = 5.754 *10^ {8} \ V/m$

Explanation:

From the question we are told that

The diameter is $d = 6.53 \mu m = 6.53*10^{-6}\ m$

The charge is $Q = -.2.55 *10^{-12} \ C$

The permittivity of free space is $\epsilon_o = 8.85* 10^{-12}\ C / V.m$

The distance considered is $d = 3.05 \mu m = 3.05 *10^{-6} \ m$

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

$E_3 = \frac{ k * |Q|}{ r^2 }$

Here $k$ is the coulomb constant with value

$k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}$

r is the radius of the sphere which is mathematically as

$r = \frac{d}{2} = \frac{6.53*10^{-6}}{2} = 3.265 *10^{-6} \ m$

$E_3 = \frac{ 9*10^{9} * |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }$

$E_3 = 2.153 *10^{9} \ V/m$

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

$E_4 = \frac{ k * |Q|}{ R^2 }$

Here R is mathematically represented as

$R = 3.265 *10^{-6} + 3.05 *10^{-6}$

=> $R = 6.315 *10^{-6}$

So

$E_4 = \frac{ 9*10^{9} * |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }$

$E_4 = 5.754 *10^ {8} \ V/m$

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