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vampirchik [111]
3 years ago
13

The driver of a 1,000 kg car travelling at a speed of 16.7 m/s applies the car's brakes when he sees a red light. the car's brak

es provide a frictional force of 8,000 n. determine the stopping distance of the car.
Physics
2 answers:
Dmitry [639]3 years ago
6 0
Energy conservation : 
<span>kinetic energy Ek = braking work W </span>
<span>mV^2 = 2Fb*x </span>
<span>x = mV^2/2Fb = 1000*16.7^2/16.000 = 17.43 meters</span>
Sauron [17]3 years ago
5 0

Answer:

1.04 meters

Explanation:

Thinking process:

Gathering the data

mass  = 1 000 kg

speed, u = 16.7 m/s

Frictional force = 8 000 N

distance, s  = ?

final velocity  = 0 (car stops)

We know that the final velocity is calculated as: v^{2} = u^{2}  + 2as

But, a = negative (declaration)

And F = ma

      8 000 = 1000 (a)\\          a = -8 m^{-2}

substituting, 0 = (16.7) - 2 (8) (s)

                    -16.7 = -16s

                         s  = 1.04 m

   Stopping distance = 1.04 meters

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Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
A ball is dropped from rest. how fast is the ball going after 3 seconds​
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Answer:

v = 29.4m/s

Explanation:

Since the ball is dropped at rest,

u = 0m/s

a = 9.81m/s²

Using

v = u + at

After 3 seconds,

v = 0 + (9.81)(3)

v = 29.4m/s

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3 years ago
A net force of 345 N accelerates a boy on a sled at 3.2 m/s^2 . What is combined mass of the sled
Daniel [21]

Answer:

Mass, m = 26.54kg

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

Fnet = Fapp + Fg

Where;

  • Fnet is the net force
  • Fapp is the applied force
  • Fg is the force due to gravitation

<u>Given the following data;</u>

Net force, Fnet = 345

Acceleration, a = 3.2m/s²

<u>To find mass;</u>

Fnet = Fapp + Fg

Fnet = ma + mg

Fnet = m(a+g)

m = Fnet/(a+g)

We know that acceleration due to gravity, g = 9.8m/s²

Substituting into the equation, we have;

m = 345/(3.2 + 9.8)

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