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balandron [24]
2 years ago
9

100 point question!! A driver begins to brake when her car is traveling at 15.0 m/s, and the car comes to a stop 4.0 s later. Ho

w much farther does the car go after the driver begins to brake, assuming constant acceleration?
Physics
1 answer:
Anni [7]2 years ago
3 0

Answer:

30 m

Explanation:

Let's list out the known variables that we are given in this question.

  • v₀ = 15 m/s
  • v = 0 m/s
  • t = 4 s

We are trying to solve for displacement in the x-direction.

  • Δx = ?

Find which constant acceleration equation contains all 4 of these variables.

  • \displaystyle \triangle x = \Big (\frac{v+v_0}{2} \Big ) t

Plug the known values into the equation and solve for delta x.

  • \displaystyle \triangle x = \Big (\frac{0+15}{2} \Big ) \cdot 4
  • \displaystyle \triangle x = \Big (\frac{15}{2} \Big ) \cdot 4  
  • \displaystyle \triangle x =\frac{60}{2}
  • \triangle x = 30

The car goes 30 m after the driver begins to brake.

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Answer:

It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. ... Once calculated, this area represents the displacement of the object.

Explanation:

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Answer:

5.38 m/s^2

Explanation:

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F = ma

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1 year ago
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5 0
2 years ago
To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where
BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

\int E.dA = \frac{q}{\epcilon_0}

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

q = \int \rho dV

q = \rho * \pi r^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r}{2 \epcilon_0}

Now if we have a situation that the point lies outside the cylinder

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q = \rho * \pi r_0^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

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7 0
3 years ago
A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.
Vaselesa [24]
The acceleration of the object which moves from an initial step to a full halt given the distance traveled can be calculated through the equation,
                                     d = v² / 2a
where d is distance, v is the velocity, and a is acceleration
Substituting the known values,
                                     180 = (22.2 m/s)² / 2(a)
The value of a is equal to 1.369 m/s²
The force needed for the object to be stopped is equal to the product of the mass and the acceleration.
                                      F = (1300 kg)(1.369 m/s²) 
                                            F = 1779.7 N
4 0
3 years ago
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