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IRINA_888 [86]
4 years ago
12

A car traveling at 30m/s screeches to a halt, leaving 55 m long skid marks. what was the cars acceleration while braking

Physics
1 answer:
Finger [1]4 years ago
6 0
For this problem we use the derived equations for rectilinear motion at constant acceleration. The equation is written below:

2ax = v² - v₀²
where 
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity

Since the car comes to a stop eventually, v=0. Substituting the values,

2a(55) =0² - 30²
a = -8.18 m/s²
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A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori
EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

x - x_o = u_xt

\mathtt{x = u_xt  \ \  \ since (x_o = 0)}  ---- (1)

the equation of the motion y is :

\mathtt{y - y_o =u_yt - 0.5 gt^2}

\mathtt{y = u_yt-4.9t^2     \ \ \  since (y_o =0)}

\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2        }

\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}

\mathtt{1 = 5.14t - 4.9t^2}

\mathtt{4.9t^2 - 5.14t +1 = 0}

By using the quadratic formula, we have;

\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}}     }

where;

a = 4.9,   b = -5.14     c = 1

= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8}  \  \ OR \  \  \dfrac{ 5.14- \sqrt{6.8196}}{9.8}}    }

= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8}  \  \ OR \  \  \dfrac{ 5.14- 2.6114}{9.8}}    }

= \mathtt{ \dfrac{ 7.7514}{9.8}  \  \ OR \  \  \dfrac{ 2.5286}{9.8}}    }

= \mathbf{ 0.791 \  \ OR \  \  0.258}    }

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

\mathtt{x = u_x(0.258)}

\mathtt{x = ucos 40^0 (0.258)}

\mathtt{x = 8 \ cos 40^0 (0.258)}

\mathbf{x = 1.581  \ m}

Thus, the child is 1.581 m far from the fence

6 0
3 years ago
Hydrometer is a devise used to measure the volume of a liquid. *<br> True<br> False
RSB [31]

Answer:

false

Explanation:

3 0
3 years ago
A shot putter heaves a 7.26kg shot with final velocity of 7.50m/s what is kinetic energy?
ivanzaharov [21]

Answer: KE=.5mv^2

KE=.5(7.62)(7.5)^2

KE=.5(7.62)(56.25)

KE=.5(428.625)

KE=214.3125 J of KE

Explanation:

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3 years ago
What is the average horizontal component of force exerted on his feet by the ground during acceleration?
Arte-miy333 [17]
Well you have to think of it like electricity go through your answer closes to that and figure it out
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Physicists. Welcome :)
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