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3 years ago

A car driving on the highway is going 72mph After 3 hours how many miles will the car have driven ?

Physics

2 answers:

Sergeu [11.5K]3 years ago

7 0

Answer:

216 miles

Explanation:

you times 72 by three

diamong [38]3 years ago

6 0

I think it is 216mph

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You're driving down the highway late one night at 22 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim

sleet_krkn [62]

speed of the car is given on the road

$v = 22 m/s$

reaction time is given as

$t = 0.50 s$

now we can find the distance that it will move during this reaction time

$d_1 = 22* 0.50 = 11 m$

now the deceleration of the car is 10 m/s^2

so the distance that it will move before stop is given by

$v_f^2 - v_i^2 = 2 a d$

$0^2 - 22^2 = 2*(-10)*d$

$d = 24.2 m$

so the total distance that it require to stop is given as

$d = 24.2 + 11 = 35.2 m$

while the deer is standing at distance 38 m

so the car will stop 2.8 m before the position of deer

3 0

3 years ago

Tammy is a forensic investigator examining a body at a crime scene. She notes that the body is stiff but still flexible. What do

wlad13 [49]

I may be wrong but does it mean it was revent? because i know shortly after someone dies your body becomes fully stiff so maybe it was recent and it's in the process off stiffening up

8 0

3 years ago

A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.

miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

$\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}$

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.