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Liula [17]
2 years ago
6

If you are driving 80 km/h along a straight road and you look to the side for 2.1 s , how far do you travel during this inattent

ive period?
Physics
1 answer:
ella [17]2 years ago
7 0
Velocity = displacement / time

First, Always make sure that your units for things are the same. In this question, we are given time in seconds and hours. So, we need to make them all use 1 unit.

I will do seconds.

To convert 80km/h to m/s, divide by 3.6.

22.2222…. m/s = displacement / 2.1s

Multiply by 2.1s on both sides

46.6666……m = displacement

You travelled 47 meters.
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Which type of cloud would you expect to be involved in some form of precipitation?
fomenos
Stratus clouds maybe
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3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
A rightward-moving truck skids to a stop from a very high speed with its wheels locked. What forces are exerted on the object?
aksik [14]
Friction, gravity, normal force, and the brakes constitute an applied force.
4 0
3 years ago
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A 1000 kg car is rolling slowly across a level surface at 1m/s, heading toward a group of small innocent children.
Ne4ueva [31]

Answer:

The force is -1620.73 N.

Explanation:

Given that,

Mass of car = 1000 kg

Velocity = 1 m/s

Distance = 2 m

Angle = 30°

We need to calculate the force

Using formula of work done

W_{net}=K_{f}-K_{i}

F\times d=K_{f}-K_{i}

Fd\cos\theta=-K_{i}

F=\dfrac{-K_{f}}{d\cos\theta}

F=-\dfrac{\dfrac{1}{2}mv^2}{d\cos\theta}

Put the value into the formula

F=-\dfrac{\dfrac{1}{2}\times1000\times(1)^2}{2\times\cos30}

F=-1620.73\ N

Hence, The force is -1620.73 N.

7 0
4 years ago
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How long will it take to travel 200 km traveling 10 m/s ?
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<span>Convert 200 km to meters which equals 200000 meters. Then take 10m/s*200000meters which equals 2000000 seconds.
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4 years ago
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