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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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Answer:

Synthesis Reaction

Explanation:

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What’s the velocity of a ball falling with 100 joules of kinetic energy and a mass of 2 kilograms? Use the formula, .
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The answer is 10 m/s
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In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal
Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

    p = m1 v1 + m2 v2

Part B.

Find the derivative

     dp / dt = m1 dv1dt + m2 dv2 / dt

     dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

     m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1        f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

     dp / dt = 0

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mrs_skeptik [129]

Answer:

t=16.67s

Explanation:

From the question, acceleration from 6m/s^2  \ to \ 0m/s^2 in 10 seconds.

Acceleration function can be written as:-

a_t=6-0.6t

From the acceleration equation, we can obtain the velocity equation:

v(t)=\int\limits^t_b {0} \, dv=\int\limits^t_0 {a(t)} \, dt\\v(t)=\int\limits^t_0 {(6-06t)} \, dt=6t-0.3t^2\\*dv=a(t)dt

We calculate velocity after 10sec from the above v(t) as 30m/s.

To obtain distance travelled after 10 seconds:-

S=\int\limits^{10}_0 {6t-0.3t^2} \, dt=|3t^2-0.1t^3|   \ *lims(10,0)\\ S=200m

Therefore 200m is for 10seconds, and next 200m at 30m/s

Total time=10+\frac{200}{30}=16.67s

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