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MatroZZZ [7]
2 years ago
15

A bicyclist covers the first leg of a journey that is d1meters in t1seconds at a speed of v1m/s and the second leg of d2meters i

n t2seconds at a speed of v2m/sIf his average speed is equal to the average of v1 and v2
then which of the following is true?
1. t1=t2
2. t1≠t2
3. d1=d2
4. d1≠d2
Physics
1 answer:
Murljashka [212]2 years ago
4 0

According to motion in straight line  t1≠t2


A biker travels d1 meters in t1 seconds at v1 m/s for the first leg and d2 meters in t2 seconds at v2 m/s for the second leg. It's possible that t1t2 if his average speed is equal to the average of v1 and v2.
An object is said to be in motion if its position in relation to its surroundings changes over time. It is a shift in an object's position over time. The only type of motion that exists is motion in a straight line.


A reference system is constantly used to describe a particle's motion. An arbitrary origin point is used to create a reference system, and a coordinate system is imagined to be connected to it. The reference system for a specific problem is the coordinate system that has been selected for it. For the majority of the problems, we typically select an earth-based coordinate system as the reference system.


To learn more about Motion in straight lines please visit -brainly.com/question/17675825
#SPJ1

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Explanation:

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The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for
pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v )  = 20.5 mi/h  = 9.164  M/S

distance ( d ) = 11.6 meter                                       ( m = mass of the car )

Work done = 0.5 m v²  = 0.5 * 9.164² * m J  = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 )   =  3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

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d = 69 .57 meter

7 0
3 years ago
Atomic math challenge will give brainly and thanks
nevsk [136]

Answer:

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8 0
3 years ago
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An airplane starts from A and flies to B at a constant speed. After reaching B it returns to A at the same speed. There was no w
Dafna1 [17]

Answer:

When there is wind it takes longer

Explanation:

With no wind, the round trip time is

t_1=\frac{d}{v}+\frac{d}{v}=\frac{2d}{v}

When we have a constant wind speed w

t_2=\frac{d}{v-w} +\frac{d}{v+w} =\frac{2vd}{v^{2} -w^{2}}

comparing the reciprocal times;

\frac{1}{t_2}=\frac{v^{2}-w^{2} }{2vd}=\frac{v}{2d}-\frac{w^{2}}{2vd}   \leq \frac{v}{2d}=\frac{1}{t_1}

This means that t1 is smaller than t2, ergo, it takes longer with wind

4 0
3 years ago
Calcular la rapidez promedio
Vlada [557]

Answer:

  v_average = 15 m / s

Explanation:

The average speed can be found in two ways,

* taking the distance traveled and divide it by the time spent

* taking the velocities in each time interval and then finding the weighted average by the time fraction

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Let's apply this last equation

               

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              v_average = 10/20 10 + 10/20 20

              v_average = 10/2 + 20/2

              v_average = 15 m / s

7 0
3 years ago
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