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blondinia [14]
2 years ago
7

In a car moving at constant acceleration, you travel 230 m between the instants at which the speedometer reads 40 km/h and 70 km

/h how many seconds does it take you to travel the 230 m ?
Physics
1 answer:
Goryan [66]2 years ago
7 0
The relationship between the distance covered, initial and final speeds, and time can be expressed through the equation,

First equation,

                    2ad = Vf² - Vi²

Substituting the known values,
                   2(a)(0.230 km) = (70 km/h)² - (40 km/h)²
The value of a from the equation is 7173.92 km/h².

Second equation,
                   d = (Vi)(t) + 0.5at²

Substituting the known values,
                    0.230 km = (40 km/h)(t) + (0.5)(7173.92 km/h²)(t²)

The value of t from the equation is 4.1818 x 10^-3 hours which is also equal to 0.2509 minutes or 15 seconds.

Answer: 15 seconds
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If the Lifted Index is positive, is the parcel warmer than, the same temperature as, or cooler than its environment?
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8 0
2 years ago
A policeman starts giving chase 60 seconds after a stolen car zooms by at 108 km/hr. At what minimum speed should he drive if he
Iteru [2.4K]

Answer:

30.93 m/s

Explanation:

Given that, the speed of stolen car is,

v_{s} =108km/hr\\v_{s} =108\times \frac{5}{18}m/s\\ v_{s} =30m/s

As policeman start chasing the stolen car after 60 seconds.

Now suppose the speed of policeman car is, v_{p}

The policeman catches the stolen car at a distance of,

S=60km\\S=60000m

Now the distance covered by the policeman in time t is v_{p}\times t

And the distane cover by the thief in stolen car in time(t+60s) is v_{s}\times (t+60sec).

And these distances are equal and they are equal to 60000 m.

Therefore,

v_{p}\times t=v_{s}\times (t+60sec)=60000m

Therfore,

v_{s}\times (t+60sec)=60000m\\30m/s\times (t+60sec)=60000m\\(t+60s)=2000s\\t=1940s

Now use this value to solve for minimum speed of policeman's car.

v_{p}\times 1940=60000\\v_{p}=30.93 m/s

Therefore minimum speed of policeman's car is 30.93 m/s.

6 0
3 years ago
You need to get to class, which is 282 meters away. You can only walk in the hallway at about 2.0 m/s (if you go any faster, you
worty [1.4K]

Answer:

Time, t = 141 seconds

Explanation:

Given that,

Distance between the class and the person, d = 282 m

Speed of the person, v = 2 m/s

We need to find the time taken by the person to get to your class. Let the time taken is t. The speed of the person is given by :

v=\dfrac{d}{t}

t=\dfrac{d}{v}

t=\dfrac{282\ m}{2\ m/s}

t = 141 seconds

So, the time taken by the person to get to your class is 141 seconds. Hence, this is the required solution.

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Answer:

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Hope it helps!!!

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