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almond37 [142]
3 years ago
12

. A driver travels 4.25 km. in two minutes. If the driver’s initial velocity is 20m/s, what is the acceleration?

Physics
1 answer:
Novosadov [1.4K]3 years ago
4 0

Answer:

32.9166667 m / s^2

Explanation:

s = 4.25km (1000m / 1km)

= 4250m

u = 20m/s

delta T = 20min (60sec / 1min)

= 1200s

Use formula s = ut + (1/2)at^2

4250m = 20m/s * 1200s + (1/2)a*1200s^2

Rearrange it to find a

a = (s-ut)  / (1/2 * t^2)

a = (4250m - 20m/s*1200s) / (1/2 * 1200s^2)

a = -32.9166667 m / s^2

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2m X 2m = 4m ^2
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3 years ago
The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train aver
luda_lava [24]

Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

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Stage 2:

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Stage 3:

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Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

So

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100

Lets find the speed at stage 1

Speed =  \frac{Distance }{Time}

Speed =  \frac{80}{1}

Speed 1= 80 km/hr

The speed at stage 2

Speed =  \frac{Distance }{Time}

Speed =  \frac{200}{2}

Speed 2  = 100 km/hr

The speed at stage 3

Speed =  \frac{Distance }{Time}

Speed =  \frac{x}{4}

Speed 3  = \frac{x}{4}

we kow that average is ,

\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100

\frac{ 80 + 100+ \frac{x}{4} }{3} = 100

\frac{ 180 + \frac{x}{4} }{3} = 100

\frac{ \frac{720 +x}{4} }{3} = 100

\frac{720 +x}{4} \times \frac{1}{3} = 100

\frac{720 +x}{12} = 100

720 +x = 100 \times 12

720 +x = 1200

x = 1200- 720

x = 480

6 0
3 years ago
If the rotation of a planet of radius 5.32 × 106 m and free-fall acceleration 7.45 m/s 2 increased to the point that the centrip
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Answer:

v = 6295.55 m/s

Explanation:

Given that,

The radius of a planet, r=5.32\times 10^6\ m

The free fall acceleration of the planet, a = 7.45 m/s²

We need to find the tangential speed of a person standing at the equator.

Also, the centripetal acceleration was equal to the gravitational acceleration at the equator.

We know that,

Centri[etal acceleration,

a=\dfrac{v^2}{r}\\\\v=\sqrt{ar}\\\\v=\sqrt{5.32\times10^6\times 7.45}\\\\v=6295.55\ m/s

So, the tangential speed of the person is equal to 6295.55 m/s.

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D) A warm front brings drizzly weather.

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