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3 years ago

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A van is traveling with an initial velocity of 12 m/s. The driver takes a time of 45 seconds to speed up to a velocity of 20 m/s

. How much distance did the van cover during this time?

1 answer:

Rufina [12.5K]3 years ago

4 0

Initial velocity=u=12m/s
Final velocity=v=20m/s
Time=t=45s

$\\ \rm\hookrightarrow Acceleration=\dfrac{v-u}{t}$

$\\ \rm\hookrightarrow Acceleration=\dfrac{20-12}{45}$

$\\ \rm\hookrightarrow Acceleration=\dfrac{8}{45}$

$\\ \rm\hookrightarrow Acceleration=0.1m/s^2$

Now

Distance=s

$\\ \rm\hookrightarrow v^2-u^2=2as$

$\\ \rm\hookrightarrow (20)^2-12^2=2(0.1)s$

$\\ \rm\hookrightarrow 400-144=0.2s$

$\\ \rm\hookrightarrow 256=0.2s$

$\\ \rm\hookrightarrow s=\dfrac{256}{0.2}$

$\\ \rm\hookrightarrow s=1280m$

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Explanation:

A ball is thrown horizontally

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The initial velocity is the horizontal component $u_{x}$

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

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→ -60 = -4.9t²

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∴ t = 3.5 seconds

* <em>The ball is in the air for 3.5 seconds </em>

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The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = $u_{x}$ t

→ t = 3.5 s , x = 100 m

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Divide both sides by 3.5

→ $u_{x}$ = 28.57 m/s

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The vertical component of the final velocity is $v_{y}$

→ $v_{y}$ = $u_{y}$ + gt

→ $u_{y}$ = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ $v_{y}$ = 0 + (-9.8)(3.5)

→ $v_{y}$ = -34.3 m/s

<em>The vertical component of the final velocity is 34.3 m/s downward</em>

The final velocity v is the resultant vector of $v_{x}$ and $v_{y}$

→ Its magnetude is $v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

→ Its direction $tan^{-1}\frac{v_{y}}{v_{x}}$

→ $v_{y}$ = 28.6 , $v_{y}$ = -34.3

Substitute this values in the rules above

→ $v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66$

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A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

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The acceleration of gravity for an object near a planet is given by

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$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

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The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

d)

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

e)

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

f)

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago