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3 years ago

10

A 675-pound trailer is sitting on an exit ramp inclined at 40o on Highway 35. How much vertical force is required to keep the tr

ailer from rolling back down the exit ramp?

1 answer:

Phantasy [73]3 years ago

8 0

Answer:

433.88 lbs

Explanation:

We are given that mass of trailer =675 pound

Angle= $40^{\circ}$

We have to find the vertical force required to keep the trailer from rolling back down the exit ramp.

The force required to keep the trailer from rolling back down the exit ramp= $wsin\theta$

Where w=675 pounds and $\theta=40^{\circ}$

Substitute the values then we get

$F=675 sin40^{\circ}$

$sin40^{\circ}=0.642787$

Substitute the value then we get

$F=675\times 0.642787$ =433.88 lbs

Hence, the force required to keep the trailer from rolling back down the exits ramp=433.88 lbs

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2 years ago

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

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a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

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Explanation:

a)

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

$g=\frac{GM}{(R+h)^2}$

where

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M is the mass of the planet

R is the radius of the planet

h is the height above the surface

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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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b)

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:

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3 years ago