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3 years ago

The Sears Tower in Chicago is approximately 444 m tall Suppose a book

Physics

1 answer:

Reptile [31]3 years ago

4 0

Answer:

A. 66.0 m/s downwards

Explanation:

The Tower has a height of 444m

The book is dropped ,finding the velocity of the book 222m above the ground, means the book will be on air for a height of 222 m .

Apply the formula for free fall in a horizontal projection as;

h= u²×sin²∅ /2g where

h= maximum height =222m

g= acceleration due to gravity =9.81 m/s²

∅ = projectile angle = 0

u = velocity of the book

Applying the formula as ;

h= u²×sin²∅ /2g

222 = u²/2*9.81

222*19.62 = u²

4355.64 = u²

√4355.64 = u

65.99 m/s = u

66 m/s downwards

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A passenger at an airport pulls a rolling suitcase by its handle. If the force used is 10 N and the handle makes an angle of 25

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Answer:

W= 1812.6 J

Explanation:

Work (W) is defined as the scalar product of force F by the distance (d) the body travels due to this force.

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3 years ago

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4 years ago

Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

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Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

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4 years ago