A government agency would likely spend most of its budget on which of the following?

The International Space Station (ISS) orbits Earth in a nearly circular orbit that is 345 km above Earth's surface.

Tems11 [23]

Answer:

1) v = 7.70 10³ m/s , 2) F = 115 N and 3) (F/W)% = 90.2%

Explanation:

1) To solve the problem let's use Newton's second law where force is gravitational force and acceleration is centripetal

F = ma.

F = G m M / r²

a = v² / r

G m M / r² = m v² / r

G M / r = v²

Let's look for the distance is the distance from the surface of the has to the station 345 103 m plus the radius of the Earth

r = Re + 345 103

r = 6.37 10⁶ + 3.45 10⁵

r = 6.715 10⁶ m

Let's calculate the speed

v = √ (6.67 10⁻¹¹ 5.98 10²⁴ / 6,715 10⁶) = √ (59,399 10⁶)

v = 7.70 10³ m/s

The speed module is constant, so we can use the uniform motion relationships

v = d / t

The distance is the length of the circle

d = 2π r

d = 2π 6.715 106

d = 42.2 10⁶ m

Let's calculate the time

t = d / v

t = 42.2 10⁶ / 7.70 10³

t = 5.48 10³ s

2) Let's use the universal gravitation equation

F = G m M / r²

F = 6.67 10⁻¹¹ 13.0 5.98 10²⁴ /(6.715 10⁶)²

F = 11.5 10¹ N

F = 115 N

3) in this for we are asked the relationship is out with the weight of the body on earth

F / W = F / mg

F / W = 115 / (13.0 9.8)

F / W = 0.902

F / W% = 90.2%

5 0

3 years ago

A distracted driver is driving towards a turn where the edge of the road leads into a 75.0 m cliff. The velocity of the vehicle

Vlada [557]

As long as the car is on the road, it moves with a constant speed of 80km/h.

As soon as the car starts to fall down the cliff, it follows a parabolic motion. It means that it still moves with constant speed along the x axis, but it also starts to move along the y axis, with constant acceleration (i.e. the acceleration due to gravity).

The good thing about parabolic motions is that the two motions along the x and y axes are completely separable.

So, first of all, we need to know how long it takes for an object to fall for 75m. The equation of a constantly accelerated motion is

$s=s_0+v_0t+\dfrac{1}{2}at^2$

Where $s_0$ is the initial position, $v_0$ is the initial speed, and $a$ is the constant rate of acceleration. In our case, we start from an initial height of 75m, an initial (vertical!) speed of zero, and our acceleration is $-g$ . So, our equation becomes

$s=75-\dfrac{g}{2}t^2$

And we want to solve for the time when $s=0$ (i.e. we want to know how long will it take for the object to reach the ground). We have

$0=75-\dfrac{g}{2}t^2 \iff 75=\dfrac{g}{2}t^2 \iff \dfrac{2\cdot75}{g}=t^2 \iff t=\sqrt{\dfrac{150}{g}}$

(I'm discarding the negative solution because it wouldn't make sense)

Now that we've used the vertical motion to find out the falling time, we can go back to the horizontal motion. We know that the car moves for a certain amount of time at a certain speed. So, we simply have to plug our values in the $s=vt$ equation, to get

$s=80\sqrt\dfrac{150}{g}}$

This is how far from the base of the cliff the vehicle lands.

3 0

3 years ago

Covalent bonds ___ conduct electricity well. Always Usually Rarely Never

mr_godi [17]

They rarely conduct electricity well.

4 0

3 years ago

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A cylinder with a mass M and radius r is floating in a pool of liquid. The cylinder is weighted on one end so that the axis of t

Blizzard [7]

Answer:

acylinder whit is yoiyes

8 0

4 years ago

Consider what happens at the moment when the block leaves the surface of the globe. Which of the following statements are correc

erma4kov [3.2K]

Answer:

Statement 1 and 3 are correct.

Explanation:

1. The mass moves downward, so the net acceleration of the block is straight downward.

2.The mass is sliding through the globe, so only the force of gravity is acting on the mass which pulls it in downward direction. The force of gravity has two components [mg sin∅] and [mg cos∅].

6 0

4 years ago

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