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-1

It's an easy method. You need to calculate m/s, when you have distance in meter and time in seconds. Just use the simple relation. 1 minutes equals 60 seconds. That means 320941.03 minutes equals 320941.03*60 s. Which is 19256461.8 seconds. Now use the formula, speed equals distance per time. i.e, v = s/t = 7720298/19256461.8 = ...


0

To find the speed in m/s, you will need to have a distance measurement in meters (which you already have), and a time measurement in seconds (which you don't). You can change a measurement from one unit to another version of that unit by multiplying by 1. Multiplying a number by 1 doesn't change the value, which is why changing units doesn't change the ...


1

If you take the ratio of the values you have, you will have the velocity in meters per minute, right? Then, to obtain the value in meters per second, you just have to convert that time from minutes to seconds. You can do that by multiplying by 60, because 1 minute is equal to 60 seconds. Therefore, your time is now $320941.03 \text{ minutes} \times ...


0

Displacement refers to the object's position relative to the observer. The "place in space" of the orange. Distance is the object's position relative to an earlier position. If you pick up an orange, and run 10 miles holding it straight out, no work gets done on the orange. But if you extended your arm 10 miles, you would have to be doing work on the ...


17

It depends on whether the force field is conservative or not. Example of a conservative force is gravity. Lifting, then lowering an object against gravity results in zero net work against gravity. Friction is non-conservative: the force is always in the direction opposite to the motion. Moving 10 m one way, you do work. Moving back 10 m, you do more work. ...


3

If you 'carry' an object 10 meters in one direction then return it back 10 meters from where you started the work done on the object is not the force you expended times distance walked. The formula you write is often misunderstood and misused. In your example, when you lift the object in a gravitational field, the work being done on the object is its weight ...


0

A trajectory or flight path is the path that a moving object follows through space as a function of time. The object might be a projectile or a satellite, for example. It thus includes the meaning of orbit—the path of a planet, an asteroid or a comet as it travels around a central mass. Read more here : http://en.wikipedia.org/wiki/Trajectory I believe that ...


1

What a strange question. Is $x(t)$ the distance measured along the track? Without defining $x$, there is nothing else to say. And if it is the distance, then it's the distance. Along the track. And the car is traveling along the track. So x is the distance. Which is what is asked. I suggest you read ...


1

You want to consider the following factors: Maximize initial energy stored in the rubber band. This means you need to be able to twist the band lots of times, and as it unwinds it must continue to produce torque. Minimize internal friction - make the mechanism that converts power from the band to kinetic energy as "direct" as possible. In fact a rubber ...


1

This can be understood in terms of vector differentiation and the dot product. Take the example that $v \perp r$. The change in the square of the displacement is $$\frac{d}{dt}r^2$$ $$=2r \cdot v$$, and if they are perpendicular, the dot product is zero.


3

Spectroscopic parallax is the technique whereby you estimate the absolute magnitude (i.e. the brightness it would have if it were placed at 10 pc) by estimating what "type" of star it is using information fro a spectrum. It can be applied to any kind of star where (a) you have a reasonable chance of determining the type of star from its spectrum and (b) ...



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