Answer: Hello! An objects speed is constant and has the units meters per second (m/s); thus, it does not change overtime. Acceleration is a rate of change where the speed does either increase or decrease overtime from its inital value; its units are meters per second second (m/s/s). I hope that helps!
Answer:
All object changes are compared with a <em>reference</em> , which is an object that appears to stay in place.
Explanation:
In scientific experiments, the changes in the experimental object are observed by comparing the changes with a reference object. In the reference object, no changes are made and conditions are kept normal in it. For example, if we want to measure the distance of two cars from a point, the point will be the reference point from which the distance shall be measured. Hence, all changes are made by comparison from a reference object or point.
Answer:
Explanation:
The work done on the object at rest is all converted into kinetic energy, so we can write
Or, re-arranging for v,
where
v is the final speed of the object
W is the work done
m is the object's mass
If the work done on the object is doubled, we have W' = 2W. Substituting into the previous formula, we can find the new final speed of the object:
So, the new speed of the object is .
We can determine a planet's orbital period and separation from its star using any detection method. The transit method can yield sizes, whereas the astrometric and doppler approaches can provide minimum masses.
We can calculate average density by combining the transit and doppler approaches. Numerous physical properties, including the semi-major axis, stellar mass, star radius, planet radius, eccentricity, and inclination, are calculated from these observable data. The mass of the planet is also calculated using the star's combined radial velocity readings.
List briefly the planetary characteristics that, in theory, can be detected with the present detection techniques. We can determine a planet's orbital period and separation from its star using any detection method. The transit method can yield sizes, whereas the astrometric and doppler approaches can provide minimum masses.
To know more about orbital period
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Answer:
29.0 g
Explanation:
The mass of the piece of gold is given by:
m = dV
where
m is the mass
d is the density
V is the volume of the piece of gold
The density of gold is
d = 19.3 g/cm^3
while the volume of the sample is equal to the volume of displaced water, so
V = 64.5 mL - 63.0 mL = 1.5 mL
And since
1 mL = 1 cm^3
the volume is
V = 1.5 cm^3
So the mass of the piece of gold is:
m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g
