Answer:
Mercury / Mars
Explanation:
For an object launched straight upward, the following SUVAT equation can be used
where
v is the final velocity
u is the initial velocity
g is the acceleration of gravity (free fall acceleration) (the negative sign is due to the downward direction of gravity)
h is the maximum height reached
At the maximum height, the velocity is zero, so v = 0. Re-arranging the equation,
So we see that for equal initial velocity (u), the maximum height reaches is inversely proportional to the acceleration of gravity. Therefore, the potato gun will reach the highest altitude in the planets with lowest acceleration of gravity, therefore Mercury and Mars (3.7 and 3.6 m/s^2).
Answer:
There is no friction between the card and the cup.
Explanation:
The question is oversimplified, and pretty sloppy.
Relative to the Earth . . .
The Moon is in an elliptical orbit around us, with a period of
27.32... days, and with the Earth at one focus of the ellipse.
Relative to the Sun . . .
The Moon is in an elliptical orbit around the Sun, with a period
of 365.24... days, and with the Sun at one focus of the ellipse,
and the Moon itself makes little dimples or squiggles in its orbit
on account of the gravitational influence of the nearby Earth.
I'm sorry if that seems complicated. You know that motion is
always relative to something, and the solar system is not simple.
Answer:
value-added tax (VAT) is a consumption tax that is levied on a product repeatedly at every point of sale at which value has been added. ... VAT is commonly expressed as a percentage of the total cost. For example, if a product costs $100 and there is a 15% VAT, the consumer pays $115 to the merchant.
Explanation:
archemdes principle states that the boyant upward force by liquid is equal to liquid displaced by solid...
mark me as brainliest ❤️
The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:
The centripetal acceleration is given by:
Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:
