This year is 60 years since I learned this stuff, and one of the things I always remembered is the formula for the distance a dropped object falls:
D = 1/2 A T²
Distance = (1/2) (acceleration) (time²)
The reason I never forgot it is because it's SO useful SO often. You really should memorize it. And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)
The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth. Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .
In 5 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (5 sec)²
D = (4.9 m/s²) (25 sec²)
D = 122.5 meters
In 6 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (6 sec)²
D = (4.9 m/s²) (36 sec²)
D = 176 meters
Answer:
a physical change
Explanation:
after the water turns to ice, it will melt and became water again making which means it's reversible this being. a physical change
Answer:
The workdone is
Explanation:
From the question we are told that
The potential difference is 
Generally the charge on
is 
Generally the workdone is mathematically represented as

=>
=>
The gravitational force <em>F</em> between two masses <em>M</em> and <em>m</em> a distance <em>r</em> apart is
<em>F</em> = <em>G M m</em> / <em>r</em> ²
Decrease the distance by a factor of 7 by replacing <em>r</em> with <em>r</em> / 7, and decrease both masses by a factor of 8 by replacing <em>M</em> and <em>m</em> with <em>M</em> / 8 and <em>m</em> / 8, respectively. Then the new force <em>F*</em> is
<em>F*</em> = <em>G </em>(<em>M</em> / 8) (<em>m</em> / 8) / (<em>r</em> / 7)²
<em>F*</em> = (1/64 × <em>G M m</em>) / (1/49 × <em>r</em> ²)
<em>F*</em> = 49/64 × <em>G M m</em> / <em>r</em> ²
In other words, the new force is scaled down by a factor of 49/64 ≈ 0.7656, so the new force has magnitude approx. 76.56 N.
Answer:
Zero
Explanation:
The work done by a force on an object is given by:

where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and the displacement of the object
In this situation, the force is the force of gravity acting on the satellite. This force always points towards the centre of the trajectory, so it is always perpendicular to the direction of motion of the satellite (since the orbit is circular), so
and
. Therefore, the work done by gravity is also zero.