Answer: both hoops have the same kinetic energy at the bottom of the incline.
Explanation:
If we assume no work done by non conservative forces (like friction) , the total mechanical energy must be conserved.
K1 + U1 = K2 + U2
If both hoops start from rest, and we choose the bottom of the incline to be the the zero reference level for gravitational potential energy, then
K1 = 0 and U2 = 0
⇒ ΔK = ΔU = m g. h
If both inclines have the same height, and both hoops have the same mass m, the change in kinetic energy, must be the same for both hoops.
The center of mass isn't affected by the explosion.
To find the answer, we need to know about the trajectory of motion at zero external force.
<h3>How is the trajectory of an object changed when the net external force on it is zero?</h3>
- When there's no net external force acting on an object, its momentum doesn't change with time.
- As its momentum doesn't change, so it continues with the original trajectory.
<h3>Why doesn't the trajectory of firework change when it's exploded?</h3>
- When a firework is exploded, its internal forces are changed, but there's no external force.
- So, although the fragments follow different trajectories, but the trajectory of center of mass remains unchanged.
Thus, we can conclude that the center of mass isn't affected by the explosion.
Learn more about the trajectory of exploded firework here:
brainly.com/question/17151547
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V= s/t = 400/4.5=800/9 (km/h)
The tiny space invader and the new space station will have equal orbital speed.
<h3>The orbital speed of satellite </h3>
The orbital speed of satellite on Earth surface is given as;
where;
- V is the orbital speed
- G is universal gravitation
- M is mass of the Earth
- r is the radius of the circular path
Since the two objects are at the same height from Earth's surface, the distance from the central point (r) will be the same. Thus, the two objects will have equal orbital speed.
Learn more about orbital speed here: brainly.com/question/22247460
Let car A's starting position be the origin, so that its position at time <em>t</em> is
A: <em>x</em> = (40 m/s) <em>t</em>
and car B has position at time <em>t</em> of
B: <em>x</em> = 100 m - (60 m/s) <em>t</em>
<em />
They meet when their positions are equal:
(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>
(100 m/s) <em>t</em> = 100 m
<em>t</em> = (100 m) / (100 m/s) = 1 s
so the cars meet 1 second after they start moving.
They are 100 m apart when the difference in their positions is equal to 100 m:
(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m
(subtract car B's position from car A's position because we take car A's direction to be positive)
(100 m/s) <em>t</em> = 200 m
<em>t</em> = (200 m) / (100 m/s) = 2 s
so the cars are 100 m apart after 2 seconds.
