Answer:
<u><em>Circular motion requires a net inward or "centripetal" force. Without a net centripetal force, an object cannot travel in circular motion. In fact, if the forces are balanced, then an object in motion continues in motion in a straight line at constant speed.</em></u>
Explanation:
I think it's 3. within an outer arm
Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.
Solution:
It is given that mass of object before explosion is,m = 10 g
Speed of object before explosion, v = 2 m/s
Let
be the masses of the three fragments.
Let
be the velocities of the three fragments.
Therefore, according to the law of conservation of momentum,


So the x- component of the velocity of the m2 fragment after the explosion is,

∴ 