Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin
1 answer:
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Answer:
-387883.3 m/s²
0.000286168546055 seconds
Explanation:
t = Time taken
u = Initial velocity = 370 m/s
v = Final velocity = 259 m/s
s = Displacement = 9 cm
a = Acceleration
The acceleration of the bullet is -387883.3 m/s²
The time taken is 0.000286168546055 seconds
Answer:
a)
1.35 kg
b)
2.67 ms⁻¹
Explanation:
a)
= mass of first body = 2.7 kg
= mass of second body = ?
= initial velocity of the first body before collision =
= initial velocity of the second body before collision = 0 m/s
= final velocity of the first body after collision =
using conservation of momentum equation
Using conservation of kinetic energy
b)
= mass of first body = 2.7 kg
= mass of second body = 1.35 kg
= initial velocity of the first body before collision = 4 ms⁻¹
= initial velocity of the second body before collision = 0 m/s
Speed of the center of mass of two-body system is given as
ms⁻¹
Answer: hello the complete question is attached below
answer :
r2 = 4r1
Explanation:
Electric field strength = F / q
we will assume the rod has an infinite length
For an infinitely charged rod
E ∝ 1/ r
considering two electric fields E1 and E2 at two different locations as described in the question
E1/E2 = r1/r2 ----- ( 2 )
<u>Calculate for r2 when E2 = E1/4 </u>
back to equation 2
E1 / (E1/4) = r1 / r2
∴ r2 = 4r1
