A bullet is fired horizontally, and at the same instant a second bullet is dropped from the same height. Ignore air resistance.
1 answer:
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Answer:
The displacement of the spring due to weight is 0.043 m
Explanation:
Given :
Mass Kg
Spring constant
According to the hooke's law,
Where force,
displacement
Here,
(
)
N
Now for finding displacement,
Here minus sign only represent the direction so we take magnitude of it.
m
Therefore, the displacement of the spring due to weight is 0.043 m
<h2>5.3 km</h2>
Explanation:
This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.
Let us denote each of the individual displacements by a vector. Consider the unit vectors as the unit vectors in the direction of East and North respectively.
By simple calculations, we can derive the unit vectors in the directions North,
South of West and
North of West respectively.
So Total displacement vector = Sum of individual displacement vectors.
Displacement vector =
Magnitude of Displacement =
∴ Total displacement =
Answer:
v = √2G / R
Explanation:
For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)
Eo = K + U = ½ m1 v² - G m1 m2 / r1
Ef = - G m1 m2 / r2
When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf
Eo = Ef
½ m1v² - G m1 / R = - G m1
/ R
v² = 2G (1 / R - 1 / Rinf)
If we do Rinf = infinity 1 / Rinf = 0
v = √2G / R
Ef = = - G m1 m2 / R
The mechanical energy is conserved
Em = -G m1 / R
Em = - G m1 / R
R = int ⇒ Em = 0