A bullet hits a wall with a velocity of 20m/s and penetrates it up to a distance of 6 cm. find the diceleration of the bullet in the wall
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velocity of bullet = 20 m/sec
bullet penetrate the 6cm thick wall and take rest after the Penetration .
e.g final velocity = 0
use kinematics equation ,
v² = u² + 2as
0 = (20)² + 2a× (6/100)
-20 × 20 = 12a/100
a = - 20 × 20 × 100/12
a = -10000/3 m/s²
a = -3333.33 m/s²
bullet penetrate the 6cm thick wall and take rest after the Penetration .
e.g final velocity = 0
use kinematics equation ,
v² = u² + 2as
0 = (20)² + 2a× (6/100)
-20 × 20 = 12a/100
a = - 20 × 20 × 100/12
a = -10000/3 m/s²
a = -3333.33 m/s²
RadhaIyer:
how s=20/100
sry now refresh it
i edit answer
ya thx....I got the same answer
Answered by
1
[tex]\text{Hey there!}
[/tex]
[tex]\math Initial velocity = 20 m/s [/tex]
Distance = 6 cm
We know,
v² = u² + 2as
Put the values:
0 = (20)² + 2a× (6/100)
-20 × 20 = 12a/100
a = - 20 × 20 × 100/12
a = -10000/3 m/s²
a = -3333.33 m/s²
[tex]\math Initial velocity = 20 m/s [/tex]
Distance = 6 cm
We know,
v² = u² + 2as
Put the values:
0 = (20)² + 2a× (6/100)
-20 × 20 = 12a/100
a = - 20 × 20 × 100/12
a = -10000/3 m/s²
a = -3333.33 m/s²
could u pls tell me aanswer of this question also..An object is dropped from rest at a height of 150 m and simultaneously
another object is dropped from rest at a height 100 m. What is the difference
in their heights after 2 s if both the objects drop with same accelerations?
How does the difference in heights vary with time?
another object is dropped from rest at a height 100 m. What is the difference
in their heights after 2 s if both the objects drop with same accelerations?
How does the difference in heights vary with time?
Please ask in brainly as another qn
I will answer!
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