Question

A particle of mass m moving along a straight line experiences force f which varies with the distance travelled as shown in figureif the velocity of the particle at x is

Answer 1

Explanation:

Increase in KE = work done by all forces

Increase in KE = work done by all forcesΔKE=F

Increase in KE = work done by all forcesΔKE=F net

Increase in KE = work done by all forcesΔKE=F net ⋅ds

Increase in KE = work done by all forcesΔKE=F net ⋅ds2

Increase in KE = work done by all forcesΔKE=F net ⋅ds21

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 2

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 −

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 2

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×(

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )=

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 2

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v 2

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v 2 =

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v 2 = m

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v 2 = m11F

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v 2 = m11F 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v 2 = m11F 0 x

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v 2 = m11F 0 x 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v 2 = m11F 0 x 0

Increase in KE = work done by all forcesΔKE=F net ⋅ds21 mv 22 − 21 m×( m2F 0 x 0 )= 21 (2F 0 +F 0 )3x 0 ⇒v 2 = m11F 0 x 0

Answer 2

Answer:

atmospheric refraction