A 1460 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at

Solution given:

North car

mass[m1]=1460kg

velocity[u1]=27 m/s

mass[m2]=2165kg

velocity [u2]=18m/s

let v be velocity after collision

we have

From the principle of conservation of linear momentum

m1u1+m2u2=(m1+m2)v

1460*27+2165*18=(1460+2165)v

v=

v=21.6m/s

the speed after the collision in 21.6 m/s.

For angle.

Tan angle =

Tan angle =

Tan angle=327.74

angle=Tan-¹(327.74)=89.82=90°

in degrees north of east is 90°

The position of the object  placed so that the image formed is a real image of the same size as the object is at a distance of twice the focal length

Explanation:

In this case, if the object is to be placed in such a way that the image thus formed by the object is real image of the same size so, the position of the object is at a distance  twice  of the focal length .

So the object is  said to be in position at a distance of two times of its focal length  i.e., it can be represented as 2

Explanation: