Answer:
The velocity of Mosquito with respect to earth will be 0.302m/s
Explanation:
V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air
V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.
Velocity of Mosquito with respect to earth will be
V(me) = V(ma) + V(ae)
We need to find the mosquito’s speed with respect to Earth in the x direction.
V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )
Angle (ae) = –90.0° − 35°=−125°
V(x, me) = 1.10 + (1.4)Cos(-125)
= 1.10 + 1.4(-0.57)
= 1.10 -0.798
= 0.302
So the velocity of Mosquito with respect to earth will be 0.302m/s
the second level will have 8 electrons.
Answer:
Part a)
F = 0.051 N
Part b)
Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.
Explanation:
Part a)
Electrostatic force between two charged spherical balls is given as

here we will have

here the distance between the center of two balls is given as

now we will have


Part b)
Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.
Explanation:
Given that,
Focal length of a converging lens, f = +25 cm
Size of the object, h = 10 cm
Object distance, u = -30 cm
We need to find the image distance and the size of the image.
Using lens formula,
, v is image distance

Magnification,

h' is the size of image

So, the image distance is 150 cm the size of image is 50 cm.