Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
.................2
put here value and we get
<h3>v = 0.155 m </h3>
so
Magnification will be here as
m =
m =
<h3>m = 1</h3>
Answer:
Explanation:
Here image distance is fixed .
In the first case if v be image distance
1 / v - 1 / -25 = 1 / .05
1 / v = 1 / .05 - 1 / 25
= 20 - .04 = 19.96
v = .0501 m = 5.01 cm
In the second case
u = 4 ,
1 / v - 1 / - 4 = 1 / .05
1 / v = 20 - 1 / 4 = 19.75
v = .0506 = 5.06 cm
So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )
Answer:
The magnitude of the frictional force is
Explanation:
From the question we are told that
The force exerted on the box is
Generally for the box to remain at rest then it means that the frictional force is greater than or equal to the force applied to move it i.e
=>
Answer:
B is the answer for the question
Answer:
The net force acting on the tennis ball while it is in contact with the racquet is 50.73 N
Explanation:
The impulse-momentum theorem said that the net impulse is equal to the change of the momentum, this is:
but the net impulse is too the net force times the change in time:
so using (2) on (1) we have:
Decomposing that on x and y components:
(See figure below) with Pfx = m*vfx= m*vf*cos(15°)=(0.058kg)(40m/s)cos(15°),
Pox = -m*vox= m*vo*cos(15°)=-(0.058kg)(30m/s)cos(15°), the same analysis to Pfy and Poy gives
Pfy=(0.058kg)(40m/s)sin(15°), Poy=-(0.058kg)(30m/s)sin(15°), using those values on (3) and (4) and solving for Fy and Fx:
So the net force acting on the tennis ball while it is in contact with the racquet is: