Answer:
1. The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.
2. The person's image is 3.38 m tall.
Explanation:
From the given question, object distance, u = 0.75 m, object height = 1.8 m, radius of curvature of the reflecting globe, r = 8 cm = 0.08 m.
f = = = 0.04 m
1. The image distance, v, can be determined by applying mirror formula:
= +
= +
- =
=
= -
⇒ v = -
= - 1.41 m
The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.
2. =
=
v =
= 3.384
v = 3.38 m
The person's image is 3.38 m tall.
Answer:
The correct answer will be "1/16".
Explanation:
The force or distance between two substances in charge seems to be inversely proportional to the square of the size of the particles.
The exerted force between "S" and "q" will be:
⇒ ...(equation 1)
The exerted force between "S" and "p" will be:
⇒ ...(equation 2)
Now,
On the comparison of "equation 1" and "equation 2", we get
⇒
Answer:
a.3.84m
b.-106.67m/s
c.947.3m/s^2
d.70.17 rad
e.2.5Hz
d.0.4secs
Explanation:
Given x=(7.8)cos[5πrad/s)t+π/3)]
a.Displacement at t=4.4
7.8cos(5π*4.4+π\3)=3.84m
b.velocity
V= dx/dr=-5π(7.8)sin(5πrad/s)t+π\3
at t=4.4
-5π(7.8)sin(5π*4.4+π\3)=-106.67m/s
c.acceleration
a=d^2x/dr^2
-(5π)^2(7.8) cos (5π*t+π\3)
at t=4.4
-(5π)^2(7.8)cos(5π*4.4+π\3)=-947.3m/s^2
d. Phase =(5πrad/s)t+π\3
At t=4.4
5π×4.4+π\3=70.17 rad
e.frequency
Given x= 7.8cos(5πt+π\3
Compare with x=Acos(2πft)
2πft=5πt
F=2.5Hz
f.T=1\f
T=1/2.5=0.4sec