Answer:
The inverse of f equals the inverse of d Subscript o Baseline plus the inverse of d Subscript I Baseline.
Explanation:
The lens equation shows the relation among focal length of the lens, image distance and object distance. It can be expressed as:
=
+ 
where: f is the focal length of the lens,
is the object distance to the lens and
is the image distance to the lens.
The lens equation can be used to determine the unknown value among the variables f ,
and
.
To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

Through the aforementioned formula we will have to

The particulate part of the rest, so the final speed would be



Now from Newton's second law we know that

Here,
m = mass
a = acceleration, which can also be written as a function of velocity and time, then

Replacing we have that,


Therefore the force that the water exert on the man is 1386.62
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
Answer:
C. 12m
Explanation:

from the graph v = 4m/s and t = 3 s
d = vt = 4 × 3 = 12 m
Answer:
D
Explanation:
The student had displaced their in the class when she left. The phone is what's displaced and student leaving equals distance.