Answer:
The focal length of eyepiece is 3.68 cm.
Explanation:
Given that,
Distance = 19 cm
Focal length = 5.5 mm
Magnification = 200
Object distance = -25 cm
We need to calculate the focal length
Using formula of magnification
Put the value into the formula
Hence, The focal length of eyepiece is 3.68 cm.
Correct answer choice is:
D. A continuous transmission of energy from one location to the next.
Explanation:
Waves include the carrier of energy without the carrier of matter. In outcome, a wave can be characterized as a change that progresses into a medium, carrying energy from one spot (its source) to different spot without carrying matter.
Calculate the pressure due to sea water as density*depth.
That is,
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa
Atmospheric pressure is 101.3 kPa
Total pressure is 94423 + 101.3 = 94524 kPa (approx)
The area of the window is π(0.44 m)^2 = 0.6082 m^2
The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx
Answer:
Dynamic flexibility
Explanation:
Dynamic flexibility can be generally defined as the ability of the body muscles and joints to move in full range of motion. High flexibility in these joints and muscles leads to the decreasing pain and injury in different parts of the body.
Proper warm up exercises are needed to be carried out that involves both the combination of controlling movements and stretching of the body, and this directly enhances the dynamic flexibility of the body.
The athletes and sports persons possesses a good dynamic flexibility of their body as they carry our different types of body exercises.
Answer:
50m
Explanation:
Given parameters:
Initial velocity = 20m/s
Acceleration = 4m/s²
Time = 10s
Unknown:
Distance traveled by the rocket = ?
Solution:
To solve this problem use the expression below;
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
final velocity = 0
Insert the parameters and solve;
0² = 20² + 2 x 4 x s
-400 = 8s
s = 50m
Disregard the negative sign because distance cannot be negative.
