The answer is B. On a sunny day, the air over a lake will be cooler than the air over the bordering land.
<span>this is a dynamometer</span>
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency, 
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f = 
f = 
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:



Answer:
a). Maximum Length L=0.929m
b). T=0.83 Hz or 1.2s
c). Longer, the effortless waling T=2.1 Hz or t=0.475s
d). t=1.2s V=0.774 
t=0.475s V=1.95 
Explanation:
Length legs=L=1.1m
angle=50
the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image
Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length
a).


Length of the step
L=0.464m*2
L=0.928m
b).
period=T

c).

The period is the inverse of the time of the motion so, the T1 is faster that the T because

d).
The speed is the relation between the distance with time so:
