Answer:
97.03%
Explanation:
The equation for volumetric expansion due to thermal expansion is as follows
V/Vo=(1+γΔT)
V=final volume
Vo=initial volume
γ=coefficient of volume expansion=3.2 × 10–5 K–1
ΔT=
temperature difference
assuming that the earth is a sphere the volume is given by
V=(4/3)pi R^3
if we find the relationship between the initial and final volume we have the following
taking into account the previous equation
r/ro=(1+γΔT)^(1/3)
r/r0=(1-3.2x10-5(3000-300))^(1/3)=
r/ro=0.9703=97.03%
Noaa. The buoys transmit some of the data on a daily basis to NOAA through a satellite in space.
To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.
Our values are given as
The angular velocity of a body can be described as a function of frequency as
PART A) The expression for the maximum angular velocity is given by the amplitude so that
PART B) The maximum acceleration on your part would be given by the expression
Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;
Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.
f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;
Here, p is the power of the lens.
Put f= 35.71 cm.
p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
