Answer:
a. Speed = 1.6 m/s
b. Amplitude = 0.3 m
c. Speed = 1.6 m/s
Amplitude = 0.15 m
Explanation:
a.
The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:
frequency =
= 0.25 Hz
The wavelength of the wave is the distance between consecutive crests of wave. Therefore,
Wavelength = 6.4 m
Now, the speed of the wave is given as:
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
<u>Speed = 1.6 m/s</u>
<u></u>
b.
Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:
Amplitude = (0.5)(0.6 m)
<u>Amplitude = 0.3 m</u>
<u></u>
c.
frequency =
= 0.25 Hz
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
<u>Speed = 1.6 m/s</u>
<u></u>
Amplitude = (0.5)(0.3 m)
<u>Amplitude = 0.15 m</u>
To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.
By definition the exchange of heat is given by

where,
m = mass
c = specific heat
= Change in temperature
Therefore the total heat exchange is given as


Our values are given as,
Total mass is
= 200lb ,however the mass of solid vegetable and water is given as,



Replacing at our equation we have,



Therefore the heat removed is 22411.2 Btu
Answer:
1.11 dioptre
Explanation:
= Distance of the image = - (125 - 2) = - 123 cm
= Distance of the object = 54 - 2 = 52 cm
= Focal length of the lens
Using the equation


cm
Power of the lens is given as


Dioptre
Answer:
c) At a distance greater than r
Explanation:
If G= Gravitational constant
M= Mass of earth
r= distance from earth center
then orbital speed is ;
v = 
==> v²=GM/r
If speed of first satellite = V₁
==> V₁² = GM/r
==> r = GM/V₁²
If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.
So our answer is c