Answer:
Surface area of the dog is changes from A to 3A
Explanation:
It is given that surface area of dog is increased by factor 3 in a period of 4 year
We have to find the change in dog's relative surface area in the given time period
Let initially the surface area is A
As the surface area is increased by a factor of 3
So surface area after 4 year = 3×A = 3A
So surface area of the dog is changes from A to 3A
complete question:
An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°. What is the distance from the base of the cliff to the point on the ground? Round to the nearest foot
Answer:
a ≈ 5281 ft
Explanation:
The observer at the top of a 462 ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°.
The angle of depression form the top of the cliff = 5°
The 5° is outside the triangle formed . To find the angle in the triangle we have to subtract 5° from 90°. 90° - 5° = 85° Note sum of an angle on a right angle is 90°.
using SOHCAHTOA principle we can solve for the distance from the base of the cliff to the point on the ground(a)
tan 85° = opposite / adjacent
tan 85° = a / 462
cross multiply
462 × tan 85° = a
a = 11.4300523 × 462
a = 5280.66 ft
a ≈ 5281 ft
The answer to the question presented above would be PAR-Q or the <span>Physical Activity Readiness Questionnaire. The purpose of this questionnaire is to identify if you have any medical risks and to know your overall medical background for future reference of prevention of complications whenever you're training.</span>
Answer: 20 grams
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
![Q=m\times c\times \Delta T](https://tex.z-dn.net/?f=Q%3Dm%5Ctimes%20c%5Ctimes%20%5CDelta%20T)
Q = Heat absorbed by liquid = 5.23 kcal = 5230 cal (1kcal=1000cal)
C = heat capacity of liquid =
Initial temperature of the liquid =
= 101 K
Final temperature of the liquid =
= 225 K
Change in temperature ,![\Delta T=T_f-T_i=(225-101)K=124K](https://tex.z-dn.net/?f=%5CDelta%20T%3DT_f-T_i%3D%28225-101%29K%3D124K)
Putting in the values, we get:
![5230=m\times 2.09cal/g^0C\times 124K](https://tex.z-dn.net/?f=5230%3Dm%5Ctimes%202.09cal%2Fg%5E0C%5Ctimes%20124K)
![m=20g](https://tex.z-dn.net/?f=m%3D20g)
Thus the sample of liquid in grams is 20