Since f=m(v^2/r),or fnet is equal to ma.
force = unknown
velocity=22m/s
radius=75m
f=m(v^2/r)
f=925(22^2/75)
f=5969.333N
Answer:
(a) 693.12 torr
(b) 68.5 kilopascals
(c) 0.862 atmosphere
(d) 1.306 atmospheres
(e) 36.74 psi
Explanation:
(a) 0.912 atm = 0.912 atm × 760 torr/1 atm = 693.12 torr
(b) 0.685 bar = 0.685 bar × 100 kPa/1 bar = 68.5 kPa
(c) 655 mmHg = 655 mmHg × 1 atm/760 mmHg = 0.862 atm
(d) 1.323×10^5 Pa = 1.323×10^5 Pa × 1 atm/1.01325×10^5 Pa = 1.306 atm
(e) 2.50 atm = 2.50 atm × 14.696 psi/1 atm = 36.74 psi
Answer:
The answer is below
Explanation:
Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.
The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph
vₐ = vₙ + vₐₙ
vₐ = vₐₙ
Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.
The direction of the wind θ is:
sin(θ / 2) = vₙ / 2vₐ
sin(θ / 2) = 70/ (2*135)
sin(θ / 2) = 0.2593
θ / 2 = sin⁻¹(0.2593) = 15
θ = 30⁰
2α = 180° - 30°
2α = 150°
α = 75°
a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.
<h2>
Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>
Explanation:
The half-life of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.
In this case, we are given the half life of two elements:
beryllium-13:
beryllium-15:
As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?
We can find it out by the following expression:
Where is the amount we want to find:
Finally:
Therefore:
The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.