I like your profile picture:)
Answer:
350 F to 100 F it take approx 87.33 min
Explanation:
given data
oven = 350◦F
cooling rack = 70◦F
time = 30 min
cake = 200◦F
solution
we apply here Newtons law of cooling
= -k(T-Ta)
= 
(T(t) -Ta)
= 
= -k(T-Ta)
-ky = -ky
T(t) -Ta = (To -Ta) 
T(t) = Ta+ (To -Ta)
put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F
so here
200 = 70 + ( 350 - 70 ) 
k = 0.025575
so here for T(t) = 100F
100 = 70 + ( 350 - 70 ) 
time = 87.33 min
so here 350 F to 100 F it take approx 87.33 min
Answer:
So the minimum force is
32.2Newton
Explanation:
To solve for the minimum force, let us assume it to be F (N)
So
F=mgsinA
But
=>>>> coefficient of static friction x (F + mgcosA
=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)
So making F subject of formula
F + 24.0 = 56.2
F = 32.2N
Answer:
I should be active for 15 hours to meet the physical activity requirement.
Explanation:
Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.
So, t = t₀/√(1 - β²)
t = 9/√(1 - (v/c)²)
= 9/√(1 - (0.8c/c)²)
= 9/√(1 - (0.8)²)
= 9/√(1 - (0.64)
= 9/√0.36
= 9/0.6
= 15 hr
So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.
