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
A fatter handle. This allows you to provide more torque to the screw, which should eventually loosen it.
Answer:
Distance covered is: 45 meters
Displacement is 15 meters to the right of where he started
Explanation:
Notice that Brady has walk a path that looks like an incomplete rectangle of height 5 meters and length 25meters, although he actually didn't cover the full length (25 meters) when getting back to the point where he started (he made just 10 meters instead of 25 after the third turn right) See attached image.
Therefore, Brady's displacement is 15 meters to the right of where he started, and the total distance he covered is :
Distance = 5m + 25m + 5m + 10m = 45m
Answer:
b) Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
Explanation:
In the given problem, if there exists a gain-of-function mutation for the given cell, there would not be the formation of cyclin E when there is the possibility of cells movement via the checkpoint of the G1/S, even when there are non-deal conditions for the division of cell. Thus, the correct option in the lists of options is the option b.
On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C
Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)
on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.
Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:
C, B, A and then D
the same idea as on Q30 applies to Q31 part b,
D,C,B then A
