Which water depth had the biggest difference in survival rates for embryos with UV-B protection versus embryos without UV-B prot
ection? Which water depth had the biggest difference in survival rates for embryos with UV-B protection versus embryos without UV-B protection? 100 cm 0 cm 10 cm 50 cm
1 answer:
Answer:
10 cm
Explanation:
water depth had the biggest difference in survival rates for embryos with UV-B protection versus embryos without UV-B protection is 10 cm . the bar graph attached in the shows that the lengths of yellow and brown bars differ the most at 10 cm.
Hence,The biggest difference between the survival rates of UV-B protected and unprotected can be seen at the depth of 10 cm.
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Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
Answer:
4 Ohms
Explanation
(This is seriously not as hard as it looks :)
You only need two types of calculations:
- replace two resistances, say, R1 and R2, connected in a series by a single one R. In this case the new R is a sum of the two:
- replace two resistances that are connected in parallel. In that case:
I am attaching a drawing showing the process of stepwise replacement of two resistances at a time (am using rectangles to represent a resistance). The left-most image shows the starting point, just a little bit "warped" to see it better. The two resistances (6 Ohm next to each other) are in parallel and are replaced by a single resistance (3 Ohm, see formula above) in the top middle image. Next, the two resistances (9 and 3 Ohm) are nicely in series, so they can be replaced by their sum, which is what happened going to the top right image. Finally we have two resistances in parallel and they can be replaced by a single, final, resistance as shown in the bottom right image. That (4 Ohms) is the <em>equivalent resistance</em> of the original circuit.
Using these two transformations you will be able to solve step by step any problem like this, no matter how complex.
