Answer:
The solutions are attached below:
Explanation:
Answer:
C. The buoyant forces are equal on the objects since they have equal mass.
Explanation:
Correct option (C) The buoyant forces are equal on the objects since they have equal mass. For floating objects, the buoyant force equals the weight of the objects. Since each object has the same weight, they must have the same buoyant force to counteract that weight and make them float.
Answer:
k = 0.1118 per min
Explanation:
Assume;
Initial number of bacteria = N0
Number of bacteria IN 'T' time = Nt
So,
![Nt=N0e^{-kt}\\\\in\ 6.2 min\\\\\\frac{N0}{2}= N0e^{-k(6.2)}\\\\ln\frac{1}{2} = -k[6.2]](https://tex.z-dn.net/?f=Nt%3DN0e%5E%7B-kt%7D%5C%5C%5C%5Cin%5C%206.2%20min%5C%5C%5C%5C%5C%5Cfrac%7BN0%7D%7B2%7D%3D%20N0e%5E%7B-k%286.2%29%7D%5C%5C%5C%5Cln%5Cfrac%7B1%7D%7B2%7D%20%3D%20-k%5B6.2%5D)
k = 0.1118 per min
Answer:
b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.
Explanation:
From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.
The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.
However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.
The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.
It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.
As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.
Answer:
b)1.08 N
Explanation:
Given that
velocity of air V= 45 m/s
Diameter of pipe = 2 cm
Force exerted by fluid F
So force exerted in x-direction
F=0.763 N
So force exerted in y-direction
F=0.763 N
So the resultant force R
R=1.079
So the force required to hold the pipe is 1.08 N.
