Answer:
a)
, b)
, c) 
Explanation:
a) The net flux through the cube is:


b) The flux through the right face is:


c) The flux through the left face is:


Explanation:
We start by using the conservation law of energy:

or

Simplifying the above equation, we get

We can rewrite this as

Note that the expression inside the parenthesis is simply the acceleration due to gravity
so we can write

where
is the launch velocity.
1250 J in 5 sec= 250 Joule(s) per second (1250/5 0
250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)
250 Watts output = 250/0.65 efficiency = 384 Watts input
1 Horsepower = 732 Watts
Motors 1 Horsepower and under are made in certain step sizes like
3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.
3/4 Horsepower is 549 Watts
1/2 Horsepower is 366 Watts
so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have:
<span>theta = arcsin(3/20) = approx. 8.63° </span>
<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>
<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>
<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
<span>= approx. 91 seconds</span>
Explanation:
A compass needle acts as a Magnet