Much energy as would Microraptor gui have to expend to fly with a speed of 10 m/s for 1.0 minutes is 486 J.
The first step is to find the energy that Microraptor must release to fly at 10 m/s for 1.0 minutes. The energy that Microraptor must expend to fly can be found using the relationship between Power and Energy.
P = E/t
Where:
P = power (W)
T = time (s)
Now, a minimum of 8.1 W is required to fly at 10 m/s. So, the energy expended in 1 minute (60 seconds) is
P = E/t
E = P x t
E = 8.1 x 60
E = 486 Joules
Thus, the energy that Microraptor must expend to fly at 10 m/s for 1.0 minutes is the 486 J.
Learn more about Microraptor gui here brainly.com/question/1200755
#SPJ4
The work done on the box by the applied force is zero.
The work done by the force of gravity is 75.95 J
The work done on the box by the normal force is 75.95 J.
<h3>The given parameters:</h3>
- Mass of the box, m = 3.1 kg
- Distance moved by the box, d = 2.5 m
- Coefficient of friction, = 0.35
- Inclination of the force, θ = 30⁰
<h3>What is work - done?</h3>
- Work is said to be done when the applied force moves an object to a certain distance
The work done on the box by the applied force is calculated as;

where;
a is the acceleration of the box
The acceleration of the box is zero since the box moved at a constant speed.

The work done by the force of gravity is calculated as follows;

The work done on the box by the normal force is calculated as follows;

Learn more about work done here: brainly.com/question/8119756
pitch goes up on approach ... Doppler effect
my hypothesis is that If you drop a piece of buttered toast, it will land butter side down.
I tested it by dropping 10 pieces of buttered toast off the table and noted on which side it landed
It could be falsified cause I just made all of this up. In essence, it's like flipping a coin, 50/50 chance so I could say that 5 landed butter up and 5 landed butter down.
Solution :
a). Using Gauss's law :
,
.........(1)
Let
in equation (1)
Therefore,
.............(2)



....................(3)
Therefore, 

.............(4)
Now differentiating the equation (4) w.r.t. 'b', we get
Thus the radius for the inner cylinder conductor is 
b). For the energy storage, substitute the radius in (4), we get

This is the amount of energy stored in the conductor.