Answer:
v = 15 m / s
Explanation:
In this exercise we are given the position function
x = 5 t²
and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval
let's look for the displacements
t = 0 x₀ = 0 m
t = 3 
= 5 3 2
x_{f} = 45 m
we substitute
v = 15 m / s
Answer:
D
The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Explanation:
Since block moves with constant speed
So, frictional force
f = FCosq
Work done by friction
W = - fd
W = - fd Cos q
The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.
Answer:
925.04 J/s
Explanation:
T = 80 C = 80 + 273 = 353 K
To = 20 c = 20 + 273 = 293 K
A = 2 m^2
Use the formula for Stefan's law
Energy radiated per second
E = 925.04 J/s
Answer:
C. 0 J
Explanation:
When an object moves in a circle, friction provides the centripetal force. Hence, it is acting towards the centre of the circle. The displacement of the object is tangential to the circle at the point of interest. Hence, the angle between the frictional force and the displacement is 90°.
Work done is given by
<em>W</em> = <em>Fd</em> cos <em>θ</em>
where <em>F</em> is the force, <em>d</em> is the displacement and <em>θ</em> is the angle between them.
<em>W</em> = <em>Fd</em> cos 90° = 0 J
Hence, the work done by friction is 0 J.
