The first one is C. Hope this helps!! If you need anymore help just message me and I will try to get back to you quickly and help in any way i can!
S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)
S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m
Distance double 720m*2=1440m
V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
Answer:
Explanation:
Total length of the wire is 29 m.
Let the length of one piece is d and of another piece is 29 - d.
Let d is used to make a square.
And 29 - d is used to make an equilateral triangle.
(a)
Area of square = d²
Area of equilateral triangle = √3(29 - d)²/4
Total area,

Differentiate both sides with respect to d.

For maxima and minima, dA/dt = 0
d = 8.76 m
Differentiate again we get the

(a) So, the area is maximum when the side of square is 29 m
(b) so, the area is minimum when the side of square is 8.76 m
Answer:
Explanation:
solution is in the attachment below
Answer:

Explanation:
We have,
The surface temperature of the star is 60,000 K
It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

Here,
= wavelength

So, the wavelength of the star is
.