Answer:
I know the first one is C.) 4J. I don't know of the answer for the second oneis suppose to be in N/m form? but I got
2,500N/m
Answer:
Explanation:
The wheel and falling student will have common acceleration .
For rotational motion of wheel
Tx r = I α , T is tension in the crank , α is angular acceleration of wheel , I is moment of inertia , r is radius of the wheel.
= I a / r
T = I a / r²
For motion of student
Mg - T = Ma , M is mass of the wheel.
Mg - I a / r² = Ma
Mg = Ma +I a / r²
Mg = (M +I / r²)a
a = Mg / (M +I / r²)
= 51 x 9.8 / ( 51 + 9.6 / .3² )
499.8 / (51+ 106.67 )
= 499.8 / 157.67
= 3.17 m / s².
If time t is taken to fall by 12 m
12 = 1/2 a t²
24 / a = t²
24 / 3.17 =t²
t²= 7.57
t = 2.75 s
velocity to reach sidewalk
v = u + at
= 3.17 x 2.75
= 8.72 m / s
Answer:
3 cm
Explanation:
I think the question have some error because if height = 2 m = 200cm.
Volume of box = l * b * h
V = 30 cm³
b = 5 cm
h = 2 cm
l = ?
V = l * b * h
30 cm³ = l * 5 * 2
l = 30 / 10
l = 3 cm
