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:
991.67 miles per day
Explanation:
Since it travels 2975 miles per day, distance traveled in a day =2975/3=991.67 miles
Answer:
m =8.81*10^{-6}grams
time t = 52.8 year
Explanation:
GIVEN DATA:
the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s
activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s
activity , 


= ( 3.7 X 10^7 )(1.663*10^8 ) / ln2
= 8.877*10^{16}
Number of moles:
n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol
mass of the CO-60 is,
m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams
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time t = -[T1/2 / ln2]*ln[N/N0]
= - [5.3 years / ln2]*ln[1x10-6/1x10-3]
= 52.8 year
Answer:
This equation is based on twin paradox - a phenomena where one of the twin travels to space at a speed close to speed of light and the other remains on earth. the twin from the space on return discovers that the one on earth age faster.
Solution:
= 10 years
v = 0.8c
c = speed of light in vacuum
The problem can be solved by time dilation equation:
(1)
where,
t = time observed from a different inertial frame
Now, using eqn (1), we get:

t = 16.67 years
The age of the twin on spaceship according to the one on earth = 25+16.67 =41.66 years