I’m prettier sure the answer is b cuz it’s not the core ok that
To solve the problem it is necessary to take into account the concepts of the kinetic equations for the description of the torque at the rate of force and distance.
By definition the torque is given by,
where,
For the problem in question the mass of the trophy is 1.64Kg and the distance of the tropeo to the board (the shoulder) is 0.655m
PART A) For part A, the torque with the given mass and the stipulated torque in the horizontal plane must be calculated as well,
For Newton's second law
PART B) For part B there is an angle of 26 degrees with respect to the horizontal, therefore to know the net torque it is necessary to know the horizontal component to the formed angle, that is,
Answer:
The activity of cobalt-60 after 72 hours is 34.895 MBq
Explanation:
A(t) = Ao(0.5)^t/t1/2
A(t) is the activity of cobalt-60 after time t
Ao is the initial activity of cobalt-60 = 35 MBq
t is time taken to reduce in activity = 72 hours = 72/24 = 3 days
t1/2 is the half-life = 0.693 ÷ decay constant = 0.693 ÷ 0.001/day = 693 days
A(72) = 35(0.5)^3/693 = 35 × 0.5^0.00433 = 35 × 0.997 = 34.895 MBq
Explanation:
Newton has given three laws. The laws of motion are as follows :
- Newton's first law
- Newton's second law
- Newton's third law
The first law of Newton states that an object at rest will remain at rest and an object at motion will remain in motion until and unless an unbalanced force acts on it. Hence, this is the required solution.