I have attached the solution as image.
Given:
m = 0.32 kg
v = 11.5 m/s
To find:
1) Kinetic energy = ?
2) Work needed to stop the ball = ?
Formula used:
1) Kinetic energy = ![\frac{1}{2} m v^{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E%7B2%7D%20%20%20)
2) Work = kinetic energy
Solution:
1)
Kinetic energy is given by,
Kinetic energy = ![\frac{1}{2} m v^{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E%7B2%7D%20%20%20)
K.E. = ![\frac{1}{2} 0.32 \times 11.5 \times 11.5](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%200.32%20%5Ctimes%2011.5%20%5Ctimes%2011.5%20)
K.E. = 21.16 Joule
Thus, option (d) is the correct answer.
2)
Work needed to stop the ball is same as the kinetic energy because energy is the capacity to do work. Since, work is opposing the movement of the ball. Thus, Work = - Kinetic energy
Work = -21.16 Joule
Масса (килограмм), длина (метр) и время (секунда) - это 3 из 7 основных единиц в системе СИ.
Базовые единицы произвольно определены как то, что они есть, поэтому они не могут быть получены.
Все другие единицы в системе СИ являются производными от 7 основных единиц, таких как скорость (метр / секунда).
Остальные 4 - это температура (Кельвин), количество вещества (моль), электрический ток (ампер) и светимость (кандела).
Answer:
A. 2.36 Newtons
Explanation:
F = GmM/d²
F = 6.673 x 10⁻¹¹(1)(5.98 x 10²⁴) / (1.3 x 10⁷)²
F = 2.36121...
Very poor question design.
mass of box... 1 significant digit
distance... 2 significant digits
mass of earth... 3 significant digits
value of G... 4 significant digits
Answer precision to 3 significant digits is not justifiable
Answer:
μ = 0.604
Explanation:
For the cat to stay in place on the merry go round, the maximum static frictional force must be equal in magnitude to that of the centripetal force.
Now, Centripetal force is given as;
Fc = mv²/r
Where r is radius and v is tangential speed and m is mass.
We also know that maximum static frictional force is given by;
F_static = μmg
Where μ is coefficient of friction
Now, equating both forces, we have;
mv²/r = μmg
Divide through by m;
v²/r = μg
Now, tangential speed can be expressed as;
v = circumference/period
Thus, v = 2πr/T
Where T is period of rotation and
2πr is the circumference of the merry go round.
Thus,
v²/r = μg is now;
(2πr/T)²/r = μg
Making μ the subject, we have;
(2πr/T)²/rg = μ
μ = [(2π x 5.4)/6]²/(5.4 x 9.8)
μ = 0.604