Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N
Answer:
4.9 x 10^-19 J, 2.7 x 10^-19 J
Explanation:
first wavelength, λ1 = 410 nm = 410 x 10^-9 m
Second wavelength, λ2 = 750 nm = 750 x 10^-9 m
The relation between the energy and the wavelength is given by
E = h c / λ
Where, h is the Plank's constant and c be the velocity of light.
h = 6.63 x 10^-34 Js
c = 3 x 10^8 m/s
So, energy correspond to first wavelength
E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J
E1 = 4.9 x 10^-19 J
So, energy correspond to second wavelength
E2 = (6.63 x 10^-34 x 3 x 10^8) / (750 x 10^-9) = 2.652 x 10^-19 J
E2 = 2.7 x 10^-19 J
The force depends on the mass of both objects and the distance between them
F = G*m1*m2/r^2
So the force has a linear connection with the mass of both objects and a quadratic connection with the distance between the center of masses
Answer:
h = 10 cm
Explanation:
The magnification produced by a mirror is given by :

h' is the image size and h is the object size
We have, h = 10 cm, m = +1
So,

So, the size off the object is same as that of the size of the image i.e. 10 cm.