A candle is 49 cm in front of a convex spherical mirror that has a focal length
1 answer:
The convex spherical mirror is the diverging mirror. The image distance is 20.4 cm. The magnification is 0.417. The image is virtual and upright.
<h3>What is magnification?</h3>Magnification is the ratio of image distance to the object distance.
Object distance is 49 cm and the focal length of convex mirror is 35 cm, then image distance is calculated by the lens maker formula.
1/f = 1/v +1/u
1/-35 = 1`/v +1/49
v= -20.4 cm
Thus, the image distance is 20.4 cm.
The magnification is given by
m = v/u
m = -20.4 / 49
m = 0.417
Thus, the magnification is 0.417.
The image formed is by virtual meeting of light rays and above the principle axis. Thus the image is virtual and upright.
Learn more about magnification.
#SPJ1
You might be interested in
Explanation:
(1). Formula to calculate the potential difference is as follows.
=
=
=
=
= 38.7 volts
Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.
(2). Now, formula to calculate the energy stored in the capacitor is as follows.
E =
=
=
Thus, the electric-field energy stored in the capacitor is .
The time when the two players will collide is 0.96 s.
The equation of motion of the two players is given as;
x1 = 0.1 m + (–3.9 m/s )t
x2 = –6.3 m + (2.8 m/s )t
The time when the two players collide, their displacement is equal or the difference in their position will be zero.
Thus, the time when the two players will collide is 0.96 s.
Learn more here: brainly.com/question/18033352
Answer:
the penny loses contact at the piston's highest point.
f = 2.5 Hz
Explanation:
Concepts and Principles
1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by
∑F = ma (1)
2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:
a_max = -w^2A (2)
3- The angular frequency w of a wave is related to the frequency f by:
w = 2π f (3)
Given Data
- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.
- The frequency of oscillation of the piston is steadily increased.
Required Data
<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston. </em>
<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle. </em>
Solution
(a)
The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.
figure 1 is attached
Apply Newton's second law from Equation (1) in the vertical direction to the penny:
∑F_y -mg= ma
Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So
0 = m(g + a)
a = —g
Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.
(b)
The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):
a = —w^2A
where a = —g at the highest point. So
g = w^2A
Solve for w:
w =√g/A
Substitute for w from Equation (3):
2πf = √g/A
Solve for f :
f = 1/2π√g/A
Substitute numerical values:
f = 1/2π√9.8 m/s^2/0.04
f = 2.5 Hz
