A 15.0 cm object is 12.0 cm from a concave mirror that has a focal length of 4.8 cm. Its image is 8.0 cm in front of the mirror.

Answer:

Explanation:

Provided:

fiber diameter d= 18 μm

screen distance D= 30 cm

wavelength λ= 560 nm

from this, we can determine the fringe width

substituting the values yield

Answer:

v_y = 12.54 m/s

Explanation:

Given values:

- Initial vertical height y_o = 10 m

- Initial velocity v_y,o = 0 m/s

- The object's acceleration in the air = a_y

- The actual time taken to reach the ground t = 3.2 s

Find:

- How to calculate the object's speed when it arrives at the ground?

Solution:

- Apply kinematic equations to find the actual acceleration of the ball when it reaches the ground:

y = y_o + v_y,o*t + 0.5*a_y*t^2

0 = 10 + 0 + 0.5*a_y*(3.2)^2

a_y = - 20 / (3.2)^2 = 1.953125 m/s^2

- Use the total energy conservation principle of the system:

E_p - W_f = E_k

Where, E_p = m*g*y_o

W_f = m*a_y*(y_i - y_f)..... Reflecting air resistance

E_k = 0.5*m*v_y^2

Thus, m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2

g*(10) - (1.953125)*(10) = 0.5*v_y^2

v_y = sqrt(157.1375)

v_y = 12.54 m/s

Answer:

3.4 x 10⁴ m/s

Explanation:

Analyze the circular path of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of an electron = 9.1 x 10⁻³¹ kg

v  = speed in the radial direction

r = radius of the circular trajectory = 2 mm = 0.002 m

q = charge of an electron = 1.6 x 10⁻¹⁹ C

For the electron’s circular movement

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Now, consider the electron's movement in a straight line:

v' = speed in linear motion

x = distance traveled horizontally = 9 mm = 0.009 m

t = duration = = = 4.5 x 10⁻⁷ sec

Using the formula

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

The resultant speed is given by

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s