A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25° with the horizontal. What is the forward speed of the disk after it has rolled 3.0 m, measured along the plane?

The forward speed of the disk is 4.07 m/s

Explanation:

Step 1: Data given

Angle = 25 °C

Distance the disk has rolled = 3.0 m

Step 2= Calculate the initial energy

Since the disk is, at moment t=0, at rest and at a height we only have potential energy

Ei = m*g*h

⇒with E = the potential energy

⇒with m = the mass of the disk

⇒with g = 9.81 m/s²

⇒with h = the height = x*sin 25° = 3*sin25°

Ei = m*g*X*sin25

Step 3: Calculate the final energy

Ef = 1/2 (mv²) + 1/2 Iw²

Ef = 1/2 (mv²) + 1/2(mR²)(v/R)²

Ef = 1/2mv² + 1/4 mv²= 3/4 mv²

Step 4: The law of conservation of energy

Ei = Ef

m*g*X*sin25 = 3/4 mv²

g*X*sin25 = 3/4 v²

Step 5: Calculate v

v = √(4/3 * g*X sin25°)

v = √(4/3 * 9.81*3 sin25°)

v = 4.07 m/s

The forward speed of the disk is 4.07 m/s

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2