The figure above shows part of a system consisting of a block at the top of an inclined plane that rests on a table, which is lo
cated on Earth. The block and plane are at rest when the block is released. In trial 1 there is no friction between the block and the plane or between the plane and the table. In trial 2 the plane is fixed to the table so it cannot move, but there is still no friction between the block and the plane. Indicate whether the speed of the block relative to the table when the block reaches the bottom
1 answer:
Answer:
Speed is greater in trial 2
Explanation:
"<em>Indicate whether the speed of the block relative to the table when the block reaches the bottom of the plane is greater in trial 1 or trial 2.</em>"
The initial potential energy of the block is converted into kinetic energy, with no work done by friction.
In Trial 1, the ramp is free to move, so both the block and the ramp have kinetic energy.
PE = KE + KE
mgh = ½ mv₁² + ½ Mv²
In Trial 2, the ramp is fixed, so only the block has kinetic energy.
PE = KE
mgh = ½ mv₂²
Setting the expressions equal:
½ mv₁² + ½ Mv² = ½ mv₂²
Therefore, v₂ must be greater than v₁.
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Answer:
0.37 cm
Explanation:
The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.
The object is at a distant of 265 cm to the lens of the eye.
From lens formula,
=
+
where: f is the focal length, u is the object distance and v is the image distance.
Thus, u = 265.00 cm and v = 2.70 cm.
=
+
=
=
⇒ f =
= 0.37
The focal length of the eye is 0.37 cm.