<h2>
Answer: B. Gravitational potential energy </h2>
Explanation:
<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field.
</em>
That is why this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.
In the case of the <u>Earth</u>, in which <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy
will be:
Where
is the mass of the object,
the acceleration due gravity and
the height of the object.
As we can see, the value of
is directly proportional to the height.
Answer:
Option b. is correct
Explanation:
An RLC electrical circuit consists of constituent components: a resistor (R), an inductor (L), and a capacitor (C). A resistor, an inductor, and a capacitor are connected in series or parallel.
The impedances of the circuit elements depend on the frequency.
Both impedance magnitudes decrease when the frequency increases
<span>On the y-axis (the bottom of the table) hope this helps</span>
Answer: Pieces of minerals, rocks, plant and animal remains.
Explanation: Pieces of minerals, rocks, plant and animal remains. whether or not patterns cause flow rates of rivers to vary. sand settles from faster-moving water;smaller costs of silt and clay that form up mud settle from slower moving water.
Answer:
20.0 cm
Explanation:
Here is the complete question
The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Solution
Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.
Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.
Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m
Now, P' = 1/u + 1/v
1/u = P'- 1/v
1/u = 55.0 D - 1/0.02 m
1/u = 55.0 m⁻¹ - 1/0.02 m
1/u = 55.0 m⁻¹ - 50.0 m⁻¹
1/u = 5.0 m⁻¹
u = 1/5.0 m⁻¹
u = 0.2 m
u = 20 cm
So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.