Answer:
a. by moving the book without acceleration and keeping the height of the book constant
Explanation:
FOR CONSTANT KINETIC ENERGY:
The kinetic energy of a body depends upon its speed according to its formula:
ΔK.E = (1/2)mΔv²
So, for Δv = 0 m/s
ΔK.E = 0 J
So, for keeping kinetic energy constant, the books must be moved at constant speed without acceleration.
FOR CONSTANT POTENTIAL ENERGY:
The potential energy of a body depends upon its height according to its formula:
ΔP.E = mgΔh
So, for Δh = 0 m/s
ΔP.E = 0 J
So, for keeping potential energy constant, the books must be moved at constant height.
So, the correct option is:
<u>a. by moving the book without acceleration and keeping the height of the book constant</u>
<span>A van is traveling on a road at a speed of 55 km/h relative to a
stationary observer on the side of the road. A girl sitting near the
driver of the van throws a paper airplane to a boy at the back of the
van with a speed of 2 km/h relative to the girl, the boy, and the van.
The speed of the paper airplane, relative to the same stationary observer
on the side of the road, is (55 - 2) = 53 km/h. No rounding is necessary.</span>
The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
</span><span>
Keisha follows the instructions for a demonstration on gas laws.
1. Place a small marshmallow in a large plastic syringe.
2. Cap the syringe tightly.
3. Pull the plunger back to double the volume of gas in the syringe.
Now, this activity is being done at the same temperature, because there is no mention of the temperature change. Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
Explanation:
Given that,
Wavelength of the light, 
(a) Slit width, 
The angle that locates the first dark fringe is given by :
(b) Slit width, 
The angle that locates the first dark fringe is given by :
Hence, this is the required solution.
