For any object to move at a constant velocity, the net force acting on the object must be ZERO.
The force of gravity on the object is 14.47 N
Explanation:
The weight of an object (which is the force of gravity experienced by an object) at a certain location is given by
where
m is the mass of the object
g is the acceleration of gravity at the location of the object
IN this problem, we have:
m = 24.52 kg (mass of the object)
(acceleration of gravity on Pluto)
Substituting, we find the force of gravity on the object:
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Answer:
D.vibrations that cause changes in air pressure
Explanation:
Sound is a type of wave.
A wave is a periodic disturbance/oscillation that trasmits energy without transmitting matter. There are two different types of waves:
- Transverse waves: in a transverse wave, the direction of the oscillation is perpendicular to the direction of motion of the wave. These waves are characterized by the presence of crests (points of maximum positive displacement) and troughs (points of maximum negative displacement). Examples of transverse wave are electromagnetic waves.
- Longitudinal waves: in a longitudinal wave, the direction of the oscillation is parallel to the direction of motion of the wave. These waves are characterized by the presence of compressions (regions where the density of particle is higher) and rarefactions (regions where the density of particle is lower). Examples of longitudinal waves are sound waves.
Sound waves, in particular, consist of vibrations of the particles in a medium - most commonly, air - that occur back and forth along the direction of motion of the wave. Because of these motion, the air will have areas of higher pressure (which correspond to the compressions), where the density of particles is higher, and areas of lower pressure (which correspond to the rarefactions), where density of particles is lower.
Answer:
c: long and thin resistor.
Explanation:
The resistance of a resistor is given by:
R = ρ*L/A
where:
R = resistance
ρ = resistivity (depends on the material)
L = length of the material
A = cross-sectional area of the material
We can see that the length is on the numerator, which means that if we increase the length, then the resistance is increased.
We also can see that the cross-sectional area is on the denominator, then if we increase the area (for example, with a ticker resistor) the resistance decreases.
Then if we want to maximize the resistance, we need to have a long and thin resistor, so the correct answer is c.
