Answer:
They will interfere to create a crest with an amplitude of 2.
Explanation:
The two waves will interfere and the resultant amplitude will be given by the principle of superposition, which states that the resultant amplitude is given by the sum (or the difference, if they meet in opposite phase) of the amplitudes of the two waves. In this case, the crest of wave C meets the through of wave D, so they are in opposite phase, therefore the resultant amplitude will be
A = A(C) - A(D) = 3 - 1 = 2
The wavelength of the sound increases.
The frequency of the sound increases.
The wave speed of the sound increases.
Answer:
The sound is silenced.
Explanation:
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Answer:4500 N
Explanation:
Given
mass of truck
mass of car
as the car push the ground with a force of 4500 N so the force on the truck should also be 4500 N as they all the force is transmitted to push the truck
and their acceleration is
Based on Newton's third law of motion, if the car is pushing against the ground with a force of 4500N, that same force is what propels the car and any object attached to it, regardless of the object's mass. Therefore, the car also applies a force of 4500N to the truck it is pushing.
The magnitude of force that a 1200-kg car applies to a 2100-kg truck with a dead battery can be calculated using the principles of Newtons's third law of motion which states that for every action there is an equal and opposite reaction. The force exerted by the car on the ground is equal to the force exerted by the ground on the vehicle. If the car's driving force against the ground is 4500N, as per Newton's third law, the ground also pushes the car back with the same force of 4500N. This force is what propels the car and any object it's connected with.
So, when the car is pushing this truck, it applies this same force of 4500N to the truck. Therefore, the magnitude of force the car applies to the truck is 4500N.
#SPJ12
A. float
B. sink
C. sink, then float
An object will float when:
A. buoyant force is equal to the weight of the object
B. buoyant force is less than the weight of the object
C. density is equal to the weight of the object
D. density is greater than the weight of the object
What is the buoyancy force on a 15 g object which displaces 60 mL of water? (Remember to change mL of water to grams and grams to kg)
A. 900 N
B. 25 N
C. 0.59 N
D. 0.25 N
An object has a density of 20 g/cm3. When placed in a cylinder, it displaces 5 mL of water. What is the mass of the object?
A. 0.25 g
B. 4.0 g
C. 100.0 g
Substance has a mass of 16.2 grams. It displaces 8.1 grams of water. What is its specific gravity?
A. 2.0
B. 5.5
D. 131.22
1) A. float
There are two forces acting on an object in the water: the weight of the object (downward) and the buoyancy (upward), which is equal to the weight of displaced water. If the weight of displaced water is greater than the weight of the object, it means that there is a net force directed upward, so the object will float.
2) A. buoyant force is equal to the weight of the object
As stated in the previous question, there are only these two forces acting on an object in the water (buoyant force and weight of the object), so if the two forces are equal, then the object is in equilibrium, so it will float.
3) C. 0.59 N
The buoyancy force is given by:
where
is the density of the liquid (water)
is the volume of displaced water
is the acceleration of gravity
Substituting numbers into the formula, we find
4) C. 100.0 g
The density of the object is 20 g/cm^3, which is greater than the density of the water (1 g/cm^3): this means that the object will sink, so its volume is equal to the volume of displaced water.
Therefore, we have:
- object's density:
- object's volume:
so, the mass of the object is
5) A. 2.0 g
The specific gravity of an object is given by the ratio between its density () and the density of a reference substance (), in this case water:
whe can rewrite each density as the ratio between mass and volume:
where the suffix o refers to the object, while the suffix w refers to the water. However, if we assume that the object is completely in the water, the two volumes are equal, so we can simplify the formula: