Answer:
Force = 10.244 Newtons
b) No of oscillations = 0.88
Explanation:
Since the block executes SHM we can write it's position as function of time as
ω is the natural frequency of the system
A is the amplitude of the system
Thus accleration of the block
Thus using the given values at t= 3.50 sec we can calculate the acceleration as
thus force can be calculated using newtons second law as
b)
Now no of oscillations can be obtained as
no of oscillations in 3.50 seconds = 3.50/3.976 = 0.88
Marbles would have the greater density. Marbles would have the greater density because the molecules inside are more tightly packed. Even though sand has the greater mass it does not have the greater density. The sand doesn't have the greater density because it could be crushed in an instant, showing that if molecules are packed together tightly they will be tougher to break. The dice, I believe, still have a greater density than the sand.
it has a high density. since density is mass divided by volume, if you are dividing a large mass by a small volume it will have a high density
Potential and kinetic energy are at play when we talk about Newton's second law of motion through the various positions in relation to the bodies involved.
<h3>What is Newton's second law of motion?</h3>
This law states that force is equal to the rate of change of momentum and is denoted as F = mv where m is mass and v is velocity.
Potential energy is the energy is possessed by a body by virtue of its position while kinetic energy is possessed by a body by virtue of its motion. Both forms of energy are influenced by forces and are equal to the total momentum.
Read more about Newton's second law of motion here brainly.com/question/2009830
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Answer:
The distance on the screen from the center of the central bright fringe to the third dark fringe is 0.831 m.
Explanation:
Given that,
Wavelength = 617 nm
Width of slit
Distance between the slit and screen L= 2.83 m
Third dark fringe m = 3
We need to calculate the distance on the screen from the center of the central bright fringe to the third dark fringe on either side
Using formula of distance
Put the value into the formula
Hence, The distance on the screen from the center of the central bright fringe to the third dark fringe is 0.831 m.