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Answer:
t = 0.657 s
Explanation:
First, let's use the appropiate equations to solve this:
V = √T/u
This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.
Where:
V: Speed of the disturbance
T: Tension of the rope
u: linear density of the rope.
The density of the rope can be calculated using the following expression:
u = M/L
Where:
M: mass of the rope
L: Length of the rope.
We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:
u = 2.31 / 10.4 = 0.222 kg/m
Now, replacing in the first equation:
V = √55.7/0.222 = √250.9
V = 15.84 m/s
Finally the time can be calculated with the following expression:
V = L/t ----> t = L/V
Replacing:
t = 10.4 / 15.84
t = 0.657 s
Answer:
We conclude that the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.
Explanation:
Given
To determine
Kinetic Energy (K.E) = ?
We know that a body can possess energy due to its movement — Kinetic Energy.
Kinetic Energy (K.E) can be determined using the formula
where
- K.E is the Kinetic Energy (J)
now substituting m = 1.75, and v = 54 in the formula
J
Therefore, the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.
Answer:
0 m/s
Explanation:
Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.
In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.
As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.
Average velocity = displacement/time
v=0/30
v = 0 m/s
Hence, the correct option is (1).
Answer:
Principal quantum numbers are denoted by the symbol 'n'. They designate the principal electron shell of the atom. The positive whole number values starting from 1,2,3 and so on..
Explanation:
All orbitals that have the same value of n are said to be in the same shell (level). For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state. The total number of orbitals for a given n value is n2.
