Answer:
2210.91 N
Explanation:
f = v/∧ = 1/2 √ T/ μ
where f= 28.1 Hz , T= tension ,
L= 2m
mass density = μ = 350÷1000/2.00
= 0.175kg/m
from f = 1/2L √ T/ μ
make T the subject of the formula
T= (f×2L)² ₓ μ
T= (28.1×2×2)² ×0.175
T=12633.76×0.175
=2210.91 N
Answer:
0.32 m.
Explanation:
To solve this problem, we must recognise that:
1. At the maximum height, the velocity of the ball is zero.
2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.
With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:
Mass (m) = constant
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) = 2.5 m/s
Height (h) =?
PE = KE
Recall:
PE = mgh
KE = ½mv²
Thus,
PE = KE
mgh = ½mv²
Cancel m from both side
gh = ½v²
9.8 × h = ½ × 2.5²
9.8 × h = ½ × 6.25
9.8 × h = 3.125
Divide both side by 9.8
h = 3.125 / 9.8
h = 0.32 m
Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.
Answer: Object B
The velocity of object A is depicted in the graph as a straight line (constant speed therefore no acceleration).
The graph indicates that the velocity of object B increases (the object is accelerating).
Decrease. Gravity is stronger here on earth than on the moon.
