Answer: 996m/s
Explanation:
Formula for calculating velocity of wave in a stretched string is
V = √T/M where;
V is the velocity of wave
T is tension
M is the mass per unit length of the wire(m/L)
Since the second wire is twice as far apart as the first, it will be L2 = 2L1
Let V1 and V2 be the speed of the shorter and longer wire respectively
V1 = √T/M1... 1
V2 = √T/M2... 2
Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1
The equations will now become
249 = √T/(m/L1) ... 3
V2 = √T/(m/2L1)... 4
From 3,
249² = TL1/m...5
From 4,
V2²= 2TL1/m... 6
Dividing equation 5 by 6 we have;
249²/V2² = TL1/m×m/2TL1
{249/V2}² = 1/2
249/V2 = (1/2)²
249/V2 = 1/4
V2 = 249×4
V2 = 996m/s
Therefore the speed of the wave on the longer wire is 996m/s
450 J / 3 s = 150 J/s = 150 watts.
Answer:
19.6 m/s
Explanation:
The parameters given are:
Mass M = 20 Kg
Force F = 285 N
Angle Ø = 30 degree
Time t = 4 seconds
Coefficient of friction = 0.72
At the plane, the weight of the box will be mgsinØ
Resolving forces at the plane, we will have:
MgsinØ + Fr = F
Where Fr = frictional force.
Fr = F - mgsinØ
Substitute all the parameters into the formula
Fr = 285 - 20 × 9.8 sin30
Fr = 285 - 98
Fr = 187 N
But for the box moving toward the top of the plane,
F - Fr = ma
Where a = V/t
Substitute all the parameters involved into the formula
285 - 187 = 20 ( V/4)
98 = 5V
V = 98/5
V = 19.6 m/s
Therefore, the speed with which the box is moving is 19.6 m/s
