Answer:
The transverse wave will travel with a speed of 25.5 m/s along the cable.
Explanation:
let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.
then, if V is the volume of the cable:
ρ = m/V
m = ρ×V
but V = A×L , where L is the length of the cable.
m = ρ×(A×L)
m/L = ρ×A
then the speed of the wave in the cable is given by:
v = √(T×L/m)
= √(T/A×ρ)
= √[2.96×10^4/(4.49×10^-3×7860)]
= 25.5 m/s
Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.
I believe the answer is Nonmaterial Culture.
Explanation:
you measure temperature in degrees celsius using a thermometer. Thermal energy is measured in joules. A larger volume of water will take longer to heat up but will store more energy than the smaller object. However, a smaller object will lose it's heat faster than a larger object. A cup of tea has less thermal energy than a swimming pool.
Answer:
The reading of the scale during the acceleration is 446.94 N
Explanation:
Given;
the reading of the scale when the elevator is at rest = your weight, w = 600 N
downward acceleration the elevator, a = 2.5 m/s²
The reading of the scale can be found by applying Newton's second law of motion;
the reading of the scale = net force acting on your body
R = mg + m(-a)
The negative sign indicates downward acceleration
R = m(g - a)
where;
R is the reading of the scale which is your apparent weight
m is the mass of your body
g is acceleration due to gravity, = 9.8 m/s²
m = w/g
m = 600 / 9.8
m = 61.225 kg
The reading of the scale is now calculated as;
R = m(g-a)
R = 61.225(9.8 - 2.5)
R = 446.94 N
Therefore, the reading of the scale during the acceleration is 446.94 N
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
