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3 years ago

What would changing the frequency of a wave do to the wave?

1 answer:

3 0

The data convincingly show that wave frequency does not affect wave speed. An increase in wave frequency caused a decrease in wavelength while the wave speed remained constant. The last three trials involved the same procedure with a different rope tension.

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Hook's law describes an ideal spring. Many real springs are better described by the restoring force (FSp)s=−kΔs−q(Δs)^3, where q

kvasek [131]

Answer

given,

k = 250 N/m

q = 900 N/m³

(FSp)s=−kΔs−q(Δs)^3

work done = Force x displacement

$W = \int {F. dx}$

limits are x = 0 to x = 0.15 m

work done

$W = \int_0^{0.15} (k x + q x^3)\ dx$

$W = [\dfrac{kx^2}{2}+\dfrac{qx^4}{4}+ C]_0^0.15$

$W = \dfrac{300\times 0.15^2}{2}+\dfrac{900\times 0.15^4}{4}$

W = 3.375 + 0.1139

W = 3.3488 J

b) % cubic term = $\dfrac{0.1139}{3.3488}$

% cubic term = $3.4\ %$

7 0

3 years ago

When resistance force is increased on a lever which of the following happens to the effort force?

Sauron [17]

When resistance force on a lever increases, nothing happens automatically.
But if you want to keep lifting the load, then YOU must increase the force of
your effort in order to make it happen.

8 0

3 years ago

A car is moving initially at 30 m/s comes gradually at stop at 900 m . What is acceleration of the car ?

ziro4ka [17]

The acceleration of the car is 0.5 meters per seconds square.

Given the following data:

Initial velocity, U = 30 m/s

Distance, S = 900 meters

Note: The final velocity (V) of the car would be zero (0) m/s when it comes to a stop.

To find the acceleration of the car, we would use the third equation of motion;

Mathematically, the third equation of motion is given by the formula;

$V^2 = U^2 - 2aS\\\\0 = 30^2 - 2a(900)\\\\0 = 900 - 1800a\\\\1800a = 900\\\\a = \frac{900}{1800}$

Acceleration, a = 0.5 $m/s^2$

Therefore, the acceleration of the car is 0.5 meters per seconds square.