So, there should be two forces acting on the refrigerator: the applied force and the friction force.

The question mentioned that the friction force was set to zero, so the only effective force now would be the applied force.

We have an applied force of 400 N to the right, this means that:
<span>The magnitude of the net force is 400, directed to the right.</span>

3 0

3 years ago

Read 2 more answers

A carpenter on the roof of a building accidentally drops herhammer. As the hammer falls it passes two windows of equal height,a)

nadezda [96]

Answer:

Explanation:

a) the speed increment of the hammer as it drops past the first window, is greater than that of the speed of the hammer as it drops past the second window. This can also be translated as saying that the hammer spent more time at the second window.

b) III

The best answer would be answer III, The hammer spends more time dropping past window 1, which I had already included in my explanation in (a) above.

4 0

3 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

4 years ago