What is the mass of an object if a net force of 8.0 N causes it to accelerate at 1.1 m/s2?

A pressure sensor was used to measure the unsteady pressure in a cylinder. The sensor output was acquired for 15 seconds at a ra

4vir4ik [10]

Answer:

Maximum frequency on power spectrum plot = 101 Hz

Explanation:

Given:

Time taken for output = 15 seconds

Frequency rate = 202 Hz

Find:

Maximum frequency on power spectrum plot

Computation:

Maximum frequency = Given frequency rate / 2

Maximum frequency on power spectrum plot = Frequency rate / 2

Maximum frequency on power spectrum plot = 202 / 2

Maximum frequency on power spectrum plot = 101 Hz

8 0

3 years ago

Q5: An ice skater moving at 12 m/s coasts

ExtremeBDS [4]

u = 0.077

Explanation:

Work done by friction is

Wf = ∆KE + ∆PE

-umgx = ∆KE,. ∆PE =0 (level ice surface)

-umgx = KEf - KEi = -(1/2)mv^2

Solving for u,

u = v^2/2gx

= (12 m/s)^2/2(9.8 m/s^2)(95 m)

= 0.077

8 0

2 years ago

An airplane undergoes the following displacements, all at the same altitude: First, it flies 59.0 km in a direction 30.0° east o

ddd [48]

Answer:

B) 71.5 [km]

Explanation:

To solve this problem we will decompose each of the directions in the x & y axes.

To solve this problem we will decompose each of the directions in the x & y axes. also for a greater understanding of the angles, you should look at the attached image, which contains the orientations for each angle (clockwise or counterclockwise).

59.0 km in a direction 30.0° east of north

$d_{1x}= 59*sin(30) = 29.5[km]\\d_{1y}= 59*cos(30) = 51.09[km]$

58.0 km due south

$d_{2y} = - 58 [km]\\$

It flies 100 km 30.0° north of west

$d_{3x}= - 100*cos(30) = -86.6[km]\\d_{3y} = 100*sin(30)= 50 [km]$

Now we sum algebraically the components

$d_{x}=29.5-86.6 = -57.1[km]\\d_{y}=51.09 -58+50=43.09[km]\\\\$

Using the Pythagorean theorem we can find the magnitude of the displacement.

$d = \sqrt{(57.1)^{2} +(43.09)^{2} } \\d= 71.53[km]$

7 0

3 years ago

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea

Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$

$f_2 = n+f_1$

Replacing $3/s$ for n and 202Hz for $f_1,$

$f_2 = 3/s + 202Hz$

$f_2 = 3/s(\frac{1Hz}{1/s})+202Hz$

$f_2 = 206Hz$

The frequency of the tightened is 205Hz

7 0

3 years ago

How does the currently accepted model of the nucleus provide evidence of the existence of the strong nuclear force?

olganol [36]

Answer:

Protons and neutrons are all attracted to each other as a result - the strong nuclear force. This is an attractive force that only has an effect over a very short range in the nucleus.

4 0

3 years ago