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3 years ago

What is the period that corresponds to a frequency of 39.5 Hz? Answer in units of s.

Physics

1 answer:

tatiyna3 years ago

5 0

The answer for the following problem is mentioned below.

Therefore the time period is 0.02 seconds.

Explanation:

Frequency:

The number of waves that pass a fixed place in a given amount of time. (or)

The number of waves that pas by per second.

The SI unit of the frequency is Hertz(Hz).

Time period:

The time taken for one complete cycle of vibration to pass a given point.

The SI unit of time period is seconds. (s)

Given:

Frequency (f) = 39.5 Hz

To calculate:

Time period (T)

We know;

According to the problem;

From the problem;

f = $\frac{1}{T}$

Where;

f represents the frequency

T represents the time period

f = $\frac{1}{39.5}$

f = 0.02 seconds

Therefore the time period is 0.02 seconds.

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A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago

A 25kg box fell 200m with an acceleration of 5 m/s2. with what force did it hit the floor when it landed?

Marizza181 [45]

According to Newton's Second Law of motion, the net force acting on the object is equal to its mass multiplied by its acceleration. In formula, it is written as

Net Force =mass * acceleration
Net force = 25 kg * 5m/s^2
Net force = 125 Newtons

6 0

3 years ago

Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th

choli [55]

Answer:

$1.368\times 10^{-20}\ J$

Explanation:

q = Charge in the potassium ion = $19e-18e$

e = Charge of electron = $1.6\times 10^{-19}\ C$

$V_2-V_1$ = Change in potential = $0-(-85.5\times 10^{-3})$

Change in electric potential is given by

$E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J$

The energy is $1.368\times 10^{-20}\ J$

3 0

3 years ago

Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista

soldi70 [24.7K]

Answer:

$v = 7793150 m/s$

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

$v = \frac{kQ}{r}$

Where $k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }$

and it is satisfy the superposition principle, thus

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}$

$v = 222.73v$

The electrical potential at 10 cm from charge 1 is:

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}$

$v = 395.44 v$

Since the work - energy theorem, we have:

$q\Delta v = \frac{mv^{2} }{2}$

where q is the electron's charge and m is the electron's mass

Therefore:

$v = \sqrt{\frac{2q\Delta v}{m} }$

$v = 7793150 m/s$

6 0

3 years ago

"A hiker starts at point P and walks 2.0 kilometers due east and then 1.4 kilometers due north. The vectors in the diagram below

Law Incorporation [45]

Here, you need to use your "Protractor" as it is given in the question, but we can calculate the value with the help of our mathematical calculation too:
[ Protractor can be use only in real life, not here ]

Draw an imaginary line from initial position to final position.
Now, In that triangle, tan x = P/B
tan x = 1.4 / 2
tan x = 0.70
x = tan⁻¹ (0.70)
x = 35 [ tan 35 = 0.70 ]

In short, Your Answer would be 35 degrees

Hope this helps!

3 0

3 years ago