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Nezavi [6.7K]
3 years ago
10

What best describes the relationship between the frequency of an oscillation and the period of the oscillation?

Physics
1 answer:
ZanzabumX [31]3 years ago
4 0

Explanation:

Let f is the frequency of an oscillation and T is the period of the oscillation. There exists an inverse relationship between the frequency and the time period of the oscillation. Mathematically, it is given by :

f=\dfrac{1}{T}

Also, f=\dfrac{\omega}{2\pi}

So,

T=\dfrac{2\pi}{\omega}

The time taken to complete one oscillation is called the period of the oscillation and the number of oscillation is called the frequency if an oscillation.

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The shape of the orbit for most comets is a(n):<br><br> circle<br> parabola<br> ellipse<br> oval
worty [1.4K]

Answer:

Ellipse

Explanation:

Most comets have an elliptical path of orbit.

6 0
3 years ago
Rafael is driving his car at 26 m/s. What is the shortest distance in which he can brake and stop if the coefficient of static f
Hitman42 [59]

Answer:

86.14 meters.

Explanation:

Step one:

Given data

velocity of car = 26 m/s

the coefficient of static friction between the tires and the road

µ = 0.4 (kinetic)

Let us take g = 9.81 m/s^2

Required

The distance x = distance in m

We know that

N = Fg = mg\\\\F_\mu  = -\mu N = \mu F_g = \mu mg\\\\KE = (1/2) mv^2

W = F*x  (Work is force times distance)

Step two:

Conservation of energy gives  

KE = W

Substituting gives  

(1/2) mv^2 = F \mu x\\\\(1/2) mv^2 = \mu mgx\\\\mv^2 = 2 \mu mgx

Solving for distance (x) gives  

x = mv^2  / 2 \mu mg

Simplifying

x = v^2 / 2 \mu g

Substitute:  

x = v^2 / 2 \mu g

x= 26^2/2*0.4*9.81

x=676/7.848\\\\x=86.14

Therefore, the minimum braking distance is 86.14 meters.

4 0
3 years ago
Mount mckinley in Alaska is America’s highest mountain at 20,320 feet. Find it’s height in kilometers. 1mi = 1.609 km= 5280. Ans
Leto [7]

Answer:

I believe the answer is:

6.193536

5 0
3 years ago
The tallest building in the world, according to some architectural standards, is the Taipei 101 in Taiwan, at a height of 1671 f
Andrej [43]

Answer:

35.14°C

Explanation:

The equation for linear thermal expansion is \Delta L = \alpha L_0\Delta T, which means that a bar of length L_0 with a thermal expansion coefficient \alpha under a temperature variation \Delta T will experiment a length variation \Delta L.

We have then \Delta L = 0.481 foot, L_0 = 1671 feet and \alpha = 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a \Delta T =\frac{\Delta L }{\alpha L_0} = 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.

Since our temperature on a cool spring day was 13.0°C, our new temperature must be T_f=T_0+\Delta T = 35.14°C

3 0
3 years ago
(a) A space vehicle is launched vertically upward from the Earth's surface with an initial speed of vi that is comparable to but
ad-work [718]

Energy Conservation Theory,

(k+v)_i=(k+v)_f \quad \text { (No air resistone)}\\

\frac{1}{2} m v_i^2-\frac{G m M_\epsilon}{R_\epsilon}=0-\frac{G m M_\epsilon}{R_\epsilon+h}

\frac{1}{2} m v_i^2-\frac{G m M_\epsilon}{R_\epsilon}=0-\frac{G m M_\epsilon}{R_\epsilon+h}

v_{e x^2}{ }^2-v_i^2=\frac{v_{e^2 R_t} R_t}{R_t t h}\\&\frac{1}{v_{B C^2-v_1^2}^2}=\frac{R_E+h}{v_{e^2 R_E} R_E}\\\\\\h=\frac{R_E V_1^2}{v_{\text {esc }}^2-v_1^{\beta^2}}

<h3>What is law of  energy conservation?</h3>

The principle of energy conservation states that energy is neither created nor destroyed.  It may change from one sort to another. Just like the mass conservation rule, the legitimacy of the preservation of energy depends on experimental perceptions; hence, it is an experimental law. The law of preservation of energy, too known as the primary law of thermodynamics

To learn more about Energy Conservation Theory, visit;

brainly.com/question/8004680

#SPJ4

7 0
2 years ago
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