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3 years ago

An object that completes 20 vibrations in 10 seconds has a frequency of

Physics

1 answer:

nika2105 [10]3 years ago

7 0

Answer:

The object has frequency of 2 Hz and time period of 0.5 s.

Explanation:

Frequency is defined as number of oscillation per second ie back and forth swings done in single second.

Here it is given that the object oscillates 20 times in 10 seconds.

So f = $\frac{20}{10}$ = 2Hz

The time period is defined as time taken by the object to complete one full oscillation.

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

T= $\frac{1}{2}$ =0.5 s

Thus the object has frequency of 2 Hz and time period of 0.5 s.

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PART 2 OF ENERGY AND FORCES UNIT TEST

katrin2010 [14]

Answer:

1. at least two charged interacting parts

2. from the electric fields of charged subatomic particles

3 an arrow released from the bow

4Electrical fields of charged particles interact, bonding those with opposite charges.

5 the interaction of the electric fields of protons and electrons

6 The energy stored in the system increases.

7 Kinetic energy increases because the magnets move in the direction of the field.

8 Iron pieces accelerate toward the magnet, and the energy stored in the system decreases.

The energy stored in the field decreases because the magnet moves in the direction of the field.

10 The energy stored increases and then decreases.

11 The wire was not connected to the source.

12 The electromagnet will become more powerful.

the rest are written, hope this helps (:

4 0

2 years ago

You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with

Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

Explanation:

For this exercise we use the constructor equation or Gaussian equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

m = $\frac{h'}{h}$ = - $\frac{q}{p}$

h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

$\frac{1}{q} = \frac{1}{f } - \frac{1}{p}$

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

h ’= h q / p

where p has a high value and is the same for all systems

h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

f = 12 cm h ’= fo 12 cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0

3 years ago

An atom of carbon has a radius of 67.0 pm and the average orbital speed of the electrons in it is about 1.3 x 10⁶ m/s.

adoni [48]

Answer :

The least possible uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

The percentage of the average speed is, 33 %

Explanation :

According to the Heisenberg's uncertainty principle,

$\Delta x\times \Delta p=\frac{h}{4\pi}$ ...........(1)

where,

$\Delta x$ = uncertainty in position

$\Delta p$ = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

$p=m\times v$

or,

$\Delta p=m\times \Delta v$ .......(2)

Equating 1 and 2, we get:

$\Delta x\times m\times \Delta v=\frac{h}{4\pi}$

$\Delta v=\frac{h}{4\pi \Delta x\times m}$

Given:

m = mass of electron = $9.11\times 10^{-31}kg$

h = Planck's constant = $6.626\times 10^{-34}Js$

radius of atom = $67.0pm=67.0\times 10^{-12}m$ $(1pm=10^{-12}m)$

$\Delta x$ = diameter of atom = $2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m$

Now put all the given values in the above formula, we get:

$\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}$

$\Delta v=4.32\times 10^{5}m/s$

The minimum uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

Now we have to calculate the percentage of the average speed.

Percentage of average speed = $\frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100$

Uncertainty in speed = $4.32\times 10^{5}m/s$

Average speed = $1.3\times 10^{6}m/s$

Percentage of average speed = $\frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100$

Percentage of average speed = 33.2 % ≈ 33 %

Thus, the percentage of the average speed is, 33 %

3 0

3 years ago

Which of the following is true about gravity?

murzikaleks [220]

B, C, and D are your answers.

6 0

3 years ago

How long would it take a drag racer to increase her speed from 10m/s to 20 m/s if her car accelerates at a uniform rate of 15 m/

weeeeeb [17]

Answer:

t = 0.67 [s]

Explanation:

To solve this problem we must use the following kinematics equation.

$v_{f} =v_{i} +(a*t)\\where:$

Vf = final velocity = 20[m/s]

Vi = initial velocity = 10 [m/s]

a = aceleration = 15 [m/s^2]

Now replacing in the equation we have:

20 = 10 + (15*t)

t = (20-10)/15

t = 0.67 [s]

7 0

3 years ago