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4 years ago

A pendulum makes 36 vibrations in 60 seconds, what is its: Period? And Frequency

Physics

1 answer:

erica [24]4 years ago

5 0

Answer:

Period = 1.67 seconds

Frequency = 0.6 Hz

Explanation:

Number of vibrations in 60 seconds = 36

Frequency is defined as vibrations per second.

So, Frequency = 36/60 = 0.6 Hz

Time period is defines as time taken for 1 vibration,

So, Time period = 60/36 = 1.67 seconds.

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The atomic mass of an atom do not contain electrons because

Dimas [21]

Answer:

Electrons are so small that it does not affect the mass of atom .

Explanation:

Electrons are much smaller in mass than protons, weighing only 9.11 × 10^-28 grams, or about 1/1800 of an atomic mass unit. Therefore, they do not contribute much to an element's overall atomic mass.

7 0

3 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the t

Nina [5.8K]

Answer:

The speed of the boxes are 1 m/s.

Explanation:

Given that,

Mass of box = 1 kg

Mass of another box = 2 kg

Suppose 1 kg box moves with 3 m/s speed.

We need to calculate the speed of the boxes

Using formula of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v$

Where, u = initial velocity

v = final velocity

Put the value into the formula

$1\times3+2\times0=(1+2)v$

$v=\dfrac{3}{3}$

$v=1\ m/s$

Hence, The speed of the boxes are 1 m/s.

4 0

3 years ago

Select the correct answer. Eli and Andy want to find out which of the two is stronger. Eli pushes a table with a force of 120 ne

sergiy2304 [10]

Acceleration of the table: B. 0.50 meters/second2

Explanation:

The problem can be solved by using Newton's second law of motion, which states that the net force acting on an object is the product of its mass and its acceleration. Mathematically:

$\sum F = ma$