All categories

1 year ago

Calculate the time period of simple pendulum whose length is 98.2cm

Physics

1 answer:

alekssr [168]1 year ago

5 0

The time period resulting in oscillations will be 1.986 seconds.

<h3>What is the period of oscillation?</h3>

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The time period of the oscillation is;

$\rm T = 2 \pi\sqrt{\frac{L}{g}} \\\\ \rm T = 2 \times 3.14 \times \sqrt{ \frac{98.2 \ \times 10^{-2} \ m}{9.81 \ m/s^2}} \\\\ T= 1.986 \ sec$

Hence the time period resulting oscillations will be 1.986 seconds.

To learn more about the time period of oscillation refer to the link;

brainly.com/question/20070798

#SPJ1

You might be interested in

The probability distribution for a

makkiz [27]

Answer:

Explanation:

6 0

3 years ago

Identify the kind of simple machine represented by each of these examples: a. A drill bit b. A skateboard ramp c. A boat oar

marusya05 [52]

Answer:

a skateboard ramp

Explanation:

6 0

2 years ago

Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high

konstantin123 [22]

L=8m
B=4m
H=300cm=3m

$\\ \bull\tt\longmapsto Volume=LBH$

$\\ \bull\tt\longmapsto Volume=8(4)(3)$

$\\ \bull\tt\longmapsto Volume=96m^3$

6 0

2 years ago

Read 2 more answers

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

Two balls, each with a mass of 0.890 kg, exert a gravitational force of 8.06 × 10−11 n on each other. how far apart are the ball

Bond [772]

This problem involves Newton's universal law of gravitation and the equation to follow would be.

F = GM₁M₂/r²

Given: M₁ = 0.890 Kg; M₂ = 0.890 Kg; F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²

Solving for distance r = ?

r = √GM₁M₂/F

r = √(6.673 x 10⁻¹¹ N m₂/Kg²)(0.890 Kg)(0.890 Kg)/ 8.06 x 10⁻¹¹ N

r = 0.81 m

6 0

3 years ago