All categories

3 years ago

1. What is the formula for the period of a pendulum and what is the main determining factor in its period?

Physics

1 answer:

Brut [27]3 years ago

7 0

Answer:

$T=2\pi \sqrt{\frac{L}{g}}$

Explanation:

A simple pendulum is a system consisting of a mass attached to a string, and oscillating in a periodic motion, back and forth, along an equilibrium position.

The period of a pendulum is the time it takes for the pendulum to complete one oscillation.

The period of a pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity

From the formula, we see that the period of a pendulum does not depend on the mass.

Therefore, the only 2 factors affecting the period of a pendulum are:

- The length of the pendulum: the longer it is, the longer the period of oscillation

- The acceleration due to gravity: the greater it is, the shorter the period of the pendulum

You might be interested in

9. Which of the following statements is true about scientific theories? (1 point)

hammer [34]

Answer: D

Explanation:

We can immediately cross out A because theories and laws are completely different

We can then cross out B because laws have to go through many trials to prove them

C is wrong because laws and hypothesis are completely different

So that leaves us with D. Hope this helps!

6 0

2 years ago

Read 2 more answers

List two fossil fuels. describe in general how fossil fuels were formed.

vazorg [7]

Fossil Fuels:

Coal and Oil

How they are formed:

Coal is formed from big plants, from a long time ago. When the big plants died, they formed layers at the bottom of swamps and the dirt began to pile up and it formed Coal.

Oil is a fossil fuel that has been formed from a large amount tiny plants and animals such as algae and zooplankton.

4 0

3 years ago

All of these are likely to speed up the rate of a reaction except decreasing the surface area. increasing the temperature. incre

Vanyuwa [196]

Answer:

decreasing the surface area.

Explanation:

The rate of reaction or the reaction rate may be defined as the speed or the rate at which the reaction occurs. It is the speed of formation of the products from the reactants. The following factors increases the rate of reaction :

1. by adding a catalyst

2. by increasing the temperature

3. by increasing the pressure

4. by increasing the surface area

5. by increasing concentration

Hence, decreasing the surface area of the reaction reduces the rate of reaction.

6 0

3 years ago

A car is traveling at a 20.0 m/s for 7.00 s and then suddenly comes to a stop over a 3 s period.

Delicious77 [7]

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ( $v_{i}$ ) = 20.0 m/s

Final velocity of the car ( $v_{f}$ ) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time) (m/s²)

Deceleration = (( $v_{f}$ ) - ( $v_{i}$ )) ÷ time (m/s²)

Deceleration = ( 0.0 - 20 ) ÷ 3 (m/s²)

Deceleration = (- 20) ÷ 3 (m/s²)

Deceleration = - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time (meter)

Distance = 20.0 × 10

Distance = 200 meters

The total distance the car travels during the period of 10 s is 200 meters

7 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago