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3 years ago

What is the frequency of motion of a 0.50 m long pendulum?

2 answers:

4 0

1. Write down the formula for "frequency of a pendulum" from your textbook.

f = (1/2pi) * sqrt(g/L)

2. Plug in g and L.

Hope this helps!

IgorC [24]3 years ago

3 0

Complicated 2-step process.

1. Write down the formula for "frequency of a pendulum" from your textbook. 

f = (1/2pi) * sqrt(g/L) 

2. Plug in g and L.

I hope this can help!

You might be interested in

Difference between diurnal and annual motion in two points

Valentin [98]

Answer:

Explanation below:

Explanation:

Annual motion describes the changes in motion of the earth around the sun. Diurnal motion can be better understood as the change in motion caused by Earths rotation at the poles.

This might not be the answer you were looking for, your question is very vague.

5 0

3 years ago

Use the worked example above to help you solve this problem. An airboat with mass 4.30 102 kg, including the passenger, has an e

umka21 [38]

Answer:

a) a = 1,865 m / s² and b) t = 8.1 s

Explanation:

a) Let's use Newton's second law to find acceleration, we can work the equation in scalar form because displacement and force have the same direction

F = m .a

a = F / m

a = 8.02 10² /4.3 10²

a = 1,865 m / s²

b) We use kinematic relationships in one dimension

vf = vo + at

vf = 0 + a t

t = vf / a

t = 15.1 / 1.865

t = 8.1 s

3 0

3 years ago

Which of the following biomes might be found at a latitude of 33° South?

cupoosta [38]

The answer is going be desert.

5 0

3 years ago

True or False:

bulgar [2K]

Answer:

False

True

False

8 0

3 years ago

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a

Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

$d = v*t \rightarrow v = \frac{d}{t}$

Where:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

$t = \sqrt{\frac{2d}{g}}$

Where:

g: is gravity = 9.8 m/s²

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s$

Now, the horizontal velocity of the rock is:

$v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s$

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.

2. To calculate the distance at which the projectile will land, first, we need to find the time:

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s$

So, the distance is:

$d = v*t = 6.67 m/s*1.43 s = 9.54 m$

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!

7 0

4 years ago