Ask question

Ask question

All categories

3 years ago

11

How tall would a tower need to be if the period of a pendulum were 30. seconds

1 answer:

Virty [35]3 years ago

4 0

We use the following expression

T = 2*pi *sqrt(l/g)

Where T is the period of the pendulum

l is the length of the pendulum

and g the acceleration of gravity

We solve for l

l = [T/2*pi]² *g = [30s/2*pi]²* 9.8 [m/s²] = 223.413 m

The tower would need to be at least 223.413 m high

You might be interested in

What happens to jetstream’s as they get closer to the equator

MAVERICK [17]

Answer:They stop because jet streams follow boundaries between hot and cold air.

Explanation:

6 0

3 years ago

Read 2 more answers

If an 800 kg roller coaster is at the top of its 50 m high track, it will have a potential energy pf 392,000 J and a kinetic ene

Viefleur [7K]

I just woke him crying crying laughing crying and laughing I’m so mad he got me blocked him lol I got a hold on her phone the answer is 0 m

5 0

3 years ago

Someone please help me

Lapatulllka [165]

False... I hope that helps ;)

6 0

3 years ago

From what does oil form?

Akimi4 [234]

C dinosaurs. Once they break down and get pressurized they turn to oil

6 0

3 years ago

If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo

lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

temperature at t = 0

$T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}$

T(0) = 95°C

temperature after half hour of cooling

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

t = 30 minutes

$T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}$

$T( 30 ) = 20 + 75 \times 0.5488$

T(30) = 61.16° C

average of first half hour will be equal to

$T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt$

$T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30$

$T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30$

$T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]$

$T = \dfrac{1}{30}[600 - 2058.04 + 3750]$

T = 76.39°C

4 0

3 years ago