Ask question

Ask question

All categories

3 years ago

15

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that

the pendulum completes 99.0 full swing cycles in a time of 128 s . What is the magnitude of the gravitational acceleration on this planet? Express your answer in meters per second per second.

1 answer:

larisa86 [58]3 years ago

4 0

Answer:

12.4 m/s²

Explanation:

L = length of the simple pendulum = 53 cm = 0.53 m

n = Number of full swing cycles = 99.0

t = Total time taken = 128 s

T = Time period of the pendulum

g = magnitude of gravitational acceleration on the planet

Time period of the pendulum is given as

$T = \frac{t}{n}$

$T = \frac{128}{99}$

T = 1.3 sec

Time period of the pendulum is also given as

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

$1.3 = 2(3.14) \sqrt{\frac{0.53}{g}}$

g = 12.4 m/s²

You might be interested in

Which of the following is an example of a projective test?

IceJOKER [234]

Some examples of projective tests are the Rorschach Inkblot Test, the Thematic Apperception Test (TAT), the Contemporized-Themes Concerning Blacks test, the TEMAS (Tell-Me-A-Story), and the Rotter Incomplete Sentence Blank (RISB).

Some examples of projective tests are the Rorschach Inkblot Test, the Thematic Apperception Test (TAT), the Contemporized-Themes Concerning Blacks test, the TEMAS (Tell-Me-A-Story), and the Rotter Incomplete Sentence Blank (RISB).

6 0

3 years ago

What is an non example of Cultural diffusion

belka [17]

Beliefs is a non example

4 0

3 years ago

Provided following are four different ranges of stellar masses. Rank the stellar mass ranges based on how many stars in each ran

elena-s [515]

Highest to lowest number:

-less than 1 solar mass

-between 1 and 10 solar masses

-between 10 and 30 solar masses

-between 30 and 60 solar masses

<h3>What is Stellar masses ?</h3>

Stellar mass is a phrase that is used by astronomers to describe the mass of a star.

It is usually enumerated in terms of the Sun's mass as a proportion of a solar mass ( M ☉). Hence, the bright star Sirius has around 2.02 M ☉.

Stellar masses are not fixed, although they change for single stars only on long periods.

Learn more about Stellar masses here:

brainly.com/question/1128503

#SPJ4

3 0

2 years ago

What does deposition mean in a sedimentary formation

Zinaida [17]

Answer:

Sedimentary rocks are types of rock that are formed by the deposition and subsequent cementation of that material at the Earth's surface and within bodies of water. Deposition means that all the sediments, soil, and rocks are all compressed (tightly pressed into each other) and create sedimentary rocks.

I hope this helps! :)

4 0

3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago