Ask question

Ask question

All categories

3 years ago

9

Astronauts on a distant planet set up a simple pendulum of length 1.20 m. The pendulum executes simple harmonic motion and makes

100 complete oscillations in 450 s. What is the magnitude of the acceleration due to gravity on this planet?

1 answer:

Ray Of Light [21]3 years ago

4 0

Answer:

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

Explanation:

Time period of simple pendulum is given by

$T=2\pi\sqrt{\frac{l}{g}}$ , l is the length of pendulum, g is acceleration due to gravity value.

We can solve acceleration due to gravity as

$g=\frac{4\pi^2l}{T^2}$

Here

Length of pendulum = 1.20 m

Pendulum executes simple harmonic motion and makes 100 complete oscillations in 450 s.

Period, $T=\frac{450}{100}=4.5s$

Substituting

$g=\frac{4\pi^2\times 1.2}{4.5^2}=2.34m/s^2$

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

You might be interested in

The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

7 0

2 years ago

What's the value of 1,152 Btu in joules? A. 1,964,445 J B. 2,485,664 J C. 987,875 J D. 1,215,360 J

OLga [1]

Answer: Option (d) is correct.

Explanation:

Given, 1,152 British thermal units

1 British thermal unit = 1055.06 joules

So, in 1,152 British thermal units there will be :

$1152\times 1055.06 Joules=1215429.12 Joules=1.21542912\times 10^{8} Joules$

Hence, from the given options the closest answer is of option (d). So, option (d) is correct.

4 0

3 years ago

Read 2 more answers

A car travels straight for 20 miles on a road that is 30 degrees north of east what is the east component of the cars displaceme

EleoNora [17]

17.3 boiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

3 0

3 years ago

A far-sighted person has a near-point of 80 cm. To correct their vision so that they can see objects that are as close as 10 cm

bekas [8.4K]

Answer:

$f = 8.89 cm$

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

so he is able to see the object 80 cm

now the distance of lens from eye is 2 cm

and the person want to see the objects at distance 10 cm

so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

$-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}$

$f = 8.89 cm$

3 0

3 years ago

Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a st

zubka84 [21]

Answer:

R=3818Km

Explanation:

Take a look at the picture. Point A is when you start the stopwatch. Then you stand, the planet rotates an angle α and you are standing at point B.

Since you travel 2π radians in 24H, the angle can be calculated as:

$\alpha =\frac{2*\pi *t}{24H}$ t being expressed in hours.

$\alpha =\frac{2*\pi *11.9s*1H/3600s}{24H}=0.000865rad$

From the triangle formed by A,B and the center of the planet, we know that:

$cos(\alpha )=\frac{r}{r+H}$ Solving for r, we get:

$r=\frac{H*cos(\alpha) }{1-cos(\alpha) } =3818Km$

6 0

2 years ago