Ask question

Ask question

All categories

bogdanovich [222]

3 years ago

8

Using the set-up seen here, Ms. Garcia places a golf ball between the globe and the flashlight. Turning off the lights in the ro

om, she shines a flashlight from behind the golf ball. The students observe the shadow of the golf ball falling on the globe. What is this setup trying to model?
A) a solar eclipse
B) the solar system
C) the idea of gravity
D) the surface of the moon

1 answer:

Murrr4er [49]3 years ago

4 0

This is a solar Eclipse.

You might be interested in

(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the ca

gulaghasi [49]

Answer:

(a) $E_{ c} = 112.5 \mu J$

(b) $E'_{ c} = 18 \mu J$

Solution:

According to the question:

Capacitance, C = $1.00\mu F = 1.00\times 10^{- 6} F$

Voltage of the battery, $V_{b} = 15.0 V$

(a)The Energy stored in the Capacitor is given by:

$E_{c} = \frac{1}{2}CV_{b}^{2}$

$E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}$

$E_{ c} = 112.5 \mu J$

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

$E'_{c} = \frac{1}{2}CV'_{b}^{2}$

$E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}$

$E'_{ c} = 18 \mu J$

6 0

3 years ago

The innermost satellite of jupiter orbits the planet with a radius of 422 × 103 km and a period of 1.77 days. what is the mass o

FinnZ [79.3K]

The mass of Jupitar is obtained from the calculations as 5.8 * 10^-14 Kg.

<h3>What is the mass of Jupitar?</h3>

There are nine planets in the solar system and the sun lies at the enter of our solar system. This is the heliocentric model of the solar system.

Given that;

T^2 = GMr^3/4π

T = period

G = gravitational constant

r = radius

M = mass of Jupitar

Now;

1 day = 86400 seconds

1.77 days = 1.77 days * 86400 seconds/1 day

= 152928 seconds

Making M the subject of the formula;

M =4πT^2/Gr^3

M = 4 * 3.142 * (152928)^2/6.67 × 10^-11 * (422 × 10^9)^3

M = 2.9 * 10^11/5.0 * 10^24

M = 5.8 * 10^-14 Kg

Learn more about mass of a planet:brainly.com/question/13851553

#SPJ1

5 0

2 years ago

In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.

Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement of the object

t is the time elapsed

$\omega$ is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

$\theta=2\pi$ rad

And the time taken is

$t=95 min \cdot 60 =5700 s$

Therefore, the angular velocity of the telescope is

$\omega=\frac{2\pi}{5700}=0.0011 rad/s$

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

$v=\omega r$

where

v is the linear velocity

$\omega$ is the angular velocity

r is the radius of the circular orbit

In this problem:

$\omega=0.0011 rad/s$ is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

$r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m$

Therefore, the linear velocity of the telescope is:

$v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s$

4 0

3 years ago

If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit

salantis [7]

Answer:

$v = r\omega$

Explanation:

If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.

Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t

The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t

So the linear speed compared to angular speed is

$\frac{v}{\omega} = \frac{2\pi r/t}{2\pi /t} = r$

$v = r\omega$

6 0

3 years ago

Einstein calculated that ripples of gravity travel at exactly the speed of _____

damaskus [11]

Answer:

299,792,458 m/s = speed of light

Explanation:

6 0

3 years ago