What is the acceleration due to gravity?

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

Single Replacement: HCI + Zn

Kay [80]

During a single replacement, the Zn and H will switch places. That means the product formed will be ZnCl and H2.

4 0

3 years ago

Visit this website and locate the element mercury, whose chemical symbol is Hg. Click on the square to read about the history, p

Irina-Kira [14]

We simply asked to name three uses for mercury.
The most common and well-known use of mercury is the production of thermometers. It's property to stay liquid at room temperature makes it ideal for a temperature indicator. However, the use of mercury is thermometers has been phased out due to health hazards.
It is also used to form an amalgam which is the result of its combination with silver or gold. Mercury has been used to mine gold and silver. This application has also been phased out.
Today's use of mercury includes mercury-vapor lamps which are the bright lamps used in high-ways.

6 0

3 years ago

Read 2 more answers

A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor

Mumz [18]

Answer:

The maximum induced emf in the rotating coil = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

To calculate the induced emf in the rotating cell,we can use the formula:

emf = NBAωsin(ωt)

For maximum induced emf, the value of sin(ωt) will be 1

$emf_max$ = NBAω ; if (A = l × w) , we have:

$emf_max$ = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

$emf_max$ = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e $emf_max$ = $N\frac{d∅}{dt}$