****PLEASE HELP**** THERE ARE TWO QUESTIONS (ITS EASY)

How long does it take for the hubble telescope to orbit earth

il63 [147K]

Answer:

95 minutes

Explanation:

According to N.A.S.A, the Hubble Space Telescope makes one orbit around Earth every 95 minutes.

"The Hubble Space Telescope is a large telescope in space. It was launched into orbit by space shuttle Discovery on April 24, 1990. Hubble orbits about 547 kilometers (340 miles) above Earth. It is the length of a large school bus and weighs as much as two adult elephants. Hubble travels about 5 miles per second: That is like traveling from the eastern coast of the United States to the western coast in 10 minutes. Hubble is solar-powered.

Hubble takes sharp pictures of objects in the sky such as planets, stars and galaxies. Hubble has made more than one million observations. These include detailed pictures of the birth and death of stars, galaxies billions of light years away, and comet pieces crashing into Jupiter's atmosphere.

Scientists have learned a lot about the universe from these pictures. Many of them are beautiful to look at."

3 0

4 years ago

Read 2 more answers

A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between

mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 × 0.00262

V = 0.4061 V

The potential difference between the plates is 0.4061 V

5 0

3 years ago

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both

svetlana [45]

Answer:

7 deg

Explanation:

$m$ = mass of the rod = $1.4 kg$

$W$ = weight of the rod = $mg = (1.4) (9.8) = 13.72 N$

$k_{L}$ = spring constant for left spring = $59 Nm^{-1}$

$k_{R}$ = spring constant for right spring = $33 Nm^{-1}$

$x_{L}$ = stretch in the left spring

$x_{R}$ = stretch in the right spring

$L$ = length of the rod = 0.75 m

$\theta$ = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

$k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m$

Using equilibrium of force in vertical direction for right spring

$k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m$

Angle made with the horizontal is given as

$\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg$

3 0

4 years ago

How fast (in rpm) must a centrifuge rotate if a particle 6.00 cm from the axis of rotation is to experience an acceleration of 1

astra-53 [7]

The acceleration experienced by the particle is given by
$a=113000 g=113000 \cdot 9.81 m/s^2$
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed $\omega$ of the particle and its distance r from the axis by the relationship
$a= \omega ^2 r$
In our problem, $r=6 cm=0.06 m$ , so we can solve for $\omega$ :
$\omega = \sqrt{ \frac{a}{r} } = \sqrt{ \frac{113000 \cdot 9.81 m/s^2}{0.06 m} }=4298 rad/s$
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to $( \frac{1}{2 \pi} )$ revolutions, while $1 s = \frac{1}{60} min$ . So we the conversion is $\omega = 4298 rad/s \cdot ( \frac{1}{2\pi} rev/rad )( 60 s/min)=41067 rpm$

4 0

4 years ago

In a physics lab experiment, a spring clamped to the table shoots a 22g ball horizontally. When the spring is compressed 21cm ,

worty [1.4K]

a square and a triangle I think

6 0

4 years ago