If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. s

PLEASE HELP! FOR BRAINLIEST!!!

Stells [14]

Answer:

I think it's C I'm so sorry if I'm wrong.

Explanation:

7 0

3 years ago

Read 2 more answers

A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th

GREYUIT [131]

Answer:

$1.26\cdot 10^7 m/s$

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

$F=qvB$

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

$v=\frac{qBr}{m}$

where in this problem we have:

$q=1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$B=208 G=208\cdot 10^{-4}T$ is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

$r=3.45 mm = 3.45\cdot 10^{-3} m$

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

So, the electron's speed is

$v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s$

6 0

3 years ago

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

Alexxx [7]

Answer:

The power of top half of the lens is 0.55 Diopters.

Explanation:

Since, the person can see an object at a distance between 34 cm and 180 cm away from his eyes. Therefore, 180 cm must be the focal length of the upper part of lens, as the top half of the lens is used to see the distant objects.

The general formula for power of a lens is:

Power = 1/Focal Length in meters

Therefore, for the top half of the lens:

Power = 1/1.8 m

<u>Power = 0.55 Diopters</u>

3 0

3 years ago

A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i

timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

$I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2$

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

$I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2$

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

$I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2$

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

$I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}$

$\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}$

$\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s$

4)Kinetic energy before:

$E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J$

Kinetic energy after:

$E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J$

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) $a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2$

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0

4 years ago

A person's center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass mm at the top of a

muminat

Answer:

$\mathbf{v_{max} = \sqrt{gL}}$