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Romashka [77]
3 years ago
9

Jack and jill are married. this year jack earned $72,000 and jill earned $80,000 and they received $4,000 of interest income fro

m a joint savings account. how much gross income would jack report if he files married-filing-separate from jill?
Physics
1 answer:
MaRussiya [10]3 years ago
4 0
<span>If Jack is filing married-filing-separate he would report $76,000 gross income as head of household.</span>
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The Hubble Space Telescope is stabilized to within an angle of about 2-millionths of a degree by means of a series of gyroscopes
Likurg_2 [28]

Answer:

The answer to the question is;

The required torque that it would take to cause the gyroscopes to precess through an angle of 1.0×10−6 degree during a 5.0-hour exposure of a galaxy is 2.44 ×10⁻¹² N·m  

Explanation:

To solve the question we first resolve the units of the given quantities as follows

The gyroscopes spin at 19,200 rpm that is 19,200 revolutions per minute

1 revolution = 2π rad and

1 minute = 60 seconds

Therefore 19,200 revolutions per minute = 2π×19,200÷60 rad/s

= 2010.619 rad/s

The angle of precess is given as 1.0×10⁻⁶ °. We convert the angle to radians as follows

360 ° = 2π radians

1 ° = \frac{\pi }{180} radians and

1.0×10⁻⁶ ° =  \frac{\pi }{180} radians × 1.0×10⁻⁶ ° = 1.745×10⁻⁸ rad

To find the torque we note that the torque is given by

Precession angular speed × The moment of inertia × angular velocity

The precession angular speed is given by \frac{Precession. Angle}{time}

The precession angle was determined in rad as 1.745×10⁻⁸ rad

The precession time is 5 hours which is equal to 5×60×60 = 18000 s

Therefore the precession velocity = \frac{1.745*10^{-8} rad}{18000 s} =  9.696×10⁻¹³ rad/s

The moment of inertia is given by

Formula for the moment of inertia of a thin walled cylinder I = m·r²

Where:

r = Radius of the gyroscope = Diameter/2 = 5.0 cm/2 = 2.5 cm = ‪0.025‬ m

m = Mass of each gyroscopes = 2.0 kg

Therefore I = m·r² = 2.0 kg × (0.025‬ m)² = 0.00125 kg·m²

Torque, τ = Ω·I·ω

Where:

Ω = Precession velocity = 9.696×10⁻¹³ rad/s

I = Moment of inertia = 0.00125 kg·m²

ω = Angular speed = 2010.619 rad/s

τ = 9.696×10⁻¹³ rad/s × 0.00125 kg·m² × 2010.619 rad/s =

2.44 ×10⁻¹² kg·m²/ s² =   2.44 ×10⁻¹² N·m  

The required torque is 2.44 ×10⁻¹² N·m.

7 0
3 years ago
Vector of magnitude 15 is added to a vector of magnitude 25. The magnitude of this sum
loris [4]

Explanation:

Given that,

Magnitude of vector A, |A| = 15

Magnitude of vector B, |B| = 25

We need to find the magnitude of this sum.

The maximum sum of the resultant vector,

R_{max}=|A_1|+|A_2|\\\\=15+25\\\\=45

The minimum sum of the resultant vector,

R_{min}=|A_1|-|A_2|\\\\=15-25\\\\=-10

So, the magnitude of this sum either 45 or -10.

6 0
3 years ago
An object has a fixed volume and a variable shape before it changes state.
Wittaler [7]

Answer:

solid to liquid :)

Explanation:

APEX i just took it ;)

4 0
3 years ago
Read 2 more answers
The battleship and enemy ships 1 and 2 lie along a straight line. Neglect air friction. battleship 1 2 Consider the motion of th
DaniilM [7]

Answer:

\frac{t_1}{t_2} = \frac{sin\theta_1}{sin\theta_2}

Explanation:

The vertical component of the initial velocities are

v_v = v_0sin\theta

If we ignore air resistance, and let g = -9.81 m/s2. The the time it takes for the projectiles to travel, vertically speaking, can be calculated in the following motion equation

v_vt - gt^2/2 = s = 0

t(v_v - gt/2) = 0

v_v - gt/2 = 0

t = 2v_v/g = 2v_0sin\theta/g

So the ratio of the times of the flights is

t_1 / t_2 = \frac{2v_0sin\theta_1/g}{2v_0sin\theta_2/g} = \frac{sin\theta_1}{sin\theta_2}

8 0
3 years ago
A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy​
AURORKA [14]

velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

velocity = 10 m/s

_________________________________

Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

KE = 30 × 100

KE = 3000 j

8 0
3 years ago
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