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VladimirAG [237]
3 years ago
12

Can you guys make a three line poem with the word time

Physics
2 answers:
Rus_ich [418]3 years ago
7 0
I'll try and write a simple rhyme.
It shouldn't take me too much time.
But first I'll stop and eat the lime
I bought this morning for a dime.
Elina [12.6K]3 years ago
3 0
Time should not be messed with
for bad things could happen
so think before you act or you'll regret it
You might be interested in
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
Light traveling through air encounter a second medium which slows the light to 2.7 x 10^8. What is the index of the medium?
kozerog [31]

Answer:

1.11

Explanation:

The index of the medium can be calculated using below formula

V= c/ n ............eqn(1)

Where V= velocity of the light is reduced to while traveling through the second medium= 2.7 x 10^8 m/s

n= index of the medium

c= speed of light= 3 x 10^8 m/s

Substitute for the values in eqn(1)

2.7 x 10^8 = (3 x 10^8 m/s)/ n

Making " n" subject of the formula, we have

n= (3 x 10^8 )/(2.7 x 10^8)

n= 1.11

Hence, the index of the medium is 1.11

5 0
3 years ago
You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t
uranmaximum [27]
Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together.  This is done because the friction force is going to have to be compensated for.  We will need that much more force than we otherwise would to achieve the desired acceleration:

F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2}  =0.0702N

The friction force will be given by the normal force times the coefficient of friction.  Here the normal force is just its weight, mg

F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N

Now the total force required is:

0.0702N+0.803N=0.873N

5 0
3 years ago
Convert 27,549 into scientific notation
dezoksy [38]
2.7549 x 10^4 is the answer I hope this helped u
7 0
3 years ago
What is -8,675,309.0 in scientific notation
Sindrei [870]
= 3.456 × 1011
(scientific notation)

= 3.456e11
(scientific e notation)

= 345.6 × 109
(engineering notation)
(billion; prefix giga- (G))

= 345600000000
(real number)
4 0
3 years ago
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