a man is trying to move a piano by himself, but it will not move, what force must they overcome to start the piano moving?

Titan is the only moon in the solar system with a significant atmosphere. which planet does titan orbit?

sveticcg [70]

"Saturn". It is largest moon of Saturn.

In short, Your Answer would be Option B

Hope this helps!

6 0

3 years ago

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

What is the meaning of the saying "the power of a lens is 1 dioptre "

timama [110]

Answer: 1 dioptre of power of a lens is defined as the unit of measuring the power of optical lens Or curved mirror which is equal to the reciprocal of the focal length. The focal length is measured in the meter. 1 dioptre is equal to 1/m where m is the focal length. Basically dioptre is the SI unit of optical power of the lens.

7 0

3 years ago

A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

$\Phi = \frac{2\pi \delta}{\lambda}$

Where

$\delta =$ Horizontal distance between two points

$\lambda =$ Wavelength

From our values we have,

$\lambda = 500nm = 5*10^{-6}m$

$\theta = 1\°$

The horizontal distance between this two points would be given for

$\delta = dsin\theta$

Therefore using the equation we have

$\Phi = \frac{2\pi \delta}{\lambda}$

$\Phi = \frac{2\pi(dsin\theta)}{\lambda}$

$\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}$

$\Phi= 1.096 rad \approx = 1.1 rad$

Therefore the correct answer is C.

6 0

3 years ago

A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after

vodka [1.7K]

Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

3 0

3 years ago