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Leni [432]
3 years ago
10

Prior to determining the experimental design, a scientist typically? A. makes observations. B. forms a hypothesis. C. performs a

n experiment. D. predicts the result of an experiment.
Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

Prior to determining the experimental design, a scientist typically forms a hypothesis. The answer is letter B. this is to prepare the scientist, the possible outcome of their research before the experimental design whether they are wrong or not.

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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
What is the velocity of a 50kg skater if her momentum is 225kg. m/s?
Gnom [1K]

|Momentum| = (mass) x (speed)

225 kg-m/s =(50kg) x (speed)

Divide each side by (50kg):  Speed=(225 kg-m/s) / (50 kg) = 4.5 m/s .

Regarding the velocity, nothing can be said other than the speed, because
we have no information regarding the direction of the object's motion.

8 0
3 years ago
Read 2 more answers
How many electrons have been removed from a positively charged electroscope if it has a net charge of 6x10-11?
faltersainse [42]

Answer:

<h3>Because one Coulomb of charge is an abnormally large quantity of charge, the units of microCoulombs (µC) or nanoCoulombs (nC) are more commonly used as the unit of measurement of charge. To illustrate the magnitude of 1 Coulomb, an object would need an excess of 6.25 x 1018 electrons to have a total charge of -1 C.</h3>

Explanation:

<h3><em><u>mark as brainliast</u></em></h3><h3><em><u>indian </u></em><em><u>genius </u></em><em><u>s</u></em><em><u>a</u></em><em><u>r</u></em><em><u>thak</u></em></h3>
4 0
3 years ago
Does anyone know this?!
Valentin [98]

Answer:

2 is the numerical answer.

Explanation:

Hello there!

In this case, according to the given information and formula, it is possible for us to remember that equation for the calculation of the average kinetic energy of a gas is:

KE=\frac{3}{2} \frac{R}{N_A} T

Whereas R is the universal gas constant, NA the Avogadro's number and T the temperature.

Which means that for the given ratio, we can obtain the value as follows:

=\frac{\frac{3}{2} \frac{R}{N_A} T_1}{\frac{3}{2} \frac{R}{N_A} T_2} \\\\=\frac{T_1}{T_2} \\\\=\frac{500K}{250K} \\\\=2

Regards!

8 0
3 years ago
An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
4 years ago
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