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lesya [120]
3 years ago
11

Write one example of a physical change and one example chemical change

Physics
1 answer:
mojhsa [17]3 years ago
6 0
Physical change = changes the physical properties (more commonly known as it's look)
Chemical change = changes the chemical properties into an entire new chemical form
Examples of physical change would be melting ice cubes or sugar cubes.
Examples of chemical change would be cooking eggs or burning paper because you're changing its chemical properties.
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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
Light of different colors is emitted from different stars because :.
Varvara68 [4.7K]

Answer:

Different temperatures

Explanation:

6 0
2 years ago
The skateboarder has a mass of 100 kg. When traveling downward from E to D, he reaches a velocity of 11 m/s. Calculate his kinet
patriot [66]

6050 J is the kinetic energy at D

<u>Explanation:</u>

In physics, the object's kinetic energy (K.E) defined as the energy it possesses during movement. It can be defined as the required work to accelerate a certain body weight in order to rest at a certain speed. When the body receives this energy as it speeds up (accelerates), it retains this energy unless speed varies. The equation is given as,

           K . E=\frac{1}{2} \times m \times v^{2}

Where,

m - mass of an object

v - velocity of the object

Here,

Given data:

m  = 100 kg

v = 11 m/s

By substituting the given values in the above equation, we get

            K . E=\frac{1}{2} \times 100 \times(11)^{2}=\frac{1}{2} \times 100 \times 121=\frac{12100}{2}=6050\ \mathrm{J}

6 0
3 years ago
Explain why, under some circumstances, it is not advisable to weld a structure that is fabricated with a 3003 aluminum alloy. Hi
Scorpion4ik [409]

The 3003 aluminum alloy is made up of 1.25% Magnesium and 0.1% Copper. This combination is designed to increase the strength of the material over other types of alloys such as those of the 1000 series. This alloy provides a medium strength and can be educated by cold work.

The alloy is not heat treatable and generally has good formability, corrosion resistance and weldability.

However, being a material that hardens by cold work, welding a 3003 Aluminum structure will cause the body to undergo recrystallization which will generate a loss in the 'resistance' of the material and the force capable of withstanding. If this aluminum will be used for structural purposes, it should not be welded. It would be better to perform the structure with a 6061 aluminum, which has similar characteristics and is not so affected by welding.

7 0
3 years ago
A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material
denis-greek [22]

The radius of the cylinder is equal to half the diameter:

r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm

The volume of the cylinder is given by:

V=\pi r^2 h=\pi (27.5 mm)^2 (54.0 mm)=1.28 \cdot 10^5 mm^3

where h is the heigth of the cylinder. Converting into meters,

V=1.28 \cdot 10^{-4} m^3

And the density of the material will be given by the ratio between the mass and the volume:

d=\frac{m}{V}=\frac{1 kg}{1.28 \cdot 10^{-4} m^3}=7812.5 kg/m^3

5 0
3 years ago
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