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4 years ago

1. Which areas of physics deal with the following?

Physics

2 answers:

Genrish500 [490]4 years ago

8 0

Answer:

1. how fast things move - mechanics

2. how the shape of a cave affects an echo - Sound & acoustics

3. which sunglasses are best for cutting the glare on a ski slope - optics

4. how the cooling system in a refrigerator works - thermodynamics

5. what lightning is - Electricity

6. how energy is produced by the sun – nuclear physics

Explanation:

Mechanics - the study of the motions of the objects

Sound - the study of the behavior of sound waves

Optics - the study of the behavior of light waves and optical instruments

Thermodynamics - the study of the relationship between heat and other forms of energy

Electricity - the study of properties of charges, their movement and their field interactions

Nuclear physics - study of atomic nucleus and its properties

lana66690 [7]4 years ago

5 0

ANSWER:

a. Kinematics.

b. Waves/Acoustics.

c. Optics/Polarization.

d. Thermal/Conservation of Energy.

e. Electricity and magnetism.

f. Renewable energy

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A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

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2 years ago

What were four of Galileo’s discoveries that were important to astronomy?

Scorpion4ik [409]

Explanation:

1. Phases of Venus: Galileo was the first astronomer to use a telescope to observe the celestial objects. Through a telescope he observed that Venus shows the phases just like the Moon. This proved the Heliocentric theory correct against the then prevalent Geocentric theory.

2. Law of Falling bodies: The acceleration due to gravity is independent of weight of the objects that means two bodies of different mass will hit the ground at the same time if dropped from the same height.

3. The uneven surface of the Moon: He observed that the surface of the Moon is uneven and rough.

4. Discovery of the 4 Moons of Jupiter

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3 years ago

If an object is placed at the center of carvature of a convance mirror the image formed is called

Natasha_Volkova [10]

Answer:

When the object is placed between centre of curvature and principal focus of a concave mirror the image formed is beyond C as shown in the figure and it is real, inverted and magnified.

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3 years ago

If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o

andrezito [222]

The answer is

18 / x = 12 / 4
12x = 72
x = 6mm

4 0

3 years ago

Read 2 more answers

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

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2 years ago