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Makovka662 [10]
3 years ago
12

A 3 Volt battery is connected in series to three resistors: 4, 6, and 2

Physics
1 answer:
DanielleElmas [232]3 years ago
7 0

Answer:

Explanation:

   

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Magnetism, reactivity, and fluorescence are three special properties used to identify minerals. Please select the best answer fr
astraxan [27]

Answer: True

Explanation: Following are the properties that are used to identify minerals:

(1) Color

(2) Hardness

(3) Magnetism : It is the property of mineral to attract or repel with other magnetic materials.

(4) Luster : It is the property that shows the surface is reflecting light or not.

(5) Reactivity: Reactivity is also responsible to identify mineral. How the mineral is reacting with acids, bases etc helps to identify type of minerals.

3 0
3 years ago
Read 2 more answers
What is the melting point of substance A?
Misha Larkins [42]

Answer:

Solids are easily recognized by their ability to retain a fixed shape and definite volume. Particles making

up a solid are held together in a rigid form. They are not free to move about or slide past one another and

the solid does not have the ability to flow. (Although the particles of a solid do not move position to position, they do have motion in that they are constantly vibrating.

To change the temperature of a solid, heat energy must be added. The amount of heat energy that changes

the temperature of 1.0 g of a solid by 1.0°C is called its specific heat (c). Each substance has its own

specific heat. The specific heat of ice is 2.1 Joules/g°C. In other words we must supply 1.0 gram of ice

with 2.1 Joules of heat energy to raise its temperature by 1.0 °C.

The general equation for calculating heat energy to change the temperature of a solid is:

Heat = Mass x Specific Heat (solid) x Temperature Change

Q = m c DT

10 g 10 g 10 g 10 g 10 g 10 g

Calculate the heat necessary to change 10 g of ice(s) at -20 °C to 10 g of ice(s) at 0°C. (A-B)

Q = mc∆T = (10 g) (2.1 J/g°C) (20°C) = 420 J

If you continue to add heat energy once the temperature of the ice reaches 0°C , the heat absorbed is called

the heat of fusion (Lf). This heat is used to cause a change of phase (from a solid to a liquid). This heat is

increasing the potential energy of the molecules of the solid. No temperature change takes place. Each

substance has its own heat of fusion. The heat of fusion for ice is 340 Joules/g. Exactly the same amount

of heat is given up when 1.0 g of water is changed to ice. This heat is called the heat of crystallization.

The general equation for calculating heat energy to change a solid to a liquid is:

Heat = Mass x Heat of Fusion

Q = m Lf

Calculate the heat necessary to change 10 g of ice(s) at 0°C to 10 g of water(l) at 0°C.(B-C)

Explanation:

Q = mLf = (10 g)( 340 J/g) = 3400 J

3 0
3 years ago
An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.6 m/s. A child on a bridge, 71 m abo
saveliy_v [14]

Answer:

21.28 m

Explanation:

height, h = 71 m

velocity of raft, v = 5.6 m/s

let the time taken by the stone to reach to raft is t.

use second equation of motion for stone

h = ut + \frac{1}{2}at^{2}

u = 0 m/s, h = 71 m, g = 9.8 m/s^2

71 = 0 + 0.5 x 9.8 x t^2

t = 3.8 s

Horizontal distance traveled by the raft in time t

d = v x t = 5.6 x 3.8 = 21.28 m

3 0
3 years ago
What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?
Ivahew [28]

The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.

Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima.  The latitude in the middle
of that intersection is 46.585° North.  <u>That's</u> the number we need.

Here's how I would do it:

-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.

-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°. 

-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.

This sets the limit of the highest in the sky that the moon can ever appear.

90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .

That doesn't happen regularly.  It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).

Depending on the time of year, that can be any time of the day or night.

The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.

In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky.  Then it's going to be somewhere near
67° above the horizon at midnight.


5 0
3 years ago
In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0
3 years ago
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