examples of sound to radiant energy

Answers

Answer 1

Answer:

Examples of Radiant Energy All Around You

The term radiant energy refers to energy that travels by waves or particles, particularly electromagnetic radiation such as heat or x-rays. Radiant energy is created through electromagnetic waves and was discovered in 1885 by Sir William Crookes. Fields in which this terminology is most often used are telecommunications, heating, radiometry, lighting, and in terms of energy created from the sun. Radiant energy is measured in joules.

Everyday Examples of Radiant Energy

Virtually anything that has a temperature gives off radiant energy. Some examples of radiant energy include:

•The heat emitted from a campfire

•Emission of heat from a hot sidewalk

•X-rays give off radiant energy

•Microwaves utilize radiant energy

•Space heaters produce radiant energy

•Heat created by the body can be radiant energy

•Lighting fixtures

√Home heating units

•Fixtures that convert solar energy to heat

•Visible light

•Gamma rays

•Radio waves

•Electricity

•A surface heated by the sun converts the energy of the light into infrared energy which is a form of radiant energy

•Cell phones utilize radiant energy to function

•Magnetic motor generators that utilize •neodymium magnets create radiant energy

•Audio signals that come to home or cars via radio waves

•Ultraviolet light

√Infrared radiation

•The light emitted from a campfire

•The light generated from a light bulb

•A heated brake disc giving off heat

•The heat from a grill used for cooking

•Water can reflect or absorb radiant energy

•Soil can absorb radiant energy

•Light from the sun

•Heat emitted from a bunsen burner

•Heat from an overused computer

•Heat caused by friction

•Heat emitted from a dryer

•The heat generated by a light bulb

•Heat generated through reflection of visible light

•A window reflects radiant energy

Heat created from a stove or oven

•Heat emitted from a washing machine


Related Questions

Steelhead trout migrate upriver to spawn. Occasionally they need to leap up small waterfalls to continue their journey. Fortunately, steelhead are remarkable jumpers, capable of leaving the water at a speed of 8.0 m/s. What is the maximum height that a steelhead can jump

Answers

Answer:

s = 3.26 m

Explanation:

Given that,

Water leaves at a speed of 8 m/s

We need to find the maximum height that steelhead can jump. Let it can jump to a height of h.

At maximum height, final speed is equal to 0. We can use third equation of motion to find the maximum height.

[tex]v^2-u^2=2as[/tex]

a = -g

[tex]-u^2=-2gs\\\\s=\dfrac{u^2}{2g}\\\\s=\dfrac{(8)^2}{2\times 9.8}\\\\=3.26\ m[/tex]

Hence, the maximum height is 3.26 m.

initial velocity 10 m/s accelerates at 5 m/s for 2 seconds whats the final velocity

Answers

Answer:

The final velocity is 20 m/s.

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

[tex]v_f=v_o+at[/tex]

The provided data is: vo=10 m/s, [tex]a=5\ m/s^2[/tex], t=2 s. The final velocity is:

[tex]v_f=10~m/s+5\ m/s^2\cdot 2\ s[/tex]

[tex]v_f=20\ m/s[/tex]

The final velocity is 20 m/s.

A 80 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 5 meters starting from rest, its speed is 6.0 m/s. Find the magnitude of the net force on the bobsled.

How do you solve this question?

Answers

Answer:

F = 288 [N]

Explanation:

To solve this problem we must use the following equation of kinematics and find the value of acceleration.

[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]

where:

Vf =  final velocity = 6 [m/s]

Vo =  initial velocity = 0 (starting from rest)

a = acceleration [m/s²]

x = distance = 5 [m]

Now replacing, we have:

[tex](6)^{2}=0+(2*a*5)\\36=10*a\\a = 3.6 [m/s^{2}][/tex]

Since we already have the value of acceleration, we can use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

[tex]F =80*3.6\\F = 288 [N][/tex]

Thorium^+2

Chemical symbol:
Atomic Number:
Mass: 232
# of protons
# of neutrons
Group #
Period #

Answers

Answer:

chemical symbol: Th

atomic number:90

protrons :90

neutrons:142

group#:4

period#: 9

Explanation:

you take the atomic weight (232.038)and subtract the atomic number to get (90) which is your neutrons

What is the rms current flowing through a light bulb that uses an average power of 60.0 W when it is plugged into a wall receptacle supplying an rms voltage of 120.0 V?

Answers

Answer:

I = 0.5 A

Explanation:

Given that,

Average power of a light bulb, P = 60 W

The rms voltage supplied is 120 V

We need to find the value of rms current flowing through the light bulb. The relation between the power, voltage and current is given by :

P = VI

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{60}{120}\\\\I=0.5\ A[/tex]

So, the rms value of current is 0.5 A.

Any living thing is called an organism,no matter if it is one-celled or many-celled. True or False?.

Answers

Answer:

I think it's most likely true.

Explanation:

any organism has the properties of a living thing, which includes cells, whether it has one cell or many

Answer:

False

Explanation:

An organism is a living thing that is a single-celled life form

PLZ HELP ILL MARK BRAINLEIST!!!!
Why did the bowling ball make a bigger
splash than the ping pong ball?

What kind of energy made that splash
happen?

Answers

Because it has more mass and the more mass a object haves the more it’s gonna do impact to someone!It made a lot of energy

Answer:

a. The bowling ball would have more kinetic energy because of its greater mass.

b. Potential energy

Explanation

a. Bowling ball has higher mass, self explanatory.

b. A high diver has lots of stored energy when they are on the diving platform. When they dive this stored energy helps make the splash when they hit the water. Stored energy is also called potential energy.

What is the flow sensitivity of a biosensor?​

Answers

Answer:

Sensitivity of biosensor

The biosensor showed good linear correlation in the wide detection range of 0.001–2000 ng/mL with good sensitivity. In addition, it retained its biosensing property for seven days with high reproducibility

Explanation:

A rotating heavy wheel is used to store energy as kinetic energy. If it is designed to store 1.00 x 106 J of kinetic energy when rotating at 64 revolutions per second, find the moment of inertia (rotational inertia) of the wheel. (Hint: Start with the expression for rotational kinetic energy.)

Answers

We know, [tex]1\ rpm = \dfrac{2\pi}{60} \ rad/s[/tex] .

[tex]64\ rpm\ is = \dfrac{2\pi}{60}\times 64\ rad/s\\\\= \dfrac{32\pi}{15}\ rad/s[/tex]

We know, kinetic energy is given by :

[tex]K.E = \dfrac{I\omega^2}{2}\\\\I = \dfrac{2(K.E)}{\omega^2}\\\\I = \dfrac{2\times 10^6}{\dfrac{32}{15}\times \pi}\\\\I = 298415.52 \ kg \ m^2[/tex]

Hence, this is the required solution.

A horse has a momentum of 1200 kg·m/s. If the horse has a mass of 313 kg, what is the speed of the horse?

Answers

Answer:

3.83 m/s

Explanation:

The speed of the horse can be found by using the formula

[tex]v = \frac{p}{m} \\ [/tex]

p is the momentum

m is the mass

From the question we have

[tex]v = \frac{1200}{313} \\ = 3.83386..[/tex]

We have the final answer as

3.83 m/s

Hope this helps you

If 710-nm and 655-nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.1 m away?

Answers

Answer:

Explanation:

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.1 m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2403.07 x 10⁻⁶ m

= 2.403 x 10⁻³ m

= 2.403 mm .

For λ = 655 nm

position = 2 λ D / d

λ = 655 nm , D = 1.1 m

d = .65 x 10⁻³

position 2 = 2 x 655 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2216.91 x 10⁻⁶ m

= 2.217 x 10⁻³ m

= 2.217 mm .

Difference between their position

= 2.403 - 2.217 = .186 mm .

While getting buff at the gym you lift a bunch of weights applying 1000N of force to lift them from the ground to a height of 2m. How much work did you do?

A. 2000 J
B. 1000J
C. -2000 J
D. -1000 J

Answers

A. 2000 J............

If Jack weighs more that Jill, and they run up the same hill, who has done more work?


If Jack and Jill weigh the same, and Jill runs up the hill in half the time as Jack, who had more power?

Answers

Answer:

jack has done more work pulling more weight and Jill has more power.

Explanation:

PLEASE ans The question's in the pictures, please don't answer what already has answers. Only answer if you can finish both pages completely PLEASE I NEED HELP :(( if ur ans is relevant I will mark brainliest

Answers

Bro it’s no picture lol

A wave travels at 175 m/s along the x-axis.If the period of the periodic vibrations of the wave is 3.00 milliseconds,then what is the wavelength of the wave?
A) 25.5 cm
B) 35.6 cm
C) 42.9 cm
D) 49.5 cm
E) 52.5 cm

Answers

Answer:  

E) 52.5 cm

Explanation:

Step one:

given data

period T= 3 milliseconds= 0.003

velocity v= 175m/s

wave lenght λ=?

Step two:

we know that f=1/T

the expression relating period and wave lenght is

v=λ/T

λ=v*T

λ=175*0.002

λ=0.525m

to cm= 0.525*100

=52.5cm

The wavelength of the wave is E) 52.5 cm

Which of the following is true?
A
The Atlantic, Pacific, Indian, Arctic, and Southern Oceans are completely separate
from each other.
B
The ocean covers about half of the Earth's surface.
с
Scientists have studied most of the ocean, but a tiny bit remains unexplored.
D
Scientists know more about the moon than they do the ocean.

Answers

I think that d is true but I’m not for sure

Answer:

options B,C,D are true

Explanation:

Please help true or false

Answers

Answer:

the answer is true.......

A solid cylinder is released from the top of an inclined plane of height 0.50 m. From what height, in meters, on the incline should a solid sphere of the same mass and radius be released to have the same speed as the cylinder at the bottom of the hill?

Answers

Answer:

Explanation:

for rolling motion down the plane acceleration is given by the following expression

a = g sinθ / (1 + k² / R²)

here k is radius of gyration and R is radius of the object rolling down .

for cylinder I = 1/2 m R²

so k² = R² / 2

k² / R² = 1/2

a = g sinθ /( 1 + 1 / 2 )

= 2 / 3 x  g sinθ

v = √ 2 a s

= √ (2 x  2 / 3 x  g sinθ s )

= √ (4  / 3 x  g h  )

= √ (4  / 3 x  g x .5  )

= √ 2g / 3

for sphere  I = 2/5  m R²

so k² = 2/5 R²

k² / R² = 2 / 5  

a = g sinθ / (1 + 2 / 5)  

= 5 / 7  x  g sinθ

v = √ 2 a s

= √ (2 x  5 / 7  x  g sinθ s )

= √ (10/7  x  g h  )

Given

√ (10/7  x  g h  ) = √ 2g / 3

10/7  x  g h  = 2g / 3

h = 14 / 30 m

= .47 m .

Collision Lab
This activity will help you meet these educational goals:

You will explain or predict phenomena by exploring qualitative relationships between variables.
You will use positive and negative numbers to represent quantities in real-world contexts.
Directions
Read the instructions for this self-checked activity. Type in your response to each question, and check your answers. At the end of the activity, write a brief evaluation of your work.
Activity
Open this collision simulator and click Introduction. You’ll use the simulator to explore and compare elastic collisions and inelastic collisions. The mass and starting velocity of the colliding objects are kept constant. Follow the instructions in each part, and then answer the questions that follow. Use the math review if you need help with adding and subtracting negative numbers.

Question 1: Elastic Collisions
In this question, you will investigate elastic (bouncy) collisions. Be sure that the slider is to the extreme right (elasticity 100%).

Part A
Click Show Values in the upper-right corner. Study the boxes on the screen. What are the mass and initial velocity of ball 1 and ball 2?

I NEED HELP!


Part B
Part B
Click Play, and watch the balls collide. Then click Pause. What are the final velocities of ball 1 and ball 2?


The number line shows the starting and ending velocities for ball 1. What’s the change in velocity of ball 1? Calculate the value mathematically, and check it using the number line.

a number line showing an ending velocity of -0.50 meter/second and a starting velocity of 1.00 meter/second

Answers

Answer:

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

Explanation:

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

What is Collision?

A collision is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.

The following are a few instances of physical encounters that scientists might classify as collisions. Legs of an insect are said to collide with a leaf when it falls on one.

Every contact of a cat's paws with the ground while it strides across a lawn is seen as a collision, as is every brush of its fur with a blade of grass.

Therefore, Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

To learn more about collision, refer to the link:

https://brainly.com/question/13138178

#SPJ2

*Urgent* I WILL GIVE BRAINLIEST
Select the answer that helps conserve the most energy.
O walking to school
O driving a car to school
Otaking the bus to school​

Answers

Answer:

walking to school

Explanation:

Driving a car to school

, and taking the bus to school​ both take up energy, unlike walking to school.

unless ur talking about energy, counting energy you produce and use to complete things, then it would be the 3rd one, taking the bus to school.

Two blocks of masses 1.0 kg and 2.0 kg, respectively, are pushed by a constant applied force F across a horizontal frictionless table with constant acceleration such that the blocks remain in contact with each other, as shown above. The 1.0 kg block pushes the 2.0 kg block with a force of 2.0 N. The acceleration of the two blocks is

0

1.0 m/s2

1.5 m/s2

2.0 m/s2

3.0 m/s2

Answers

Answer:

1.0 m/s^2

Explanation: happy to help :)

Answer: [tex]1\ m/s^2[/tex]

Explanation:

Given

Masses of the block are [tex]m_1=1\ kg[/tex]  and

[tex]m_2=2\ kg[/tex]

Force applied by [tex]1\ kg[/tex] block on [tex]2\ kg[/tex] block is [tex]2\ N[/tex]

From the free body diagram of [tex]2\ kg[/tex] block, the net force on

[tex]\therefore m_2a=2\\\\\Rightarrow 2\times a=2\\\\\Rightarrow a=\dfrac{2}{2}\\\\\Rightarrow a=1\ m/s^2[/tex]

Thus, the acceleration of two blocks is [tex]1\ m/s^2[/tex]

Learn more: https://brainly.com/question/2361110

The fact that our preconceived ideas contribute to our ability to process new information best illustrates the importance of: the serial position effect. O repression iconic memory . semantic encoding . retroactive interference .

Answers

Answer:

It’s a

Explanation:

Don’t actually put that i needed the points mb

Which of the following is not an example of work being done on an object?

Pushing on a rock that will not move

Paddeling a canoe down a river

Lifting a bag of groceries

Throwing a ball across a field​

Answers

Answer:

Lifting a bag of groceries

Answer:

paddeling a canoe down a river :D or throwing a ball across a field

Explanation:

A child pulls a wagon across the grass so that it accelerates using a force of 50 N at an angle of 42 degrees above the ground. The loaded wagon has a mass of 12 kg. If the coefficient of friction between the wagon and grass is 0.64. What is the acceleration of the wagon? Describe the motion of the wagon.

Answers

Answer:

[tex]-1.398\ \text{m/s}^2[/tex]

Decelerating or slowing down

Explanation:

F = Force = 50 N

[tex]\theta[/tex] = Angle force is being applied = [tex]42^{\circ}[/tex]

[tex]\mu[/tex] = Coefficient of friction = 0.64

m = Mass of wagon = 12 kg

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Normal force is given by

[tex]N=mg-F\sin\theta[/tex]

Frictional force is given by

[tex]f=\mu N\\\Rightarrow f=\mu (mg-F\sin\theta)[/tex]

The force balance is given by

[tex]F\cos\theta-f=ma\\\Rightarrow \dfrac{F\cos\theta-\mu (mg-F\sin\theta)}{m}=a\\\Rightarrow a=\dfrac{50\times \cos42^{\circ}-0.64(12\times 9.81-50\times\sin42^{\circ})}{12}\\\Rightarrow a=-1.398\ \text{m/s}^2[/tex]

The acceleration of the wagon is [tex]-1.398\ \text{m/s}^2[/tex]. The negative sign indicates that the wagon is decelerating or slowing down.

The acceleration of the wagon is  [tex]-1.398 \;\rm m/s^{2}[/tex].

Given data:

The magnitude of pulling force is, F = 50 N.

The angle of inclination is, [tex]\theta = 42^{\circ}[/tex].

The mass of wagon wheel is, m = 12 kg.

Coefficient of friction between wagon and grass is, [tex]\mu =0.64[/tex].

The given problem is based on the concept of frictional force. The standard expression for the frictional force is,

[tex]f= \mu \times N[/tex]

Here, N is the normal force and its value is,

[tex]N=mg-Fsin \theta[/tex]

And the net force acting on wagon is,

[tex]F' = Fcos\theta -f\\\\ma = Fcos\theta -(\mu(mg-Fsin \theta))\\\\a = \dfrac{Fcos\theta -(\mu(mg-Fsin \theta))}{m}[/tex]

Here, a is the acceleration of wagon.

Solving as,

[tex]a = \dfrac{50 \times cos42 -(0.64(12 \times 9.8-(50 \times sin42)))}{12}\\\\a=-1.398 \;\rm m/s^{2}[/tex]

Thus, we can conclude that the acceleration of the wagon is  [tex]-1.398 \;\rm m/s^{2}[/tex].

Learn more about the frictional force here:

https://brainly.com/question/1714663

Which of the following is the recommended amount of fats per meal for male clients

Answers

Answer:

44 grams- 55 grams through the whole day. Probably about 14.6 grams per meal.

Explanation:

Answer:

2 thumbs (ISSA Guide)

Explanation:

A (10.0+A) g ice cube at -15.0oC is placed in (125 B) g of water at 48.0oC. Find the final temperature of the system when equilibrium is reached.
Specific heat of ice: 2.090 J/g K
Specific heat of water: 4.186 J/gK
Latent heat of fusion for water: 333 J/g

Answers

Answer: Final temperature is 34.15°C.

Explanation: When two objects have different temperature, they will exchange heat energy until there is no more net energy transfer between them. At that state, the objects are in thermal equilibrium.

So, when in equilibrium, the total heat flow must be zero, i.e.:

[tex]Q_{1}+Q_{2}=0[/tex]

In our case, there will be a change in state of ice into water, so total heat flow will be:

[tex]m_{1}c_{1}(T_{f}-T_{i})+m_{2}c_{2}(T_{f}-T_{i})+mL=0[/tex]

where

m₁ is mass of ice

m₂ is mass of water

c₁ is specific heat of ice

c₂ is specific heat of water

[tex]T_{f}[/tex] is final temperature

[tex]T_{i}[/tex] is initial temperature

L is latent heat fusion

Temperature is in Kelvin so the transformation from Celsius to Kelvin:

For ice:

T = -15 + 273 = 258K

For water:

T = 48 + 273 = 321K

Solving:

[tex]21(2.09)(T_{f}-258)+158(4.186)(T_{f}-321)+21(333)=0[/tex]

[tex]43.89T_{f}-11323.62+661.4T_{f}-212305.55+6993=0[/tex]

[tex]705.3T_{f}=216636.17[/tex]

[tex]T_{f}=[/tex] 307.15K

In Celsius:

[tex]T_{f}=[/tex] 34.15°C

Final temperature of the system when in equilibrium is 34.15°C

which of the following to all food chains depend on in an ecosystem

Answers

Answer:

The sun is the ultimate source of energy for all food chains. Through the process of photosynthesis, plants use light energy from the sun to make food energy. Energy flows, or is transferred through the system as one organism consumes another.

The students look through the side of the aquarium.

They notice that the image of the tongs appears to break as the tongs enter the water.

Which property of light are the students observing in this situation?

Answers

Answer:

light refraction

Explanation:

I need help understanding this question, so I know the arrow is traveling 80 meters per second, but it was launched from a starting point of 32 meters. I know for a fact an arrow does not have any thrust left at around 3 seconds of being in the air.

I just need someone to explain the questions and provide an answer to each.

Answers

Answer:

a) h(g) = 358,53 m

b) t = 8,16 s

c) t(t) = 16,71 s

Explanation:

Equations for vertical shooting are:

Vf = V₀ -  g * t   ;     h  =  V₀*t -  (1/2)*g*t²  ;    Vf² = V₀² - 2*g*h

And at maximum heigt Vf = 0 then

0 = V₀ - g * t

t = V₀/g                     V₀  = 80 m/s      and   g = 9,8 m/s²

t  =  80 / 9,8   (s)

t = 8,16 s

Then 8,16 s is the time to get maximum height

If we plug t = 8,16 (s) in equation  h  =  V₀*t -  (1/2)*g*t²

we get:     h (max)  =  (80)*8,16 - 0,5*9,8*(8,16)² (m)

h (max) = 652,8  -  326,27 m

h (max) = 326,53 m

Then relative to ground that height becomes

h(g) = 326,53 + 32

h(g) = 358,53 m

In order to get the time the arrow is in the air we proceed as follows:

a) for the arrow to be at the launched point will take the same time that from the launched point to the maximum height, and after that we have to find out the time the arrow takes from 32 m down to the ground level

Then  

t(t) = 8,16 + 8,16 + tₓ     (2)

Where tₓ  is the time from 32 m height to ground

h  =  V₀*tₓ -  (1/2)*g*tₓ²  but since the arrow now is going down then we change the sign of the second term on the right side of the equation

32 = (80)*tₓ  +  0,5 * 9,8 * tₓ²     Note that when the arrow is at 32 m height the speed is again V₀ = 80 m/s

32 = 80*tₓ  + 4,9*tₓ²

A second-degree equation for tₓ, solving it

4,9*tₓ² + 80*tₓ - 32 = 0

t₁,₂ = -80 ± √ 6400 + 627,2 / 9,8

t₁,₂ =( - 80 ± 83,8 ) / 9,8

there is not a negative time therefore we dismiss such solution and

t₁ = 3,8 / 9,8

t₁ = 0,39 s

And

t(t) = 8,16 + 8,16 + 0,39  s

t(t) = 16,71 s

if a person has a mass of 60 kg and a velocity of 2 m/s what is the magnitude of his momentum

Answers

Answer:

120 kg m/s

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 60 × 2

We have the final answer as

120 kg m/s

Hope this helps you

Other Questions
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