please help me out with these !! 50 points would greatly appreciate it.

Please Help Me Out With These !! 50 Points Would Greatly Appreciate It.
Please Help Me Out With These !! 50 Points Would Greatly Appreciate It.
Please Help Me Out With These !! 50 Points Would Greatly Appreciate It.
Please Help Me Out With These !! 50 Points Would Greatly Appreciate It.
Please Help Me Out With These !! 50 Points Would Greatly Appreciate It.

Answers

Answer 1

Answer:

Its nymber 2

Explanation:


Related Questions

how can you rewrite the force formula (f=ma) to solve the acceleration?​

Answers

a=0.5 Nkg=0.5 kg⋅m/s2kg=0.5 m/s2

The force formula can be rewritten  to solve the acceleration as:

acceleration = force/mass.

What is acceleration?

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

What is force?

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

From Newton's 2nd law of motion, we can write that:

Force = mass × acceleration

acceleration = force/mass.

Hence, the force formula can be rewritten  to solve the acceleration as:

acceleration = force/mass.

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Explain two reasons why astronomers are continually building larger and larger telescopes. Explain two reasons why astronomers are continually building larger and larger telescopes. Larger telescope mirrors have a larger surface area and can therefore collect more light, which makes faint objects bright enough to detect. Also, larger telescope mirrors produce more scattering of light due to diffraction, which contributes to better angular resolution. Larger telescope mirrors have a larger surface area and can therefore collect more light, which contributes to better angular resolution. Also, larger telescope mirrors produce more scattering of light due to diffraction, which makes faint objects bright enough to detect. Larger telescope mirrors have a larger surface area and can therefore collect more light, which contributes to better angular resolution. Also, larger telescope mirrors produce less scattering of light due to diffraction, which makes faint objects bright enough to detect. Larger telescope mirrors have a larger surface area and can therefore collect more light, which makes faint objects bright enough to detect. Also, larger telescope mirrors produce less scattering of light due to diffraction, which contributes to better angular resolution.

Answers

Answer:

* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected

* arger mirrors decreases the angle of dispersion giving a better resolution of the bodies

Explanation:

Refracting telescopes get bigger every day for two main reasons.

* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected

* the diffraction process for circular apertures is given by

               θ = 1.22 λ / D

where d is the diameter of the mirror, therefore having larger mirrors decreases the angle of dispersion giving a better resolution of the bodies

Help please ! Ill give brainliest !! ☁️✨

Answers

Answer:
Force Meter > used to measure force
Milli > Prefix that means 1/1,000
Centi > Prefix that means 1/100
Kilo > Prefixes that means 1,000
Thermometer > used to measure Temperature


Explanation: uh just trust

The study of heat is ____?

Answers

Explanation:

thermodynamics is the study of heat.

Answer The study of heat and its relationship to useful work is called thermodynamics and involves macroscopic quantities such as pressure, temperature, and volume without regard for the molecular basis of these quantitie

Explanation:

If an ocean wave has a wavelength of 2 m and a frequency of 1 wave/s, then its speed is m/s Enter the answer Check it CRATCHPAD Improve this questic 트​

Answers

Answer:

2m/s

Explanation:

v=f×wavelength

v=2×1

=2m/s

I have a massive rock weighing 3,000 Newtons but I can only accelerate it to 500 m/s2 what is its mass?

Answers

Answer:

6 kg

Explanation:

F=ma

F is Force(newtons)

m is mass(kg)

a is acceleration(m/s^2)

Plug in the numbers

3000 = m(500)

divide both sides by 500 to cancel out the 500.

3000/500=6

6 = m

6kg is the mass

If a 500-pound object is moved 200 feet how much work is being done?
a. 200 FT LB
b. 500 FT LB
c. 1000 FT LB
d. 100,000 FT LB

Answers

Answer:

D

Explanation:

Work = Distance x Mass

work done = 100,000 FT LB

What is work done ?

Work is done whenever a force moves something over a distance or The work done by a force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.

Work done = force * displacement

given :

force = 500 pound

displacement = 200 feet

work done = 500 * 200 = 100,000 FT LB

correct option is d. 100,000 FT LB

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At what latitude are there almost no differences between the seasons? Explain
why this occurs?

Answers

Answer:

The four-season year is typical only in the mid-latitudes. The mid-latitudes are places that are neither near the poles nor near the Equator. The farther north you go, the bigger the differences in the seasons.

Explanation:

hope this helps have a good day :) ❤

At a latitude equal to zero degrees there is little seasonal variation. This phenomenon is due each day the Sun's rays strike the Earth's surface at approximately the same angle near the Equator.

The Equator is the line of (zero degrees) latitude around the middle of the Earth.

Moroever, the intensity of solar radiation and therefore also the temperature at the Earth's surface largely depends on the angle of incidence of the Sun's rays.

At 0° latitude, there is a very little seasonal variation because all days the Sun's rays strike the Earth's surface at approximately the same angle. At the Equator, the Sun's rays strike the Earth's surface at an angle of 90°, causing warmer temperatures compared to higher latitudes.

In additon, at 0° latitude, all days also have the same number of hours of light and dark (i.e., approximately 12 hours of sunlight).

In conclusion, at a latitude equal to zero degrees (i.e., at the Equator) there is little seasonal variation. This phenomenon is due each day the Sun's rays strike the Earth's surface at approximately the same angle near the Equator.

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What is the period, in seconds, that corresponds to each of
the following frequencies: (a) 10 Hz, (b) 0.2 Hz, (c) 60 Hz?

Answers

Answer:

0.1s,5s,0.017s

Explanation:

T=1÷frequency

Answer:

a =

✔ 6

The period is

✔ 2 seconds.

b =

✔ pi

Explanation:

Graph the function using the graphing calculator. Find the least positive value of t at which the pendulum is in the center.

t =

✔0.5 sec

To the nearest thousandth, find the position of the pendulum when t = 4.25 sec.

d =

✔ 4.243 in.

You are inside the Great Hall, 15 m from the north wall with the doors to the RMC, and centered between two open doors that are 3 m apart. Someone is blairing a 200 Hz tone outside the Great Hall so that it enters the doors as a plane wave. You hear a maximum intensity in your current position. As you walk along the direction of the wall with the doors (but maintain a distance 15 m from the wall), how far will you walk (in m) to hear a minimum in the sound intensity

Answers

Answer:

Δr = 0.425 m

Explanation:

This is a sound interference exercise, the expression for destructive interference is

          Δr = (2n + 1) λ / 2

in this case the movement is in the same direction as the sound, therefore the movement is one-dimensional

let's use the relationship between the speed of sound and its frequency and wavelength

          v = λ f

          λ = v / f

the first minium occurs for n = 0

           Δr = λ / 2

            Δr = v / 2f

            Δr = [tex]\frac{340}{2 \ 400}[/tex]

             Δr = 0.425 m

this is the distance from the current position that we assume in the center of the room


The kinetic theory states that the higher the temperature, the faster the

Answers

Answer: the higher the kinetic energy

Explanation:

20 points!!!! A 2,00ON steel rod that is 5 meters long is placed in a corner between the floor and a wall, and balanced at an angle using a cord attached to the wall The rod is balanced such that its top end is 2.38 meters away from the wall, The cord is 40 cm long, and it is attached to the wall at a height of 75 cm above the floor. The diagram to the right shows the situation If the lower end of the rod does not slip from the corner, what is the tension in the cord?

Answers

Answer:

WE NEED TO ADD ALL 40+2.38 +75+5

Explanation:

PLSE GIVE SOME POINTS DUDE

8) A train enters a curved horizontal section of the track at a speed of 100 km/h and slows down with constant deceleration to 50 km/h in 12 seconds. If the total horizontal acceleration of the train is 2 m/s2 when the train is 6 seconds into the curve, calculate the radius of curvature of the track for this instant.

Answers

Answer:

the radius of curvature of the track for this instant is 266 m

Explanation:

Given that;

The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s

velocity of the train at t=12 s is;

[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s

now, we calculate the deceleration of the train

[tex]V_{t=12}[/tex]  = u + at

13.89 = 27.77 + [tex]a_{t}[/tex]12

[tex]a_{t}[/tex] = (13.89 - 27.77) / 12

[tex]a_{t}[/tex] = -13.88 / 12

[tex]a_{t}[/tex] = - 1.1566 m/s²

Now, the velocity of the train at 6 seconds is;

[tex]V_{t=6}[/tex]  = u + at

[tex]V_{t=6}[/tex]  = 27.77 + ( - 1.1566 m/s²)6

[tex]V_{t=6}[/tex]  = 27.77 - 6.9396

[tex]V_{t=6}[/tex]  = 20.83 m/s

The acceleration at t=6 s is;

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

we substitute

2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]

4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²

4 = 1.3377 +  ([tex]a_{n}[/tex])²

([tex]a_{n}[/tex])² = 4 - 1.3377

([tex]a_{n}[/tex])² = 2.6623

[tex]a_{n}[/tex] = √2.6623

[tex]a_{n}[/tex]  = 1.6316 m/s²

Now the radius of curve is;

a = V² / p

[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² /  [tex]a_{n}[/tex]

[tex]p_{t=6}[/tex] = ( 20.83 m/s )² /  1.6316 m/s²

[tex]p_{t=6}[/tex] = 433.8889 / 1.6316

[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m

Therefore;  the radius of curvature of the track for this instant is 266 m

Which of the following variables can be measured in joules?
A. momentum
B. Energy
C. Power
D. Work

Answers

Answer:

The variables that can be measured in joules are

B. Energy

D. Work

Hope it will help :)

An electron moves from point i to point f, in the direction of a uniform electric field. During this motion:Group of answer choicesthe work done by the field is positive and the potential energy of the electron-field system increasesthe work done by the field is negative and the potential energy of the electron-field system increasesthe work done by the field is positive and the potential energy of the electron-field system decreasesthe work done by the field is negative and the potential energy of the electron-field system decreasesthe work done by the field is positive and the potential energy of the electron-field system does not change

Answers

Answer:

the work done by the field is positive and the potential energy of the electron field system decreases

Explanation:

This exercise asks to find the work and the potential energy of an electron in an electric field.

Work is defined by

         W = F .d = F d cos θ

         

the electric force is

          F_e = q E

         W = q E d cos θ

         

since the charge of the electron is negative the force is in the opposite direction to the electric field

          W = - e E d

we select the direction to the right is positive, point i is to the left of point f,

therefore the work moving from point i to point F has two possibilities

* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated

* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too.  The electron moves away from this point, this is in accordance with the indicated movement

 

In the latter case the electric field lines go from f to i point, therefore the Work is positive

Now let's examine the potential energy

            ΔU = - q E .d

so we see that this definition is related to work,

            ΔU = -W

Therefore, as the work is positive, the power energy must decrease

When reviewing the different answers, the correct ones are:

the work done by the field is positive and the potential energy of the electron field system decreases

The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.

An electron moves from point i to point f, in the direction of a uniform electric field, then  the potential energy of the electron can be calculated s given below.

[tex]\Delta V=-qEd[/tex]

Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.

The work done by the electron in the uniform electric field can be calculated as,

[tex]W = F\times d \times cos\theta[/tex]

Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and  for uniform electric field, the value of [tex]\theta[/tex] is zero.

Hence  [tex]W=F\times d\times 1\\W=F \times d[/tex]

Electric force  [tex]F = q E[/tex]

By substituting the value of electric force on the above formula,

[tex]W = qEd[/tex]

Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.

[tex]W = -\Delta V[/tex]

The work done by the electron is negative and the potential energy of the electron increases.

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Analyze the data to identify the mathematical relationship between the
amplitude and energy of a mechanical wave. If mechanical wave A has an
amplitude of 4 cm and mechanical wave B has an amplitude of 5 cm, what
will be the relationship between the energy carried by the two waves?
Amplitude
Energy
A. Wave A has about 1.25 times more energy than wave B.
ОО
B. Wave A has about 1.6 times more energy twan wave B.
C. Wave B has about 1.6 times more energy than wave A.
O D. Wave A has about 1.15 times more energy than wave B.

Answers

Answer:

Its C

Explanation:

Because I got it wrong for you

Wave B has about 1.6 times more energy than wave A.

What is energy?

Energy is the ability or capability to do tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.

The amplitude and energy of a mechanical wave. If mechanical wave A has an amplitude of 4 cm and mechanical wave B has an amplitude of 5 cm wave B has about 1.6 times more energy than wave A.

Wave B has about 1.6 times more energy than wave A.

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What is the potential energy of an object 20 m in the air with a
mass of 600 kg?

Answers

Answer:

Ep = 117600 J

Explanation:

Data:

Mass (m) = 600 kgHeight (h) = 20 mGravity (g) = 9.8 m/s²Potential Energy (Ep) = ?

Use formula:

Ep = m * g * h

Replace:

Ep = 600 kg * 9.8 m/s² * 20 m

Multiply operations, and units:

Ep = 117600 J

What is the potential energy?

The potential energy is 117600 Joules.

Rank the four fundamental forces from strongest to weakest. Use 1 to indicate the strongest force and 4 to indicate the weakest force. The gravitational force: The electromagnetic force: The strong nuclear force: The weak nuclear force:

Answers

Answer:

4

2

1

3

Explanation:

Be safe, lovelies <3

A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target?

Answers

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

We have that the speed  must be at the speed below if they want to just hit their target

From the Question we are told that

Distance [tex]d=50m[/tex]

Height [tex]h=70m[/tex]

Constant Velocity [tex]v= 7.0 m/s[/tex]

Generally the equation for the time  is mathematically given as

[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]

Therefore

The velocity required to make horizontal movement is

[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]

Given that

Velocity on the Vertical axis is

[tex]v_y=7m/s[/tex]

Velocity on the  horizontal axis is

[tex]v_x=5m/s[/tex]

Therefore resultant speed

[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]

[tex]v_r=8.6023m/s[/tex]

In conclusion

[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target

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a 90 kilogram dog runs across the dog park at a speed of 6.5 meters per second. what is the magnitude and direction of the average force required to stop the dog in .85 seconds?

Answers

Answer:

am not sure about the answer

Explanation:

you need to find out the amount of force it's going in for example 10n or 100n then you need to times it the distance then devide by the time

What is the average speed of a car that travels 60 meters in 2
seconds?

Answers

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

The electric potential inside a living cell is lower than the potential outside. Suppose the electric potential difference between the inner and the outer cell wall is 0.095 V, a typical value. To maintain the internal electrical balance, the cell pumps out sodium ions. How much work must be done to remove a single sodium ion (charge e)

Answers

Answer:

1.52 × 10⁻²⁰ J

Explanation:

The electrical potential difference is defined as the amount of work done in carrying a unit charge from one point to another point in an electric field. Electric potential difference is measured in volts. It is given by the formula:

ΔV = ΔU / q

ΔV is electric potential difference between the two points, ΔU is the work done and q is the unit charge.

Given that ΔV = 0.095 V, q = 1.6 × 10⁻¹⁹ C. Hence:

ΔU = ΔV.q

ΔU = 0.095 V * 1.6 × 10⁻¹⁹ C

ΔU = 1.52 × 10⁻²⁰ J

The electric potential difference is the amount of effort done in an electrical field to shift a unit charge from one spot to another.The electric potential difference will be 1.52×10⁻²⁰J.

What is the electric potential difference ?

The electrical potential difference is the amount of effort done in an electrical field to shift a unit charge from one spot to another.

Traditional current flows from positive to negative terminals, signifying positive charge transfer in that direction.

The given data in the problem is

[tex]\triangle V[/tex] is the electric potential difference between the inner and the outer cell wall = 0.095 V

[tex]\triangle U[/tex] is the amount of work done

q is the charge on the electron =1.6×10⁻¹⁹C

Electric potential difference is given by the formula

[tex]\triangle V=\frac{\triangle U}{q} \\\\ \triangle U=\triangle Vq\\\\\triangle U=0.095 V\times1.6\times10^{-19}\\\\ \triangle U=1.52\times10^{-20}J[/tex]

Hence the electric potential difference will be 1.52×10⁻²⁰J.

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Did I hear correctly that the speed of light is different in deep space observation?

Answers

Answer:

Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.

The plates of a parallel plate capacitor each have an area of 0.40 m2 and are separated by a distance of 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated. Determine the magnitude of the electric field between the capacitor plates.

Answers

Answer:

 E = -1.5 10⁵ N / C

Explanation:

In a capacitor the electric field is uniform between the blades, therefore we use the expression

                   V = - E s

                   E = - V / s

let's calculate

                   E = - 3000 / 0.02

                   E = -1.5 10⁵ N / C

the sign indicates that the field and the potential are opposite, when one increases the gold decreases

There is a very long straw of charge that is uniformly charged in electro static equilibrium. It has a charge per unit length of 4.0E-9 C/m (4.0 nC/m) and a radius of 0.5 m. What is the strength of the electric field a distance of 10.0 m from its center outside the straw

Answers

Answer:

2880 N/c

Explanation:

Given that:

Charge per unit length ; λ = 4 * 10^-9

radius, r = 10

Radius, R = 0.5m

Using the relation :

2λr / 4πE0R²

Columb's constant, k = 1/4πE0 =. 9* 10^9Nm²/C²

Hence, we have :

2λrk/ R²

(2 * 4 * 10^-9 * 10 * 9 * 10^9) / 0.5^2

(720 ÷ 0.25)

= 2880 N/c

Which is the weakest of the four fundamental forces?
strong nuclear
weak nuclear
electromagnetic
gravitational

Answers

Answer:

Gravitational

Explanation:

gravitational

Answer:

Gravitational

Explanation:

In order from strongest to weakest.

Strong nuclear

Electromagnetic

Weak nuclear

Gravitational

help? its a short question

Answers

Answer:

i think its ancestor

Explanation:

sry if im wrong

Answer:

scientists compare organisms DNA to support the theory that all species share a common Ancestor.

An 80 N rightward force is applied to a 10 kg object to accelerate it to the right.
The object encounters a friction force of 50 N.

Answers

net force = 30 N

mass = 8.16 kg

acceleration = 3.68 m/s²

Further explanation

Given

80 N force applied

mass of object = 10 kg

Friction force = 50 N

Required

Net force

mass

acceleration

Solution

net force

Net force = force applied(to the right) - friction force(to the left)

Net force = 80 - 50 = 30 N

mass

Gravitational force(downward) : F = mg

m = F : g

m = 80 : 9.8

m = 8.16 kg

acceleration

a = F net / m

a = 30 / 8.16

a = 3.68 m/s²

Explain why your PE and KE are usually not both high at the same time (If PE is high then usually KE is low)

Answers

might help:

an object can have both kinetic and potential energy at the same time. for example, an object which is falling, but has not reached the ground has kinetic energy because it is moving downwards, and potential energy because it is able to move downwards even further than it already has. as an object falls its potential energy decreases, while its kinetic energy increases. the decrease in potential energy is exactly equal to the increase in kinetic energy.

A warm hockey puck has a coefficient of restitution of 0.50, while a frozen hockey puck has a coefficient of restitution of only 0.35. In the NHL, the pucks to be used in games are kept frozen. During a game, the referee retrieves a puck from the cooler to restart play but is told by the equipment manager that several warm pucks were just put into the cooler. To check to make sure he has a game-ready puck, the referee drops the puck on its side from a height of 2 m. How high should the puck bounce if it is a frozen puck

Answers

Answer:

the required height is 0.2449 m only

Explanation:

Given the data in the question;

Initial height = 2m

so speed of the puck before hitting the ground will be;

u² = 2gh

Initial speed u_ball = √2gh

u_ball = √( 2 × 9.8 × 2 )

u_ball =  √39.2

u_ball = 6.26 m/s

given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only

R = - (v_ball - v_ground / u_ball - u_ ground)

so

0.35 = - (v_ball - 0 / 6.26 - 0)

0.35 = -v_ball / - 6.26

-v_ball = 0.35 × (- 6.26)

-v_ball = -2.191 m/s

v_ball = 2.191 m/s

to get the height;

v² = 2gh

h = v² / 2g

we substitute

h = (2.191)² / 2×9.8

h = 4.800481 / 19.6

h = 0.2449 m

Therefore, the required height is 0.2449 m only

Other Questions
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