While an object near the earths surface is in free fall, its
A) velocity increases
B) acceleration increases

Answers

Answer 1

Answer:

a

Explanation:

The rate of change of an object's location with relation to a reference point is its velocity, which is dependent on time. when an object is dropped from space at rest (t = 0) under the influence of gravity, the velocity of the object changes and increases with time while the acceleration decreases.


Related Questions

Which sequence shows the chain of energy transfers that create surface currents on the ocean?

Answers

Answer:

The correct answer is A. The sun is the energy source of the surface currents in the ocean

The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .

What is surface current ?

Surface currents are currents that are located in the upper feet of the ocean , they are simply how water moves from one place to another . Pattern of surface current are determined by wind direction .

Surface currents are formed by global wind system that are fueled by energy from the sun . Because of heating effect of sun , the earth's atmosphere gets warmed up . As we know , warm air is lighter then cool air , it rises up and create low pressure near the equator because of this wind causes surface currents the ocean .

hence , The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .

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Which of the following choices is not an example of climate?
0000
San Diego has mild, warm temperatures and sea breezes year-round.
Anchorage has short, cool summers and long, snowy winters.
It will be 78° on Friday in Clovis.
Florida is tropical, with a significant rainy season.

Answers

Answer:

Florida is tropical, with a significant rainy seson

Which item will be shipped third?

—-

Answers

Answer:

I know it's groceries

Explanation:

electronics ship before clothing

electronics ship after groceries

urgent items are first so

order:

1.) A/Electronics

2.) Clothing/B

3.) Groceries(since groceries aren't urgent)

thing is it's C or D I'm leaning to D since it says it ships last but i dont know so if I'm wrong sorry.

Solar System - Scaling. When you look at Neptune in a telescope, you are actually looking into the past as the light has to travel from Neptune to your eyes. If the speed of light is ~300,000 km/s, how far back into the past are you looking (or put another way, how long does it take light to travel from Neptune to your eyes on Earth)

Answers

Answer:

Distance from sun to Neptune = 4.495E9 km

Time for light to travel = 4.495E9 / 3E5 sec = 14,980 sec

That is from sun to Neptune time fof light = 250 min

Time for light to travel from sun to earth is about 8 min

So the time from Neptune would be 242 to 258 min depending on position of Neptune - Note that Neptune is about 30X as far from the sun as earth and

250 min / 8 min is roughly 30

The uniform motion of kinematics allows us to find the time it takes for light to arrive from Neptune to Earth, which varies between:

          t₁ = 1.45 10⁴ s and t₂₂= 1.55 10⁴ s

depending on the relative distance of the two planets

given parameters

The speed of light c = 300,000 km / s = 3 10⁸ m / s The distance from Neptune to Sum

to find

The time when light arrives from Neptune to Earth

They  velocit of an electromagnetic wave is constant, so we can use the uniform motion relationships

             v = d / t

             t = d / v

where v is the speed of light, d the distance and y time, in this case the speed of the wave is the speed of light (v = c)

We look in the tables for the distances and the rotation periods around the sun

                           distance ( m)         period (s)

Sun Neptunium     4.50 10¹²             5.2 10⁹

Sun - Earth             1.5    10¹¹              3.2 10⁷

With the data of the period it is observed that the rotation of Neptune is much greater than that of Eart rotation around the sun, for which we will assume that Neptunium is fixed in space and the Earth may be in its aphelion or perihelion, maximum approach o away distance from the sun, consequently we calculate the time for the two cases:

Maximum approach

positions relative distance from the dos Plantetas is

         Δd = [tex]x_{Neptuno - Sum} - x_{Earth - Sum}[/tex]d  

     

         Δd = 4.50 10¹²  - 1.5 10¹¹

         Δd = 43.5 10¹¹ m

the time it takes for Neptune's light to reach Earth is

        Δt = [tex]\frac{ 43.5 \ 10^{11} }{3 \ 10^8}[/tex]  

        Δt = 14.5 10³ s

        Δt = 1.45 10⁴ s

       

We reduce to hours

        Δt = 1.45 10⁴ s (1 h / 3600 s) = 4.03 h

Maximum away

         Δd = [tex]x_{Neptune - Sum} + x_{Neptune-Sum}[/tex]  

         Δd = 4.50 10¹² + 1.5 10¹¹

         Δd = 46.5 10¹¹

The time is

         Δt = [tex]\frac{46.5 \ 10^{11}}{ 3 \ 10^8}[/tex]  

         Δt = 15.5 10³

         Δt = 1.55 10⁴ s

We reduce to hours

         Δt = 1.55 10⁴ s (1 h / 3600 s) = 4.31 h

In conclusion, the time it takes for light to arrive from Neptune to Earth varies between:

          t₁ = 1.45 10⁴ s and t₂ = 1.55 10⁴ s

depending on the relative distance of the two plants

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A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?

Answers

Answer:

Charge of the particle is 1 coulomb.

Explanation:

Force, F:

[tex]{ \bf{F=BeV}}[/tex]

F is magnetic force.

B is the magnetic flux density.

e is the charge of the particle.

V is the velocity

[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]

After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving

Answers

This question is incomplete, the complete question is;

A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.

After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving

Answer:

the second moth is moving at 5.062 m/s

Explanation:

Given the data in the question;

Using doppler's effect

[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )

f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )

frequency reflected from the moth,

Now, moth is the source and the bat is the receiver

f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )

hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )

we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.

given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.

we substitute

55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[  ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )

1.02996 =  ( 343 + u₂ ) / ( 343 - u )

( 343 + u₂ ) = 1.02996( 343 - u )

343 + u = 353.27628 - 1.02996u

u + 1.02996u = 353.27628 - 343

2.02996u = 10.27628

u = 10.27628 / 2.02996

u = 5.062 m/s

Therefore, the second moth is moving at 5.062 m/s

find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4​

Answers

The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = Aeσ[tex]T^{4}[/tex]

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.

So that;

[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]

     = 3.8556 x [tex]10^{-8}[/tex]

     = 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex]

Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

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A Ball A and a Ball B collide elastically. The initial momentum of Ball A is -2.00kgm/s and the initial momentum of Ball B is -5.00kgm/s. Ball A has a mass of 4.00kg and is traveling at 2.50 m/s after the collision. What is the velocity of ball B if it has a mass of 6.50kg?

Answers

The velocity of B after the collision is obtained as -2.6 m/s.

What is the principle of conservation of momentum?

Now we now that the  principle of conservation of momentum states that the momentum before collision is equal to the momentum after collision.

Thus;

(-2.00kgm/s) + ( -5.00kgm/s) = ( 4.00kg * 2.50 m/s) + ( 6.50kg * v)

-7 = 10 + 6.5v

-7 - 10 = 6.5v

v = -7 - 10 /6.5

v = -2.6 m/s

Hence, the velocity of B after the collision is obtained as -2.6 m/s.

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The cells lie odjacent to the sieve tubes​

Answers

Answer:

Almost always adjacent to nucleus containing companion cells, which have been produced as sister cells with the sieve elements from the same mother cell.

The timing device in an automobile’s intermittent wiper system is based on an RC time constant and utilizes a 0.500 micro F capacitor and a variable resistor. Over what range must R be made to vary to achieve time constants from 2.00 to 15.0 s?

Answers

Answer:

check 2 photos for answer

check 2 photos for answer

​Determine usando ecuación de Bernoulli la Presión P1 necesaria para mantener la condición mostrada dentro del sistema mostrado en la figura, sabiendo que el aceite tiene un s.g =0.45 y el valor de d=90mm.

Answers

Answer:

PlROCA

Explanation:

A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects

Answers

Answer:

P = -1 D

Explanation:

For this exercise we must use the equation of the constructor

       / f = 1 / p + 1 / q

where f is the focal length, p and q is the distance to the object and the image, respectively

The far view point is at p =∞  and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image  

        [tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]

         f = 1 m

         P = 1/f

          P = -1 D

An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.​

Answers

Answer:

[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].

The unit of both sides of this equation are [tex]\rm s[/tex].

Explanation:

The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].

The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].

On the right-hand side of this equation:

[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].

[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].

Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].

When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity

Answers

The answer is terminal force

Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B​

Answers

Explanation:

Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]

The sum of the two vectors is

[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]

[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]

The difference between the two vectors can be written as

[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]

[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]

What star is known as the "cold planet"?

Answers

Explanation:

OGLE-2005-BLG-390Lb.

PSR B1620-26 b. Surface Temperature: 72 Kelvin. ...

Neptune. Surface Temperature: 72 Kelvin. ...

Uranus. Surface Temperature: 76 Kelvin. ...

Saturn. Surface Temperature: 134 Kelvin. ...

Jupiter. Image Courtesy: NASA. ...

OGLE-2016-BLG-1195Lb. Surface Temperature: Unknown

Which of the units of the following physical quantities are derived

Answers

Answer:

where is the attachment

Explanation:

The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the gravitational force between them ? ans. 2.668× 10-9N​

Answers

[tex]F = 5.93×10^{13}\:\text{N}[/tex]

Explanation:

Given:

[tex]m_1= 2×10^{16}\:\text{kg}[/tex]

[tex]m_2= 4×10^{22}\:\text{kg}[/tex]

[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]

Using Newton's universal law of gravitation, we can write

[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]

[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]

[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]

What is an internal resistance?

Answers

Explanation:

some thing inside a resistor

A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)

Answers

Answer:

[tex]E=35921.96N/C[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=0.321mm[/tex]

Charge Density [tex]\mu=0.100[/tex]

Distance [tex]d= 5.00 cm[/tex]

Generally the equation for electric field is mathematically given by

[tex]E=\frac{mu}{2\pi E_0r}[/tex]

[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]

[tex]E=35921.96N/C[/tex]

a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10.2 m​

Answers

Answer:

Explanation:

Here's the info we have:

initial velocity is 20 m/s;

final velocity is our unknown;

displacement is -10.2 m; and

acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation

v² = v₀² + 2aΔx and filling in accordingly to solve for v:

[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex]  Rounding to the correct number of sig fig's to simplify:

[tex]v=\sqrt{400+2.0*10^2}[/tex] to get

v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.

So either 20 m/s or 24 m/s

15 . A scientist who studies the whole environment as a working unit .

Botanist
Chemist
Ecologist
Entomologist

Answers

Answer:

Ecologist.

Your answer is Ecologist.

(Ecologist) is a scientist who studies the whole environment as a working unit.

cyclist always bends when moving the direction opposite to the wind. Give reasons​

Answers

When he bends he kinda off his feet and light but if it’s not i’m so sorry this is just my thinking.

Give examples of motion in which the directions of the velocity and acceleration vectors are (a) opposite, (b) the same, and (c) mutually perpendicular​

Answers

Answer:

a) When moving body applies brake then velocity and acceleration would be in opposite direction

b) When body starts to increase velocity then velocity and acceleration would be in same direction

c) When body is circulating then velocity and acceleration would be perpendicular to each other

Explanation:

a) When body applies brake then its velocity starts decreasing, in this case its acceleration would try to stop the moving body. So direction of velocity would be same as direction of motion of body but direction of acceleration would be in opposite direction

b) When body starts to increase velocity, its acceleration would make the body to move faster. So direction of velocity would be the direction of motion of body and acceleration would also be in same direction

c) When body moves in circular path then its acceleration would be towards centre of circle and velocity would try to snap the body out of circle to straight line which in tangent to circle.

A force of 1000N is used to kick a football of mass 0.8kg find the velocity with which the ball moves if it takes 0.8 sec to be kicked.​

Answers

The velocity of the ball is 100m/s

The first step is to write out the parameters;

The force used to kick the ball is 1000N

The mass of the ball is 0.8 kg

Time is 0.8 seconds

Therefore the velocity can be calculated as follows

F= Mv-mu/t

1000= 0.8(v) - 0.8(0)/0.8

1000= 0.8v- 0.8/0.8

Cross multiply both sides

1000(0.8) = 0.8v

800= 0.8v

divide both sides by the coefficient of v which is 8

800/0.8= 0.8v/0.8

v= 1000m/s

Hence the velocity is 1000m/s

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Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.

Answers

We have that the spring constant is mathematically given as

[tex]k=2.37*10^{11}N/m[/tex]

Generally, the equation for angular velocity is mathematically given by

[tex]\omega=\sqrt{k}{m}[/tex]

Where

k=spring constant

And

[tex]\omega =\frac{2\pi}{T}[/tex]

Therefore

[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]

Hence giving spring constant k

[tex]k=m((\frac{2 \pi}{T})^2[/tex]

Generally

Mass of earth [tex]m=5.97*10^{24}[/tex]

Period for on complete resolution of Earth around the Sun

[tex]T=365 days[/tex]

[tex]T=365*24*3600[/tex]

Therefore

[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]

[tex]k=2.37*10^{11}N/m[/tex]

In conclusion

The effective spring constant of this simple harmonic motion is

[tex]k=2.37*10^{11}N/m[/tex]

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Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3 ​

Answers

Answer:

d = 30kg/cm³

Explanation:

d = m/v

d = 1080kg/(3cm*4cm*3cm)

d = 30kg/cm³

Is it true that as we gain mass the force of gravity on us decreases

Answers

Answer:

No. As we gain mass the force of gravity on us does not decrease

a beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a focal length of 40 cm

Answers

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

u

1

v

1

=

f

1

+

u

1

v

1

=

20

1

+

12

1

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

u

1

v

1

=

f

1

+

u

1

v

1

=

−16

1

+

12

1

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?

Answers

Answer:

[tex]v_1=87.40m/s[/tex]

Explanation:

From the question we are told that:

Mass of arrow [tex]m=52g[/tex]

Mass of rock [tex]m_r=1.50kg[/tex]

Height [tex]h=0.47m[/tex]

Generally the equation for Velocity is mathematically given by

 [tex]v = \sqrt{(2gh)}[/tex]

 [tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]

 [tex]v= 3.035m/s[/tex]

Generally the equation for conservation of momentum is mathematically given by

 [tex]m_1v_1=m_2v_2[/tex]

 [tex]0.052kg * v = 1.5 * 3.03m/s[/tex]

 [tex]v_1=87.40m/s[/tex]

Other Questions
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