A girl throws a 2 kg rock with a net force of 6 N. What is the acceleration of the rock?

8 m/s2
0.33 m/s
2 3 m/s2
12 m/s2

Answers

Answer 1
it’s 12 m/s2 i believe but i’m nit sure
Answer 2

Answer:

3 m/s2

Explanation:

Took the test


Related Questions

Two cats jump off a roof that is 23m off the ground. Cat A jumps directly up with a velocity of 7m/s, and cat B jumps directly down with a velocity of 7m/s. How far up does cat A go? How fast does cat A go just before it hits the ground? How fast does cat B go just before it hits the ground? How much longer is cat A in the air than cat B?

Answers

(a) The height reached by cat A is 2.5 m.

(b) The speed of cat A before it hits the ground is 36.4 m/s.

(c) The speed of cat B before it hits the ground is 22.39 m/s.

(d) The time spent in air by cat A than cat B is 1.43 seconds.

Time of motion of cat A

The time of motion of cat A is calculated as follows;

h = vt - ¹/₂gt

-23 = 7t - 4.9t²

4.9t² - 7t - 23 = 0

solve the quadratic equation using formula method;

a = 4.9, b = -7, c = - 23

t = 3.0 seconds

Time of motion of cat B

The time of motion of cat B is calculated as follows;

h = vt - ¹/₂gt

23 = 7t + 4.9t²

4.9t² + 7t - 23 = 0

solve the quadratic equation using formula method;

a = 4.9, b = 7, c = -23

t = 1.57 seconds

Height reached by cat A

h = u²/2g

h = (7²)/(2 x 9.8)

h = 2.5 m

Speed of cat A before it hits the ground

v = u + gt

v = 7  + 3(9.8)

v = 36.4 m/s

Speed of cat B before it hits the ground

v = u + gt

v = 7  + 1.57(9.8)

v = 22.39 m/s

Time spent in air by cat A than cat B

Time difference = 3 s - 1.57 s = 1.43 seconds

Thus, the height reached by cat A is 2.5 m.

The speed of cat A before it hits the ground is 36.4 m/s.

The speed of cat B before it hits the ground is 22.39 m/s.

The time spent in air by cat A than cat B is 1.43 seconds.

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Bonus: What is the velocity of an 8 kg lead shot-put if it has 484 J of energy? Type here to search e​

Answers

Answer:

11m/s

Explanation:

Given parameters:

Mass of the lead shot = 8kg

Energy of the shot  = 484J

Unknown:

Velocity of the shot  = ?

Solution:

To solve this problem, we apply the kinetic energy formula;

        K.E  = [tex]\frac{1}{2}[/tex] m v²  

m is the mass

v is the unknown

 Now, insert the parameters and solve;

        484  = [tex]\frac{1}{2}[/tex] x 8 x v²

        484  = 4v²

           v² = 121

            v  = 11m/s

Calculate the wavelength of an orange light wave with a frequency of 5.085 x 10^14 Hz. The speed of light is 3.0 x 10^8 m/s.

^ meaning a power

Answers

Answer:

5.9 x 10⁻⁷m

Explanation:

Given parameters:

Frequency = 5.085 x 10¹⁴Hz

Speed of light  = 3.0 x 10⁸m/s

Unknown:

Wavelength of the orange light  = ?

Solution:

The wavelength can be derived using the expression below;

            wavelength  = [tex]\frac{v}{f}[/tex]

v is the speed of light

f is the frequency

            wavelength  = [tex]\frac{3 x 10^{8} }{5.085 x 10^{14} }[/tex]   = 5.9 x 10⁻⁷m

1. Which is one characteristic of a SMART fitness goal?
A. Endurance
B. Intensity
C. Attainable
D. Believable

Answers

Answer:

I would say D. Believable. Sorry if its wrong.

Explanation:

It’s D for sure !!!!!!!

In this illustration, the magnetic field is strongest
HELP ME PLS PLS PLS

Answers

The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center SO THE ANSWER IS “NEAR EITHER POLE•

Consider these pictures in answering the question below.
с
D
Which diagram might represent the initial vertical component of the velocity of a ball thrown upward at an angle?
a
Ob
CU O
c
Od
Оe
E

Answers

Answer:

the answer would be D because not only is it going up but it's going up as an angle

A crane used 250,000 Joules of work to move a beam to the top of a building in 20 seconds. How much power did the crane use?

a) 5,000,000 W
b) 5,000 W
c) 50 J
d) 12,500 W

Answers

Answer:

12500W

Explanation:

Given parameters:

Work done  = 250000J

Time taken  = 20s

Unknown:

Power of the crane = ?

Solution:

Power is the defined as the rate at which work is being done;

  Mathematically;

        Power = [tex]\frac{work done}{time }[/tex]

 insert the parameters and solve;

      Power  = [tex]\frac{250000}{20}[/tex]   = 12500W

A man applies a force of 500 N to push a truck 10 m down the street. How much work has been done? Please answer!

Answers

Answer:

5000 J

Explanation:

W=F*D

W=500N*10m

w=5000 J

When a man applies a force of 500 N to push a truck 10 m down the street. Then the work of 5000 J has been done.

What is Work ?

Work done is the amount energy gained (loosed) in bringing the body from initial position to final position. It is denoted by W and its SI unit is joule(J). i.e. Work(W) is force(F) times displacement(s). W=F× s When a body is displaced with 1 newton of force by 1 m, then we can say that work has been done on the body by 1 joule.

Writing for it's dimension, W=F× s

Force has dimension [L¹ M¹ T²]

Displacement has dimension [L¹]

multiplying both the dimensions Force and Displacement we get, dimension of Work [L² M¹ T²].

Given,

F = 500 N

s = 10 m

Work = Fs = 500×10 = 5000 J

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what is work please give ans​

Answers

work= force x displacement :)

A 33 kg box sits at rest on a tabletop.
Draw and clearly label all the forces acting on the box;
Calculate the normal force.

Answers

Answer:

323.4N

Explanation:

Given parameters:

Mass of the box = 33kg

Unknown:

Normal force on the body  = ?

Solution:

The normal force of a body is the vertical force the body exerts on another body.

It is expressed as;

        Normal force  = mass x acceleration due to gravity

Acceleration due to gravity  = 9.8m/s²

Normal force  = 33 x 9.8  = 323.4N

The movie "The Gods Must Be Crazy" begins with a pilot dropping a bottle out of an airplane. A surprised native below, who thinks it is a message from the gods, recovers it. If the plane from which the bottle was dropped was flying at a height of 500 m, and the bottle lands 400 m horizontally from the initial dropping point, how fast was the plane flying when the bottle was released? Draw a 2-dimensional motion map for the velocities and another for the acceleration.

Answers

Answer:

1) The Speed of the plane is approximately 39.596 m/s

2) Please find attached the required velocity and acceleration graphs

Explanation:

1) The height at which the plane was flying, h = 500 m

The location the bottle lands from the initial dropping point = 400 m

The time it takes the bottle to land is given by the equation for free fall as follows;

h = 1/2·g·t²

Where;

g = The acceleration due to gravity = 9.8 m/s²

t = √(h/(1/2·g) = √(500/(1/2×9.8) ≈ 10.102

The Speed of the plane = The horizontal velocity of the bottle = 400/10.102 ≈ 39.596

The Speed of the plane ≈ 39.596 m/s

2) Please find attached the velocity graph for the vertical and horizontal velocities of the bottle and the acceleration graph for the vertical motion of the bottle, created with Microsoft Excel

C: Calculate the kinetic energy of a car with a mass of 1563 kilograms that is traveling at 42 kilometers per hour.
HELP PLEASEE

Answers

Hello!

C. KE ≈ 9120.105 J

D. m = 151.45 kg

Question C:

Calculate kinetic energy using the formula:

[tex]KE = \frac{1}{2}mv^{2}[/tex]

Substitute in the given mass. We have to convert kilometers to meters in order to solve.

[tex]\frac{42km}{1hr} * \frac{1hr}{3600sec} *\frac{1000m}{1km} = 11.67 m/s[/tex]

Use the equation above:

KE = 1/2(1563)(11.67)

KE ≈ 9120.105 J

Question D:

Plug in the given Kinetic Energy and velocity to solve. Convert kilometers/hour to meters/second:

[tex]\frac{7km}{1hr} * \frac{1hr}{3600sec} *\frac{1000m}{1km} = 1.94 m/s[/tex]

[tex]KE = \frac{1}{2}mv^{2}[/tex]

285 = 1/2(m)(1.94²)

570 = (1.94²)m

151.45kg = m

physical properties of light, such as its amplitude and wavelength, and the human perception of color

Answers

The study of the physical properties of light, such as its wavelength and amplitude, and the human perception of color is called psychophysics. It comprises two major areas: discrimination and absolute.

The study of wavelength and amplitude, interaction between physical stimuli and their consequential effects such as perceptions and sensations is called psychophysics.

This concept is applied in psychophysics tasks where extensive research and studies are performed on sensory and other visual information.

To put it in simple words, psychophysics is vital for measuring the perception of a subject through stimulus detection experiments.

Stimuli detection is done either through the method of limits, adjustment technique, or constant stimuli method.

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A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is the direction of its final velocity? Answer in direction (deg)

Answers

Answer:

(3504 m/s, 66°)

The final velocity of the missile is approximately 3504 m/s at an angle of 66 degrees.

Explanation:

We are going to use the horizontal and vertical components of this object since it is in projectile motion.

We are given the initial velocity of the missile; 1350 m/s at an angle of 25 degrees. We are also given the displacement in the x-direction: 23,500 meters at an angle of 55 degrees. The total time of this projectile in motion is 10.20 seconds.

Using this information, we can break up the motion of the projectile into its horizontal and vertical components and solve for the unknown variables [tex]a[/tex] and [tex]v_f[/tex].

Horizontal direction (x):

List out the known variables:

[tex]v_i=1350[/tex] [tex]v_f=?[/tex] [tex]\triangle x=23500[/tex] [tex]a=?[/tex] [tex]t=10.20[/tex]

Since we have initial velocity, displacement, and time, we can use one of the kinematic equations for constant acceleration that contains these variables, including acceleration, and solve for a:

[tex]\triangle x=v_it + \frac{1}{2}at^2[/tex]

We are solving for acceleration in the x-direction, so this equation should be in terms of the x-direction:

[tex]\triangle x_x=(v_i)_x t +\frac{1}{2}a_x t^2[/tex]

Let's break up the displacement into its horizontal (x) component:

[tex]23500cos(55)[/tex]

Let's break up the initial velocity into its horizontal component:

[tex]1350cos(25)[/tex]

Plug these values into the equation and 10.20 seconds for t.

[tex]23500cos(55)=1350cos(25) \cdot t + \frac{1}{2}a_x (10.20)^2[/tex]

Solve for [tex]a_x[/tex].

[tex]23500cos(55)=12479.85823+52.02a_x[/tex] [tex]999.1880243=52.02a_x[/tex] [tex]a_x=19.20776671[/tex]

The acceleration in the x-direction is about 19.21 m/s².

Now, we can use this acceleration and solve for the final velocity in the x-direction using this constant acceleration kinematic equation:

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

Use this equation in terms of the x-direction:

[tex](v_f)_x=(v_i)_x+a_xt[/tex]

Plug the known values into the equation and solve for [tex](v_f)_x[/tex].

[tex](v_f)_x=1350cos(55)+19.20776671 \cdot 10.20[/tex] [tex](v_f)_x=1419.434733[/tex]

The final velocity in the x-direction is about 1419.43 m/s.

Vertical direction (y):

This process will be the same as solving for acceleration and final velocity in the x-direction, except this time we will be using the vertical (y) components.

[tex]\triangle x_y=(v_i)_y t +\frac{1}{2}a_y t^2[/tex]

Vertical component of initial velocity:

[tex]1350sin(25)[/tex]  

Vertical component of displacement:

[tex]23500sin(55)[/tex]

Plug known values into the equation and solve for [tex]a_y[/tex].

[tex]23500sin(55)=1350sin(25) \cdot (10.20) + \frac{1}{2}a_y (10.20)^2[/tex] [tex]23500sin(55)=5819.453464 +52.02a_y[/tex] [tex]13430.61958=52.02a_y[/tex] [tex]a_y=258.181845[/tex]

The acceleration in the y-direction is about 258.18 m/s².

Now let's use the same equation we used previously to solve for the final velocity in the y-direction.

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

Use this equation in terms of the y-direction:

[tex](v_f)_y=(v_i)_y+a_y t[/tex]

Plug known values into the equation and solve for [tex](v_f)_y[/tex].

[tex](v_f)_y=1350sin(25)+258.181845 \cdot 10.20[/tex] [tex](v_f)_y=3203.989472[/tex]

The final velocity in the y-direction is about 3203.99 m/s.

Final velocity and direction:

The magnitude of the final velocity is the square root of the final velocity in the horizontal and vertical directions squared and added together.

[tex]|v|=\sqrt{(v_f)^2_x + (v_f)^2_y}[/tex]

Plug the final velocity in the x and y-directions into the equation.

[tex]|v|=\sqrt{(1419.434733)^2+(3203.989472)^2}[/tex] [tex]|v|=\sqrt{2014794.961+10265548.54}[/tex] [tex]|v|=\sqrt{12280343.5}[/tex] [tex]|v|=3504.332105[/tex]

The magnitude of the final velocity is about 3504 m/s. We can solve for the direction of the final velocity by using arccos to find the angle that is formed with the x-axis.

[tex]\displaystyle\theta = cos^-^1 \big{(} \frac{v_x}{|v|}\big{)} }[/tex]  

Plug in the x-component of the initial velocity and the magnitude of the final velocity of the projectile into the equation.

[tex]\displaystyle \theta =cos^-^1 \big{(}\frac{1419.434733}{3504.332105}\big{)}}[/tex]  [tex]\displaystyle \theta = cos^-^1 \big{(} .4050514308 \big{)}}[/tex] [tex]\displaystyle \theta = 66.1056499[/tex]

The direction of the final velocity is about 66 degrees.

Need quick response

Are my answers for 1-8 correct and if not what is the correct answer for the ones that are wrong and why?
Also need answers for 9 and 10 since I’m not sure what the answers are and I want an explanation for both of those.

Answers

Answer:

I'll give you all the answers, I know them all, I just did this. I need $1 via cas ap tho since it's hard

which of these properties is the best measure of a star's brightness?(1 point) absolute magnitude absolute magnitude age age size size apparent magnitude

Answers

Apparent magnitude refers to the properties which best measures a star's brightness and is denoted as option D.

What is Apparent magnitude?

This is a term commonly used in astronomical system which talks about the brightness of a star based on its composition such as hydrogen and helium.

it is also determined by how far or bright it is from the planet such as earth when they are viewed by individuals which is regarded as the property in which astronomers use to define its brightness.

This is therefore the reason why apparent magnitude was chosen as the correct choice.

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Answer:

A: Absolute magnitude

Explanation:

I Just did it and the correct answer is ABSOLUTE MAGNITUDE

Please help! Didn’t have right subject so i chose one that may be similar.
Tides are caused by


the rotation of the Earth

the orbit of the Moon

gravitational pull of just the Sun

gravitational pull of the Sun and Moon

Answers

Answer:

Moon's Orbit

Explanation:

Answer:

saaaaaaaaaaaay it please

Explanation:

All battery powered devices, run off of what type of power?

Answers

Answer:

Electrical energy

Explanation:

electrical energy Essentials. A battery is a device that stores chemical energy and converts it to electrical energy. The chemical reactions in a battery involve the flow of electrons from one material (electrode) to another, through an external circuit. The flow of electrons provides an electric current that can be used to do work.

A rectangular tank is filled with water to a depth of 4m. Calculate the pressure at the bottom of the tank. PLEASEEE HELP!! IM BEGGING!!!! PLEASEEEEE

Answers

For this we have the hidrostatic equation:

[tex]\boxed{P = \rho gh}[/tex]

Being:

P = Pressure

ρ = Density of liquid (water) = 1000 kg/m

g = Gravity = 9,81 m/s²

h = Height = 4 m

Replacing acording formula:

[tex]\boxed{P = 1000 \ kg/m ^{3} * 9,81 \ m/s^{2} * 4 \ m}[/tex]

Resolving:

[tex]\boxed{P = 39240 \ Pa}[/tex]

Result:

The pression at the bottom is 39240 Pa.

How much work is needed to move an object 20 meters with 40 newtons of force?

a) 800 J
b) 850 N
c) 875 W
d) 825 m

Answers

Answer:

a) 800 J

Explanation:

Work is said to be done when a force induces movement over a distance in an object. The amount of work done (W) is calculated by multiplying the force by the distance traveled.

That is;

W = F × d

Where;

W = work done (J or N/m)

F = force (N)

d = distance (m)

In the information given in this question, F = 40N, d = 20m

Hence,

Work done = 40N × 20m

Work done = 800N/m or 800 J

Marcus is hanging a porch swing on his front porch. The first thing he has to do is attach two springs to the ceiling. Then, he fastens the swing to the springs so that they can stretch and support the weight of the swing. Which two types of potential energy do the springs have?

Answers

Answer:

Gravitational potential energy and Elastic potential energy

Explanation:

The two types of potential energy the springs described in the question have are Gravitational potential energy and Elastic potential energy.

Firstly, it must be noted that potential energy is the energy stored/present in a stationary (non-moving) object (or object at rest).

Gravitational potential energy is the potential energy as a result of the vertical position of an object due to force of gravity acting to pull it down. The gravitational potential energy is as a result of the springs been attached to the ceiling which causes the force of gravity to want to pull it down. This gravitational force is further increased when the swing is attached causing the weight of swing to act in the direction of the force of gravity.

Elastic potential energy is the potential energy stored in an elastic object or energy stored in an object that can be stretched or compressed. This is the energy that would always be found in a spring, so when the springs have been attached to the ceiling and the swing has been fastened to them, there elastic potential energy would reduce as a result of the kinetic energy that must have been used to stretch the springs.

At what seed is Jay traveling at 5 minutes into walk? At what time in his walk does he have the greatest speed?

Answers

The velocity of Jay after five minutes is 1 Km/min. The greatest speed occurred at 45 mins.

What is the velocity time graph?

The velocity time graph is a representation of the movement of a body a body from one point to another. It is the plot of the velocity on the vertical axis against the time on the horizontal axis. Ideally this graph would be divided into three portions;

The region of uniform accelerationThe region of constant velocity The region of uniform deceleration

We can be able to read off from the graph, the instantaneous velocity of the object at any time. This could be done by tracing the time against the velocity axis and extrapolating according from the graph as shown.

Recall that the acceleration of the body is obtained as the slope of the velocity time graph at any point and the acceleration could be uniform or non uniform as judged from the velocity time graph.

Having said all these, it is clear that the velocity of Jay after five minutes is 1 Km/min.

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Heat flows from __________ energy to __________ energy.
A. higher, lower
B. lower, higher

Answers

Answer:

A. higher, lower

Explanation:

Hope this helps!     :3

plz mark as brainliest!

A student exerts a force of 35 newtons to move a cart loaded with lab equipment to the prep room 20 meters. If it takes the student 10 seconds to push the cart to the prep room, how much power did the student use?

a) 700 W
b) 2.5 Nm
c) 62.5 J/s
d) 70 W

Answers

huhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Mendeleev recognized the periodic law. True or False​

Answers

Answer: True

Explanation:

What is the displacement of the object from 12 seconds to 16 seconds

Answers

Answer:

The displacement of the object is -8 m

Explanation:

Displacement

The displacement of a moving object can be calculated as the area under (or above) the graph of velocity vs time.

If the area is below the y-axis, then the displacement is negative. Otherwise is positive.

It's important to differentiate displacement from distance. Displacement takes into consideration the direction of the movement. Distance does not and it's always positive.

From the graph provided, we can see the velocity from t=12 s from t=16 s is negative, and the displacement will also be negative.

The displacement is calculated as the area of the triangle with base b=16-12= 4 seconds and height = -4 m/s, thus:

[tex]\displaystyle D = \frac{4*(-4)}{2}=-\frac{16}{2}=-8\ m[/tex]

The displacement of the object is -8 m

Why does surface water disperse when touched with soap?

Answers

The end of the detergent molecule which attaches to fat (grease) repels water molecules. It is known as hydrophobic, meaning "water fearing." By attempting to move away from the water molecules, the hydrophobic ends of the detergent molecules push up to the surface. This weakens the hydrogen bonds holding the water molecules together at the surface. The result is a break in the surface tension of the water.

Answer:

In a soap-and-water solution the hydrophobic (greasy) ends of the soap molecule do not want to be in the liquid at all. Those that find their way to the surface squeeze their way between the surface water molecules, pushing their hydrophobic ends out of the water. This separates the water molecules from each other.

Explanation:

Question 24 (2 points)
What is power? (2 points)
O a
O b
O c
Od
The magnitude of a force needed to move an object
How much work can be done in a given time
The distance over time that an object moves
The energy needed to create work

Answers

A because of the keyword “magnitude”

24. A soccer ball is kicked straight up into the air with a velocity of 10 m/s. How much time does it take the ball to reach its maximum height? (g = 10 m/s^2)
A 2 sec
B 100 sec
C 1.4 sec
D 1 sec​

Answers

Answer:

D. Time t = 1 [s]

Explanation:

To solve this problem we must use the following equation of kinematics.

[tex]v_{f}=v_{o}-g*t[/tex]

where:

Vf = final velocity = 0 (when the ball reaches the maximum height its elevation won't be increased since the velocity is zero)

vi = initial velocity = 10 [m/s]

g = gravity acceleration = 10 [m/s²]

t = time [s]

Now replacing:

[tex]0 = 10-10*t\\10 = 10*t\\t = 1 [s][/tex]

energy released in a chemical reaction comes from ?

Answers

Chemical reactions often involve changes in energy due to the breaking and formation of bonds. Reactions in which energy is released are exothermic reactions, while those that take in heat energy are endothermic.
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