John throws a ball with a velocity of 30 m/s at an angle of 60 degrees. What is the horizontal component of the velocity?

a 30 m/s
b 0 m/s
c 25.9 m/s
d 15 m/s

Answers

Answer 1

The horizontal component of the velocity is equal to: D. 15 m/s.

Given the following data:

Velocity = 30 m/sAngle = 60°

To determine the horizontal component of the velocity:

The horizontal component of the velocity represents the influence of velocity  in displacing an object or projectile in the horizontal direction.

Mathematically, the horizontal component of velocity is given by the formula:

[tex]V_x = Vcos(\theta)[/tex]

Substituting the given parameters into the formula, we have;

[tex]\\\\V_x = 30cos(60)\\\\V_x = 30 \times 0.5[/tex]

Horizontal component, Vx = 15 m/s

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Related Questions

Read the excerpt from The Building of Manhattan.

A temporary, small narrow-gauge track system was installed on each floor as it was needed. This enabled the material to be moved from the truck at ground level onto dump carts, raised by elevator to the designated floor, wheeled onto the track, and moved quickly to the exact spot needed. Turntables built into the track allowed the carts to be shifted about in any direction.

Which question can be answered only if an illustration accompanies the text?

What did turntables built into the track allow?
How were dump carts raised to designated floors?
When was a track system installed on a floor?
What does a turntable built into a track loo

Answers

Answer: this would be D.) What does a turntable built into a track look like

Explanation:

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is ____? ​

Answers

Let, the maximum height covered by projectile be [tex]\sf{H_m}[/tex]

[tex]\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }} [/tex]

Projectile is thrown with a velocity = v Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be [tex] \sf{v_0}[/tex]

[tex]\qquad[/tex]______________________________

Then –

[tex]\qquad[/tex] [tex]\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }[/tex]

Now, the vertical component of velocity of projectile at the height half of [tex] \sf{h_m}[/tex] will be –

[tex]\qquad[/tex] [tex]\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }[/tex]

Therefore, the vertical component of velocity of projectile at this height will be–

☀️[tex]\qquad[/tex][tex] \pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }[/tex]

Answer:

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is v sintheta / √2

10 time 20 minus 5 and yeah that’s all i need

Answers

Answer:

[tex]10 \times 20 - 5 \\ \\ = 200 - 5 \\ = 195[/tex]

assume that the hubble constant is 72 km/s/mpc. if a galaxy is 100mpc away, how fast is it moving away from us?

Answers

The galaxy is 7200 km/s moving away from us.

A Hubble is a large telescope in space.

The question can be solved using the formula below

⇒ Formula

v = Cd........................ Equation 1

⇒ Where:

v = Speed of the galaxyC = Hubble constantd = Galaxy's distance

⇒ Given:

C = 72 km/s/mpcd = 100 mpc

⇒ Substitute these values into equation 1

v = 72(100)v = 7200 km/s.

Hence, the galaxy is 7200 km/s moving away from us.

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how to find change in kinetic energy wfrom joules

Answers

Answer:

What do you mean? Can you please explain yourself better?

Samantha's teacher asks her to describe motion in one dimension. This type of motion is always

Answers

A motion in one dimension is a type of motion that is always along a straight line and has a constant acceleration.

Motion refers to a change in the location (position) of an object or physical body with respect to a reference point, especially due to the action of an external force.

In Science, the motion of an object or physical body is described in terms of the following parameters:

SpeedForceAccelerationDistanceTime

Furthermore, there are three (3) forms of motion based on dimension and these include:

1. Motion in one dimension

2. Motion in two dimension

3. Motion in three dimension

A motion in one dimension (one-dimensional motion) is also referred to as rectilinear or linear motion. Also, it is always along a straight line in any direction and characterized by constant acceleration.

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how do you find average velocity (vaverage) from acceleration (a) and time (t)?

Answers

Answer:

1) Average velocities:

Given the magnitudes of 2 velocity vectors, in different time instants, the average velocity is defined as the difference of the 2 velocity magnitudes, in absolute value, divided by the interval of time occuring between the measurement of the velocity magnitudes.

2) Average acceleration

Given the magnitudes of 2 acceleration vectors, in different time instants, the average acceleration is defined as the difference of the 2 acceleration magnitudes, in absolute value, divided by the interval of time occuring between the measurement of the acceleration magnitudes.

As you see the definitions are pretty similar.

Note: I didn't copy-paste this answers from any other site.

A car travels 80 km at an avarage speed of 40 km/h. It travels the remaining distance in 3h. What is its dis placement if the average speed of the car is 30 km/h.? And explain

A) 100 km

B)120 km

C)150km

D)180km

Answers

When traveling with average speed 40 km/h, the car would cover a distance of 80 km in time t such that

40 km/h = (80 km) / t   ⇒   t = 2 h

so the total travel time is 2 h + 3 h = 5 h.

Average speed is defined as the total distance traveled divided by the time it took to cover that distance. So if the overall average speed was 30 km/h, then

30 km/h = (80 km + x) / (5 h)   ⇒   x = 70 km

where x is the distance traveled in the last 3 h of the trip.

Then the total displacement of the car is 80 km + 70 km = 150 km.

Define the following:
Variable
Data
Control
Technology
Hypothesis
Physical Science
Experiments

Answers

Answer:

Variable- not consistent or having a fixed pattern; liable to change

Data- facts and statistics collected together for reference or analysis

Control- a group or individual used as a standard of comparison for checking the results of a survey or experiment

Technology- the application of scientific knowledge for practical purposes, especially in industry

Hypothesis- a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation

Physical Science- the sciences concerned with the study of inanimate natural objects, including physics, chemistry, astronomy, and related subjects

Experiments- a scientific procedure to make a discovery, test a hypothesis, or demonstrate a known fact

Explanation:


An airplane initially travels at 12 m/s when passing the "acceleration line." The airplane then
accelerates to 9 m/s2 until reaching its take off velocity of 40.0 m/s. What is the displacement of
the plane during the acceleration?

Answers

The displacement of  the plane during the acceleration is equal to 80.89 meters.

Given the following data:

Initial velocity = 12 m/sFinal velocity = 40 m/sAcceleration = 9 [tex]m/s^2[/tex]

To determine the displacement of  the plane during the acceleration, we would use the third equation of motion;

[tex]V^2 = U^2 + 2aS[/tex]

Where:

V is the final speed.U is the initial speed.a is the acceleration.S is the displacement traveled.

Substituting the given parameters into the formula, we have;

[tex]40^2 =12^2 + 2(9)S\\\\1600=144+18S\\\\18S=1600-144\\\\18S=1456\\\\S=\frac{1456}{18}[/tex]

Displacement, S = 80.89 meters.

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please help me !! i’ll mark brainliest if you’re right!

Answers

Option 2nd is the answer

A car has a kinetic energy of 4.33x10^5 J when traveling a speed of 25 m/s. What is the mass ?

Answers

4.33x50

10^5 would be 50 so 50/25

What is the difference between folk dance and ballroom dance

Answers

Answer:

folk dancing is old music and ball room is young dancing

Please HELPPPPPP THIS IS TIMED

what would the conversion of energy look like from start to finish of a person running?

Answers

Answer:

When a person runs, their body must convert potential energy into kinetic energy. Potential energy is the energy stored within a system.When a person runs, their body must convert potential energy into kinetic energy. Potential energy is the energy stored within a system. Potential energy is used when the system uses kinetic energy to move in a horizontal direction.

In the human body, potential energy is stored in the form of chemical energy. The chemical energy comes from the food that a person consumes throughout the day. The body needs a certain amount of calories (the energy from food) in order to perform certain activities. If the runner has not consumed enough calories throughout their day, they will run out of potential energy and become tired. This is because the body is not very efficient at retaining energy. Energy cannot be created or destroyed, but it can go elsewhere. As the person runs, most of their stored energy is released in the form of thermal energy. This is why people get hot and start to sweat when they do physical activity such as running. The body is heating up because it is literally burning the calories that it has consumed in order to keep moving in a horizontal direction. The runner will sweat, because sweating is the body's natural cooling mechanism. If the runner did not have the ability to sweat, the conversion of potential energy (the chemical energy) into kinetic energy which is released as thermal energy would cause the runner's body to overheat. The chemical energy that the runner consumes in the form of calories is also released in the form of sound energy. Every time the runner's foot hits the ground, energy is leaving the runner's body as sound waves emit from the impact of the runner's foot on the ground. Because energy is being released from the runner's body with every step they take, it is important for the runner to consume enough chemical energy in the form of calories prior to their run. The runner's body needs a substantial amount of calories as a reserve so that they will have more to burn as their potential energy is released throughout the run.

Thermal energy is measured in calories. Calories are released from a given item as it burns. The amount of calories that are in something depends directly on the amount of chemical bonds that are broken and formed as it burns. For example, when a piece of wood burns, 3000 calories of thermal energy are released per gram. When an apple is burned however, it releases about 600 calories of thermal energy. Therefore, it is reasonable to assume that there is more energy available from breaking the atomic bonds in wood than from breaking the atomic bonds in an apple.

One calorie is defined as the amount of thermal energy needed to raise the temperature of one gram of water one degree Celsius. Calories burn very slowly in the human body, and as they do, kinetic energy becomes available to the runner. 1 calorie is the equivalent to 4.186 Joules of energy. So, the more calories that the runner consumes prior to running, the more energy they will have available to them throughout the run. The runner�s energy can also be measured in the form of watts, or electrical power. One calorie also translates to about 4.186 watts. So, if the runner has 500 calories available to them, they are capable of producing over 2000 watts of electrical power.Kinetic energy is equal to one half of the runner's mass times their velocity squared (KE=1/2mv^2). So, if the runner has a mass of 60 kg and wants to run at a rate of 9m/s, they will use about 2,430 Joules of energy. The runner is not able to change their mass, but they can increase or decrease their use of kinetic energy by increasing or decreasing their velocity. If the runner has not consumed a lot of chemical energy throughout their day, it would be wise for them to decrease their velocity as to decrease their kinetic energy and therefore use less of their stored potential energy.

how are mass, distance and gpe related

Answers

Answer:

It is direct proportionality. The greater the mass, the greater is the gravitational potential energy. The equation for GPE is : GPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above the ground. As you can see GPE is directly proportional to mass, and height. KT.

Explanation:

Gravitational potential energy is a function of both the mass of your system and the mass of the thing generating the gravity field around your system.

The relationship is linear, which means that if you multiply or divide one of the masses by some number but leave everything else the same, you multiply or divide the potential energy by the same number. A 3kg mass has three times the gravitation potential energy of a 1kg mass, if placed in the same location.

classroom and draw an approxim object Three forces are acting on an object (Figure 1.32) which is in equilibrium. Determine force A 1200 N Force A 51 39 42.00 1400 N Figure 1.32 Three forces, acting on​

Answers

I’m not good with physics but I’m good with the theory what goes up must come down

☆ Correct and rewrite the following statements:
_____
e). electroplating is based on magnetic effect of electricity.

f). an electric bulb glows due to the chemical effect of electricity.
_______________________

#No spam
#No Internet answers, please #Explanation needed​

Answers

Answer:

e)Question:-Electroplating is based on magnetic effect of electricity.Answer:-It's incorrect sentence As we know that, while electroplating there are two rods, one becoming cathode and another one becoming anode. During electroplating, when electricity is given through the rods, the anode produces positive ions which gets received at the cathode rod and negative ions produced at cathode collects at anode rod which results to the plating of the desired substance. Here, it is a chemical effect of electricity resulting to plating.

So correct sentence is:

Electroplating is based of chemical effect of electricity.

f)Question:-An electric bulb glows due to the chemical effect of electricity.Answer:-It's incorrect sentence As we know that, in an electric bulb there is a filament. When electricity is given in that filament of the electric bulb, the filament gets intensely heated up and results to the emission of light. Due to this heating of the filament due to electricity the electric bulb glows.

So correct sentence is:

An electric bulb glows due to the heating effect of electricity.

soda is an example of which of these?

liquid–liquid solution

gas–liquid solution

solid–liquid solution

solid solution

Answers

B. Gas-liquid
solution


What happens to a liquid in a U-tube manometer when placed in a vacuum

Answers

Answer:

With a greater pressure applied to the left side of a U-tube manometer, the liquid lowers in the left leg and rises in the right leg. The liquid moves until the unit weight of the liquid, as indicated by h, exactly balances the pressure.

Answer:

Here

Explanation:

With a greater pressure applied to the left side of a U-tube manometer, the liquid lowers in the left leg and rises in the right leg. The liquid moves until the unit weight of the liquid

A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below.

B) What equilibrant force is required to keep the block at rest once it does?


C) If you were to rearrange the three forces as you like, what maximum force could be applied to slow the block?


D) How much time would it take for the block to stop under maximum force conditions?

Answers

Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

 

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

      -10 + 12 sin 60 + Fₓ = 0

        Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

       12 cos 60 - 6 + F_y = 0

        F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

Form of coordinates F = -0.39 i ^ N Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

       [tex]F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}[/tex]  

       F = 0.39N

We use trigonometry for the angle.

       [tex]tan \theta = \frac{F_y}{F_x}[/tex]

       tan θ=  0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

        θ = 180 + θ'

        θ = 180 + 0

        θ = 180º

C) if the three forces can be moved and the maximum force occurs when they are all linear.

          10+ 6 + 6 + F = 0

          F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

          [tex]a = \frac{F}{m}[/tex]  

          a = [tex]\frac{24}{6}[/tex]  

         a =  4 m / s²

The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

        v = v₀ - a t

Final velocity when stopped is zero

         t = [tex]\frac{0-v_o}{a}[/tex]

         t = 100/4

         t = 25 s

In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:

    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

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A water balloon is hovering directly above the line join points ANB which are 4.6 km apart if the angles of elevation to the balloon from point a to B or 28.8° and 52.2 respectively find the altitude of the balloon

Answers

Answer:

Drawing the triangle:

H / x = tan 52.2 = 1.29

H / (4.6 - x) = tan 28.8 = .550

H = 1.29 x

H = .55 * 4.6 - .55 x

1.84 x = 2.53        combining equations

x = 1.38

4.6 - 1.38 = 3.22

Total base of triangle = 1.38 + 3.22 = 4.6

H / x = tan 52,2 = 1.29

H = 1.29 * 1.38 = 1.78 height of triangle

Check:

1.78 / 3.22 = tan 28.9    

This agrees with the given value of 28.8

A net force of 79 N causes a mass to accelerate at a rate of 3 m/s2. Determine the mass

Answers

Force= mass x acceleration

79=mass(3)

m=79/3

m=26.3 kg

There can be multiple equal & opposite forces acting on a singular object?
True or False

Answers

Answer:

true

Explanation:

negative positive horizontal vertical gravity, friction and more

I couldn't find one .

Answers

Start from the top and go 7 down then go to the right 9 times and there you should fine the letter "one"

Construct an explanation for the fact that we don’t see the same amount of energy coming out at the end of these transformations as we did at the beginning. *

Answers

Answer:

When energy is transformed from one form to another, or moved from one place to another, or from one system to another there is energy loss.

Explanation:

Hope this helps! :)

Describe the direction in which each object will accelerate. If the object won't accelerate, write "no acceleration." Explain what might be happening to each object in terms of its overall motion.

Answers

Is this photo apart of this question? Or you need answer for just the sitting part

A 263 g ball of clay falls onto a vertical spring.

The spring constant k = 2.52 N/cm.

The clay sticks to the spring and as the spring compresses a distance of 11.8 cm the clay momentarily comes to rest.
Determine the speed of the clay just before it hit the spring in [m/s]?

Answers

As the spring is compressed, it performs

-1/2 (252 N/m) (0.118 m)² ≈ -1.75 J

of work on the clay, while gravity does

(0.263 kg) (9.80 m/s²) (0.118 m) ≈ 0.304 J

on it. Then the total work W performed on the clay ball is approximately -1.45 J.

By the work-energy theorem, W is equal to the change in the clay ball's kinetic energy ∆K. If v is the speed with which it hits the spring, then

W = ∆K

-1.45 J = 0 - 1/2 mv²

Solve for v :

-1.45 J = -1/2 (0.263 kg) v²

v² ≈ 11.0 m²/s²

v ≈ 3.32 m/s

The Earth has a mass of 6.0 x 1024 kg, and the Moon has a mass of 7.3 x 1023
kg. They are separated by a distance of about 3.84 x 105 kilometers. How
much gravitational force do the two bodies exert on one another?
0 1.0 x 1021 N
O 2.0 x 1021 N
o 3.0 x 1021 N.
O 4.0 x 1021 N

Answers

The amount of gravitational force the two bodies exert on one another is [tex]2.0 \times 10^{27} \;Newton[/tex].

Given the following data:

Mass of Moon = [tex]7.3 \times 10^{23}\; kg[/tex]Mass of Earth = [tex]6.0 \times 10^{24}\; kg[/tex]Radius = [tex]3.84 \times 10^5\; kilometers[/tex]

Scientific data:

Gravitational constant = [tex]6.67\times 10^{-11}[/tex]

To determine the amount of gravitational force the two bodies exert on one another, we would apply Newton's Law of Universal Gravitation:

Mathematically, Newton's Law of Universal Gravitation is given by the formula:

[tex]F = G\frac{M_1M_2}{r^2}[/tex]

Where:

F is the gravitational force.G is the gravitational constant.M is the mass of object.r is the distance between centers of the masses.

Substituting the given parameters into the formula, we have;

[tex]F = 6.67\times 10^{-11} \times \frac{7.3 \times 10^{23} \; \times \;6.0 \times 10^{24}}{(3.84 \times 10^5)^2}\\\\F = 6.67\times 10^{-11} \times \frac{4.38 \times 10^{48} }{1.48 \times 10^{11} }\\\\F=\frac{2.92 \times 10^{38} }{1.48 \times 10^{11}} \\\\F= 2.0 \times 10^{27} \;Newton[/tex]

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will give brainlyiest what is the main need or want this wave technology meets to efficiently communicate with others to hear and see others in real time to efficiently store and share music with others to speak with and see others who are far away

Answers

Answer: To see objects surrounding more clearly

Explanation:

Resultant of two froce 3N nad 4N acting ar apoit is 5N show that the two froce are 2 perpendicular to each other​

Answers

Two forces – A = 3N and B = 4NResultant of these forces, R = 5N

[tex]\qquad[/tex] ☀️We are asked to prove that two forces are perpendicular to each other.

Formula to calculate magnitude of resultant 'R' of two forces 'A' and 'B' when they are at an angle 'p' to each other.

[tex]\qquad[/tex][tex]\purple{ \longrightarrow \bf R^2= A^2 + B^2 + 2 AB cos p} [/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf 5^2= 3^2 + 4^2+ 2 \times 3\times 4\times cos p[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf 25 = 9 + 16 + 24 cos p [/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf 25 = 25 + 24 cos p[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf 25 - 25 = 24 cos p[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf 0 = 24 cos p[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf cos p = 0[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf cos p = cos 90°[/tex]

[tex]\qquad[/tex][tex]\purple{ \longrightarrow \sf p = 90°}[/tex]

Hence, Proved that –

Forces A = 3N and B = 4N are Perpendicular to each other.

[tex]\qquad[/tex]_________________________

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