Solve this chemical equation: CH3CH2OH+__O2=CO2+__H2O

._.Answer:.

ijji._. ji

Explanation:.

Answer:

Explanation:

For range o a projectile , the formula is as follows

R = u² sin2Ф / g where u is initial velocity of throw , Ф is angle of throw and g is acceleration due to gravity .

Here u = 14 m /s

R = 14² sin2Ф  / 9.8

R = 20 sin2Ф

Now R will have maximum value when sin2Ф has maximum value .

Maximum value of sin2Ф = 1

sin2Ф = 1  = sin 90°

Ф = 45°

So when throw is aimed at 45° , range will be maximum .

Answer:

With or without.

Explanation:

According to the National Electrical Code (NEC), here the air conditioner disconnecting means is not within sight from the equipment, the provision for locking or adding a lock to the disconnecting means shall be on the switch or circuit breaker and remain in place with or without the lock installed. Thus, this is in accordance with section 110.25 of the National Electrical Code (NEC).

Visible and infrared light.

Answer:

The answer is below

Explanation:

a) Coulomb's law of electric force for charges at rest states that Like charges repel each other while unlike charges attract one another.

Therefore since the microbes has small positive charges, the microbe would be repelled by the positively charged electrodes and attracted by the negative charged electrodes.

Hence, the microbes would collect on the negatively charged electrodes.

b) The microbes can easily removed from the negative electrode for analysis by discharging the electrode from the source. Thereby making the electrode to be incapable of attracting the microbe.

Answer:

The more massive an object is, the greater the gravitational force. Since gravitational force is inversely proportional to the distance between two interacting objects, more separation distance will result in weaker gravitational forces.

Explanation:

hope this help:)

Answer:

Explanation:

The acceleration up the incline is 1.00 m/s²

Net force acting on two masses = total mass x acceleration

= 350 x 1 = 350 N

weight acting down the plane = m g sinФ

= 350 x 9.8 x sin30 = 1715 N

Friction force acting down the plane = mg cosФ x μ where μ is coefficient of friction

= 350 x 9.8 x cos30 x .2 = 594N

Net force acting on total mass

= F - 1715 - 594 = 350 , where F is required force

F = 2659 N .

Answer:

the height of the bridge above the water is 49.6 m.

Explanation:

Given;

initial velocity of the stone, u = 15 m/s

time of motion of the stone, t = 2 s

The height of the bridge above the water is calculated from the following kinematic equation as follows;

h = ut + ¹/₂gt²

h = (15 x 2) + ¹/₂(9.8)(2²)

h = 30 + 19.6

h = 49.6 m

Therefore, the height of the bridge above the water is 49.6 m.

Answer:

a increases

Explanation:

as distance between two objects increases the gravitational force decreases so when distance decreases the gravitational force increases

Answer:

the required charged is 7.06 × 10⁻¹³ C

Explanation:

Given that;

Radius = 30 microns = 30 × 10⁻⁶

Speed v = 20 m/s

length x = 1.6 cm = 0.016 m

spacing d = 1.0 mm = 0.001 m

Voltage V = 1500 V

from the question, the electric field between the plates is uniform and equal to Voltage divided by the distance between the plates.

Electric field E = V/d

E = 1500 V /  0.001 m

E = 1.5 × 10⁶ V/m

Mass of ink drop m = pv

m = 10³ kg/m³ × [tex]\frac{4}{3}[/tex]πr³

m = 1000 kg/m³ × [tex]\frac{4}{3}[/tex]π × (30 × 10⁻⁶)³

m = 1.131 × 10⁻¹⁰ Kg

Time taken to travel t =  x / sped

t = 0.016 m / 20 m/s

t = 0.0008 s

From the kinematic equation

to form the top of a letter 3 mm ( 0.003 m )high

y = [tex]\frac{1}{2}[/tex]at²

2y = at²

a = 2y/t²

we substitute

a = (2 × 0.003 m) / (0.0008 s)²

a =  9375 m/s²

Now Force F = Eq = ma

so

q = ma / E

we substitute

q = ( 1.131 × 10⁻¹⁰ Kg × 9375 m/s² ) / ( 1.5 × 10⁶ V/m )

q = 7.06 × 10⁻¹³ C

Therefore, the required charged is 7.06 × 10⁻¹³ C

Answer:

using W=1/2kW2

k=22N/m w=2.9

w=1/2×22×2.9×2.9

w=92.51Joules

Approximately 93J answer is C

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so [tex]V_{max1}[/tex] = A × ω

[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s

[tex]V_{max1}[/tex] =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

[tex]V_{max2}[/tex] = A' × ω

[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69

[tex]V_{max2}[/tex] = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Answer:

Explanation:

Given the following

Ax = 9,

Ay = 12,

Bx = 15

By = 20

Get A and B

A = √9²+12²

A= √81+144

A = √225

A = 15

Get B;

B = √15²+20²

B = √225+400

B = √625

B = 25

get /A+B/

A+B = 15+25

/A+B/ = /40/

Hence the value of /A+B/ is 40

Answer:

meteor

Explanation:

A asteroid stays still and a meteor goes fast

Answer:

meteor

Explanation:

or some people call it a shooting star

Answer:

Explanation:

The attraction weakens. Two objects that are farther apart are not drawn together as strongly as if they were close together.

Answer:

They are used in periscopes, for signalling, in kaleidoscopes, to see round dangerous bends, in meters, as mirror tiles, in a sextant, in an overhead projector, an SLR camera, car wing mirrors, in microscopes and as reflecting number plates to mention only some!

Explanation:

Hope it is helpful....

Answer:

two uses are:

Answer:

[tex]T_2 = 0.592[/tex]

Explanation:

Given

[tex]T_1 = 1.48s[/tex]

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

[tex]\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}[/tex]

Take LCM

[tex]\frac{1}{C_{up}} = \frac{1+1}{C}[/tex]

[tex]\frac{1}{C_{up}}= \frac{2}{C}[/tex]

Cross Multiply

[tex]C_{up} * 2 = C * 1[/tex]

[tex]C_{up} * 2 = C[/tex]

Make [tex]C_{up}[/tex] the subject

[tex]C_{up} = \frac{1}{2}C[/tex]

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

[tex]C_1 = 2 * C_{up}[/tex]

[tex]C_1 = 2 * \frac{1}{2}C[/tex]

[tex]C_1 = C[/tex]

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

[tex]C_p = C+C[/tex]

[tex]C_p = 2C[/tex]

So, the total capacitance (C2) is:

[tex]\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}[/tex]

[tex]\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}[/tex]

Take LCM

[tex]\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}[/tex]

[tex]\frac{1}{C_2} = \frac{5}{2C}[/tex]

Inverse both sides

[tex]C_2 = \frac{2}{5}C[/tex]

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

[tex]T\ \alpha\ C[/tex]

Convert to equation

[tex]T\ =kC[/tex]

Make k the subject

[tex]k = \frac{T}{C}[/tex]

[tex]k = \frac{T_1}{C_1} = \frac{T_2}{C_2}[/tex]

[tex]\frac{T_1}{C_1} = \frac{T_2}{C_2}[/tex]

Make T2 the subject

[tex]T_2 = \frac{T_1 * C_2}{C_1}[/tex]

Substitute values for T1, C1 and C2

[tex]T_2 = \frac{1.48 * \frac{2}{5}C}{C}[/tex]

[tex]T_2 = \frac{1.48 * \frac{2}{5}}{1}[/tex]

[tex]T_2 = \frac{0.592}{1}[/tex]

[tex]T_2 = 0.592[/tex]

Hence, the time constance of (b) is 0.592 s

Answer:

C the metal handle because it is a good conductor

Answer:

D.

Explanation:

Although the metal handle will last longer, if heated up enough it could burn her hand.

Answer:

A.) glass water air

Explanation:

hope this helps :) have a great day!!

Answer:

A.) glass - water - air

good luck, i hope this helps :)

This question is incomplete, the complete question is;

Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).

how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of  charge and the distance between the charges. There are many correct answers

Answer:

Given the data in question;

Dipole moment P = 1 × 10⁻⁹ C.m

now dipole pointing to the right;

               P→

[tex]_{-\theta }[/tex] (-) ---------------->(+) [tex]_{+\theta }[/tex]

               d

so let distance between the dipoles be d

∴ P = d[tex]\Theta[/tex]

Let [tex]\Theta_{1}[/tex] = 1 nC

so

P = d[tex]\Theta[/tex]

1 × 10⁻⁹ =  1 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (1 × 10⁻⁹)

d = 1 m

Also Let [tex]\Theta_{2}[/tex] = 2 nC

so

P = d[tex]\Theta[/tex]

1 × 10⁻⁹ =  2 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (2 × 10⁻⁹)  

d = 0.5 m

Also Let [tex]\Theta_{3}[/tex] = 3 nC

so

P = d[tex]\Theta[/tex]

1 × 10⁻⁹ =  3 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (3 × 10⁻⁹)

d = 0.33 m

such that;

charge                 distance

1 nC                        1.00 m      

2 nC                       0.50 m

3 nc                        0.33 m

4 nC                       0.25 m

5 nC                       0.20 m      

Answer:

F = 228.24 N

Explanation:

       where pf = final momentum = m*vf

                  p₀ = initial momentum = m*v₀

       [tex]m*v_{f} - m*v_{o} = -2*v_{o} -m*v_{o} = -3*m*v_{o} = -3*0.3kg*12.68m/s = -11.41m/s (2)[/tex]

       [tex]F_{net} = \frac{-11.41m/s}{0.05s} = -228.24 N (3)[/tex]

Answer:

Explanation:

Kinetic energy of the mass of 10 kg

= 1/2 m v² , m is mass and v is velocity .

= .5 x 10 x 20²

= 2000 J

The opposing force stops it . so work done by opposing force will be equal to this energy and it will be negative .

So energy transfer will be  - 2000 J .

= 2 kJ .

If distance travelled by mass is d , force 25 N will have a displacement of d . so work done by force of 25 N

= 25 x d

25 d = 2000

d = 80 m .

Hence system travels a distance of 80 m .

Answer:

vegetables and legumes or beans

fruit

lean meats and poultry, fish, eggs, tofu, nuts and seeds, legumes or beans

grain (cereal) foods, mostly wholegrain or high cereal fibre varieties

milk, yoghurt, cheese or alternatives, mostly reduced fat.

Explanation:

Foods are grouped together because they provide similar amounts of key nutrients. For example, key nutrients of the milk, yoghurt, cheese and alternatives group include calcium and protein, while the fruit group is a good source of vitamins, especially vitamin C.

Answer:

It is a measure of the electric force per unit charge on a test charge.

Explanation:

The magnitude of the electric field is defined as the force per charge on the test charge.

Since we define electric field as the force per charge, it will have the units of  force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).

Because from the definition of the electric field intensity, it can be defined as force per unit charge. The correct answer is option D

Electric field is the region of space where electric force can be felt. It can also be expressed as electric field intensity E. Mathematically, it can be expressed as;

E = F/q or E = V/d

From the question, the statements that is true concerning the magnitude of the electric field at a point in space is " It is a measure of the electric force per unit charge on a test charge "

Because from the definition of the electric field intensity, is can be defined as force per unit charge.

Therefore, option D is the right answer.

Learn more about Electric Field here : https://brainly.com/question/14372859

in the service producing sector

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          [tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_i m_i[/tex]

in this case we have two bodies

          x_{cm} = [tex]\frac{1}{M}[/tex] (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = [tex]\frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )[/tex]

          x_{cm} = [tex]\frac{1}{M} ( m_1 v_1 + m_2 v_2 )[/tex]

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = [tex]\frac{m_1}{m_1+m_2}[/tex]  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         [tex]v_{cm o}[/tex] = [tex]\frac{1}{m_1 +m_2}[/tex] (m₁ vo)

final

         [tex]v_{cm f}[/tex] = [tex]\frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o )[/tex] ( v) = v

         v_{cm f} =  [tex]\frac{m_1}{M^2} v_o[/tex]

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = [tex]\frac{1}{M} = \frac{1}{m_1 +m_2}[/tex]

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = [tex]\frac{m_1 -m_2}{m1 +m2 } v_o[/tex]

             v_{2f} = [tex]\frac{2 m_1}{m-1+m_2} vo[/tex]

now let's find the velocity of the center of mass

initial

          [tex]v_{cm o}[/tex] = [tex]\frac{1}{M}[/tex] m₁ v₀

final

          [tex]v_{cm f}[/tex] = [tex]\frac{1}{M}[/tex]  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = [tex]\frac{1}{M}[/tex] [  [tex]m_1 \frac{m_2}{M}[/tex] + [tex]m_2 \frac{2 m_1}{M}[/tex] ] v₀

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = [tex]\frac{1}{M}[/tex] M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]