What word chemical equation describes this chemical reaction?

Answer : sodium + chlorine → sodium chloride

Answers

Answer 1
Sodium + chlorine = sodium chloride

Related Questions

Which of the following represents the velocity time relationship for a falling apple?

Answers

Answer "a" would be correct.

Answer:

d

Explanation:

There's an acceleration from gravity, thus the velocity is becoming faster and faster as it reaches the ground. Thus its D

Brainliest please~

There are two beakers of water on the table. We can compare the average kinetic energy of the water molecules in the two beakers by measuring their

A temperatures.

B volumes.

C densities.

D masses.

Answers

Answer: masses

Explanation:

Trust me

A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium are in the tank? Express your answer in moles.

Answers

Answer:

82.25 moles of He

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 10 L

Mass of He = 0.329 Kg

Temperature (T) = 28.0 °C

Molar mass of He = 4 g/mol

Mole of He =?

Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
Take the coefficient of expansion to be 25 x 10-6 (oC)-1 . Write the answer in meters with three significant figures

Answers

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

A mass-spring system oscillates with an amplitude of 4.20 cm. If the spring constant is 262 N/m and the mass is 560 g, determine the mechanical energy of the system.

Answers

Answer:

[tex]M.E=41J[/tex]

Explanation:

From the question we are told that:

Amplitude [tex]a=4.20cm[/tex]

Spring Constant [tex]K=262N/m[/tex]

Mass [tex]m=560g[/tex]

Generally the equation for mechanical energy is mathematically given by

[tex]M.E=\frac{1}{2}km^2[/tex]

[tex]M.E=0.5*262*0.56^2[/tex]

[tex]M.E=41J[/tex]

Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 35 cm while traveling through air. What are the
(a) frequency and
(b) wavelength as the signal travels through 3-mm-thick window glass into your room?

Answers

Answer:

(a) 8.57 x 10^8 Hz

(b) 23.3 cm

Explanation:

Wavelength = 35 cm = 0.35 m

speed =3 x10^8 m/s

Let the frequency is f.

(a) The relation is

speed  = frequency x wavelength

3 x 10^8 = 0.35 x f

f = 8.57 x 10^8 Hz

(b) refractive index of glass  is 1.5

The relation for the refractive index and the wavelength is

wavelength in glass= wavelength in air/ refractive index.

Wavelength in glass= 35/1.5 = 23.3 cm

A ​12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall​ (above the fire​ truck) if the ladder makes an angle of with the horizontal

Answers

Complete Question

A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall​ (above the fire​ truck) if the ladder makes an angle of  

40° 16' with the horizontal.

Answer:

 [tex]d=8.01m[/tex]

Explanation:

From the question we are told that:

Length of ladder [tex]l=12.5m[/tex]

Angle [tex]\theta=40° 16'=20.26 \textdegree[/tex]

Generally the Trigonometric equation for distance d it goes up the wall is mathematically given by

 [tex]d=l sin \theta[/tex]

 [tex]d=12.5 sin 40.26[/tex]

 [tex]d=8.01m[/tex]

What is (a) the x component and (b) the y component of the net electric field at the square's center

Answers

Answer:

What is (a) the x component and (b) the y component of the net electric field at the square's center

The gravitational field strength due to its planet is 5N/kg What does it mean?

Answers

Answer:

The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.

Weight can be calculated using the equation:

weight = mass × gravitational field strength

This is when:

weight (W) is measured in newtons (N)

mass (m) is measured in kilograms (kg)

gravitational field strength (g) is measured in newtons per kilogram (N/kg)

recognizing forms of energy

Answers

Answer:

hi the question isn't obvious and need a photo I guess

The question is missing

vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?​

Answers

Answer:

45 × 8 units = A + B as formular

A hot air balloon is a sphere of volume 2210 m3. The density of the hot air inside is 1.13 kg/m3, while the air outside has a density of 1.29 kg/m3. The balloon itself has a mass of 240 kg. What is the TOTAL NET force acting on the balloon?
[?]N

Answers

The total net force acting on the balloon will be 24498 Newtons

Given that

Volume of the balloon = 2210 cubic meter

Density of the air inside the balloon = 1.13  kg/m3

What will be the net force exerted on the balloon ?

Here force on the balloon will be equal to the weight of the air displaced by balloon

[tex]F= mass of air displaced\times gravity[/tex]

[tex]F= Density \times volume \times gravity[/tex]

[tex]F=1.13 \times 2210 \times 9.81[/tex]

[tex]F=24498 N[/tex]

The total net force acting on the balloon will be 24498 Newtons

To know more about buoyancy force follow

https://brainly.com/question/117714

Electrical resistance is a measure of resistance to the flow of _?____

Answers

Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.

Hope this helps!!!!

Answer:

electric current

Explanation:

The answer is electric current

Based on the information in the table, what
is the acceleration of this object?

t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2

Answers

Answer:

Option A. –5 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (v₁) = 9 m/s

Initial time (t₁) = 0 s

Final velocity (v₂) = –6 m/s

Final time (t₂) = 3 s

Acceleration (a) =?

Next, we shall determine the change in the velocity and time. This can be obtained as follow:

For velocity:

Initial velocity (v₁) = 9 m/s

Final velocity (v₂) = –6 m/s

Change in velocity (Δv) =?

ΔV = v₂ – v₁

ΔV = –6 – 9

ΔV = –15 m/s

For time:

Initial time (t₁) = 0 s

Final time (t₂) = 3 s

Change in time (Δt) =?

Δt = t₂ – t₁

Δt = 3 – 0

Δt = 3 s

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Change in velocity (Δv) = –15 m/s

Change in time (Δt) = 3 s

Acceleration (a) =?

a = Δv / Δt

a = –15 / 3

a = –5 m/s²

Thus, the acceleration of the object is

–5 m/s².

A body of mass 4kg is moving with a velocity of 108km/h . find the kenetic energy of the body.​

Answers

Answer:

KE = 2800 J

Explanation:

Usually a velocity is expressed as m/s. Then the energy units are joules.

[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]

v = 30 m / sec

KE = 1/2 * 4 * (30)^2

KE =2800 kg m^2/sec^2

KE = 2800 Joules

A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km

Answers

Answer:

20 seconds

Explanation:

We are given 2 givens in the first statement

v0=0 and a=5

And we are trying to find time needed to cover 1km or 1000m.

So we use

x-x0=v0t+1/2at²

Plug in givens

1000=0+2.5t²

solve for t

t²=400

t=20s

need help pleaseee,question is in the pic​

Answers

Explanation:

For engine 1,

Energy removed = 239 J

Energy added = 567 J

[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]

For engine 2,

Energy removed = 457 J

Energy added = 789 J

[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]

For engine 3,

Energy removed = 422 J

Energy added = 1038 J

[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]

So, the engine 2 has the highest thermal efficiency.

What is significant about the primary colors of pigments?
They can be mixed together to make almost any other color.
Any two primary colors of pigments combine to make white pigment.
Each primary color of pigment absorbs all other colors.
Any two primary colors of pigments combine to make black pigment.

Answers

Answer:

They can be mixed together to make almost any other color.

Explanation:

All the three primary colors can mix to form white color.

Blue and red mix to form a black color.

what is time taken by radio wave to go and return back from communication satellite to earth??​

Answers

Answer:

Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.

Explanation:

hope it help

What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5

Answers

Answer:

2) when acceleration triples force triples,  3) a diagram with dynamic friction force in the opposite direction of movement of the car

4)  a = 2.44 ft / s², 5)  fr = 894.3 N

Explanation:

In this exercise you are asked to answer some short questions

2)  Newton's second law is

         F = m a

when acceleration triples force triples

3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.

The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car

4) let's use the scientific expressions

          v = v₀ - a t

as the car stops v = 0

          a = v₀ / t

let's reduce the magnitudes

          v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s

          a = 58.667 / 24

          a = 2.44 ft / s²

5) let's use Newton's second law

           fr = m a

We must be careful not to mix the units, we will reduce the acceleration to the system Yes

           a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²

           fr = 1200  0.745

           fr = 894.3 N

1. A block of mass m = 10.0 kg is released with a speed v from a frictionless incline at height 7.00 m. The
block reaches the horizontal ground and then slides up another frictionless incline as shown in Fig. 1.1. If the
horizontal surface is also frictionless and the maximum height that the block can slide up to is 26.0 m, (a) what
is the speed v of the block equal to when it is released and (b) what is the speed of the block when it reaches
the horizontal ground? If a portion of length 1 2.00 m on the horizontal surface is frictional with coefficient
of kinetic friction uk = 0.500 (Fig. 1.2) and the block is released at the same height 7.00 m with the same
speed v determined in (a), (c) what is the maximum height that the block can reach, (d) what is the speed of the
block at half of the maximum height, and (e) how many times will the block cross the frictional region before
it stops completely?
1 = 2.00 m (frictional region)

Answers

Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).

At point A, the block has total energy

E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²

E (A) = 686 J + 1/2 (10.0 kg) v₀²

At point B, the block's potential energy is converted into kinetic energy, so that its total energy is

E (B) = 1/2 (10.0 kg) v₁²

The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,

E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J

Throughout this whole process, energy is conserved, so

E (A) = E (B) = E (C) = E (D)

(a) Solve for v₀ :

686 J + 1/2 (10.0 kg) v₀² = 2548 J

==>   v₀19.3 m/s

(b) Solve for v₁ :

1/2 (10.0 kg) v₁² = 2548 J

==>   v₁22.6 m/s

Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:

• net horizontal force:

∑ F = -f = ma

• net vertical force:

F = n - mg = 0

where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :

n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N

f = µn = 0.500 (98.0 N) = 49.0 N

==>   - (49.0 N) = (10.0 kg) a

==>   a = - 4.90 m/s²

The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that

v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)

==>   v₂² = 490 m²/s²

and thus the block has total/kinetic energy

E (C) = 1/2 (10.0 kg) v₂² = 2450 J

(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so

2450 J = (10.0 kg) (9.80 m/s²) h

==>   h = 25.0 m

(d) At half the maximum height, the block has speed v₃ such that

2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²

==>   v₃15.7 m/s

The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by

v = v₁ + at = 22.6 m/s - (4.90 m/s²) t

The block comes to a rest when v = 0 :

0 = 22.6 m/s - (4.90 m/s²) t

==>   t ≈ 4.61 s

It covers a distance x after time t of

x = v₁t + 1/2 at ²

so when it comes to a complete stop, it will have moved a distance of

x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m

(e) The block crosses the rough region

(52.0 m) / (2.00 m) = 26 times

The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane: y1(x, t) = (8.20 mm) sin(4.00πx - 430πt) y2(x, t) = (8.20 mm) sin(4.00πx + 430πt), with x in meters and t in seconds. An antinode is located at point A. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?

Answers

Answer:

Explanation:

From the information given:

The angular frequency ω = 430 π rad/s

The wavenumber k = 4.00π which can be expressed by the equation:

k = ω/v

4.00 =  430 /v

v = 430/4.00

v = 107.5 m/s

Similarly: k  = ω/v = 2πf/fλ

We can say that:

k = 2π/λ

4.00 π = 2π/λ

wavelength λ = 2π/4.00 π

wavelength λ = 0.5 m

frequency of the wave can now be calculated by using the formula:

f = v/λ

f = 107.5/0.5

f = 215 Hz

Also, the Period(T) = 1/215 secs

The time at which particle proceeds from point A  to its maximum upward displacement  and to its maximum downward displacement  can be computed as t = T/2;

Thus, the distance(x) covered by each wave during this time interval(T/2) will be:

x = v * t

x = v * T/2

x = λ/2

x = 0.5/2

x =  0.25 m

what is conservation energy?

Answers

Explanation:

Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant

hope it is helpful to you

When you hammer a nail into wood, the nail heats up. 30 Joules of energy was absorbed by a 5-g nail as it was hammered into place. How much does the nail's temperature increase (in °C) during this process? (The specific heat capacity of the nail is 450 J/kg-°C, and round to 3 significant digits.

Answers

Answer:

13.33 K

Explanation:

Given that,

Heat absorbed, Q = 30 J

Mass of nail, m = 5 g = 0.005 kg

The specific heat capacity of the nail is 450 J/kg-°C.

We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:

[tex]Q=mc\Delta T\\\\\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{30}{0.005\times 450}\\\\=13.33\ K[/tex]

So, the increase in temperature is 13.33 K.

Three 15-Ω and two 25-Ω light bulbs and a 24 V battery are connected in a series circuit. What is the current that passes through each bulb?
1) 0.18 A
2) 0.25 A
3) 0.51 A
4) 0.74 A
5) The current will be 1.6 A in the 15-Ω bulbs and 0.96 A in the 25-Ω bulbs.

Answers

Answer:

I = 0.25 A

Explanation:

Given that,

Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.

In series combination, the equivalent resistance is given by :

[tex]R=R_1+R_2+R_3+....[/tex]

So,

[tex]R=15+15+15+25+25\\\\=95\ \Omega[/tex]

The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.

V = IR

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A[/tex]

So, the current of 0.25 A passes through each bulb.

12) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen, what things could you do to help yourself (before calling me over to assist you?)

Answers

Answer:

See the answer below

Explanation:

After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem

1. Drop a few drops of immersion oil on the slide and view again under high the power objective.

2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.

The north pole of magnet A will __?____ the south pole of magnet B

Answers

Answer:

A will attract

B will repare

A 64-ka base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.

Required:
a. How much mechanical energy is tout due to friction acting on the runner?
b, How far does he slide?

Answers

Answer:

Explanation:

From the given information:

mass = 64 kg

speed = 3.2 m/s

coefficient of friction [tex]\mu =[/tex] 0.70

The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:

[tex]W = \Delta K.E[/tex]

[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]

[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]

Thus, the mechanical energy touted = 327.68 J

According to the formula used in calculating the frictional force

[tex]F_r = \mu mg[/tex]

= 0.70 × 64  kg× 9.8 m/s²

= 439.04 N

The distance covered now can be determined as follows:

d = W/F

d = 327.68 J/  439.04 N

d = 0.746 m

Explain what a circuit breaker is and how it helps protect your house?

Answers

Explanation:

A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by overcurrent/overload or short circuit. Its basic function is to interrupt current flow after protective relays detect a fault.

Circuit breakers have been designed to detect when there is a fault in the electricity, so it will “trip” and shut down electrical flow. ... This detection is key to preventing surges of electricity that travel to appliances or other outlets, which can cause them to break down

A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.

Answers

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]

             a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

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