To calculate the final enthalpy of the overall chemical equation, which step must occur?

Answer:

The tub turns 37.520 revolutions during the 25-second interval.

Explanation:

The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:

[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)

Then, we expand the previous expression by kinematic equations for uniform accelerated motion:

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)

Where:

[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.

[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.

And each acceleration is determined by the following formulas:

Acceleration

[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)

Deceleration

[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)

Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.

If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:

[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]

[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]

[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]

[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]

[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]

[tex]\Delta n = 37.520\,rev[/tex]

The tub turns 37.520 revolutions during the 25-second interval.

Answer:

26280

Explanation:

In current time, good telescope can measure redshift to a galaxy in 10 minutes.  

Thus, in one year that has on an average 365 days, the total time taken to  measure redshifts is = ( 365 *12 *60) minute  

= 262800 minutes .

Hence, the number of redshifts observed in a year = (262800/10) = 26280

Answer: divide by zero, or square root of a negative

Explanation: If your question is how to get infinity as an answer to a problem, that generally means that the answer is undefined or doesn't exist.

A couple of ways to get that...

Answer:

i believe the answer is D

Explanation:

Hope it works

Answer:

Explanation:

Answer: 0

Explanation: Trust

The magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.

The magnetic force on a charged particle moving through a magnetic field can be calculated using the formula:

F = q * v * B * sin(θ)

Where:

F is the magnetic force,

q is the charge of the particle (in this case, 4.88 x 10^(-6) C),

v is the velocity of the particle (in this case, 265 m/s),

B is the magnetic field strength (in this case, 0.0579 T),

θ is the angle between the velocity vector and the magnetic field vector (in this case, 90 degrees).

Plugging in the values:

F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * sin(90°)

Since sin(90°) is equal to 1, the equation simplifies to:

F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * 1

Calculating the value:

F = 6.47 x 10^(-4) N

Therefore, the magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.

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Answer:

     v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

         ρ g y₁ = ½ ρ v₂²

         v₂ = [tex]\sqrt {2g \ y_1}[/tex]

let's calculate

         v₂ = √( 2 9.8 250)

         v₂ = 70 m / s

Answer:

The explosion is set off by an electrostatic spark. When the mixture ignites, the rapid increase in temperature brings about a huge increase in gas pressure. If the burning vapour were to be confined the resulting rise in pressure could destroy the chamber with a loud explosion.

Explanation:

Explanation:

Total voltage = 9 × 2 = 18v

Resistance = 82 Ω

Ohm's law::

V = IR

18v = 82 Ω × I

18v /82 /Ω = I

18/82 Ampere is the current

Answer:

c. physiological reactions; convey emotional states to other members of the species

Explanation:

Darwin believed that emotional expressions began as physiological reactions that came to have evolutionary value because they convey emotional states to other members of the species. These reactions are used by many different types of species to convey various things.

Complete Question:

Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.

Answer:

Resistance = 9.95 Ohms

Explanation:

Given the following data;

Length = 0.65 m

Radius = 0.25 mm to meters = 0.00025 m

Resistivity = 3 * 10^{-6} ohm-metre.

To find the resistance of the wire;

Mathematically, resistance is given by the formula;

[tex] Resistance = P \frac {L}{A} [/tex]

Where;

First of all, we would find the cross-sectional area of the wire.

Area of circle = πr²

Substituting into the equation, we have;

Area  = 3.142 * (0.00025)²

Area = 3.142 * 6.25 * 10^{-8}

Area = 1.96 * 10^{-7} m²

Now, to find the resistance of the wire;

[tex] Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}} [/tex]

[tex] Resistance = 3 * 10^{-6} * 3316326.531 [/tex]

Resistance = 9.95 Ohms

Answer:

Explanation:

This is an exercise in kinematics, the speed they give is in two dimensions, let's work on each component

X axis

initial velocity v₀ₓ = 3 m / s in a time of t = 12 s, the velocity is doubled, the final velocity is vₓ = 6 m / s

acceleration is

           vₓ = v₀ₓ + aₓ t

           aₓ = [tex]\frac{v_x - v_{ox}} {t}[/tex]

           aₓ = 6 - 3/12

           aₓ = 0.25 m / s²

the distance traveled is

           vₓ² = v₀ₓ² + 2 aₓx x

           x = vx² - vox² / 2a

           x = 6² - 3² / 2 0.25

           x = 54 m

 Y axis

we look for acceleration

          v_y =     v_{oy} + a_y t

          a_y = [tex]\frac{v_y - v_{oy} }{t}[/tex]

          a_y = [tex]\frac{8 -4} {12}[/tex]

          ay = 0.3333 m / s²

the distance is

          v_y² = v_{oy}² + 2 a⁷y

          y = vy² - voy² / 2 0.25

          y = 8² - 4² / 2 0/3333

          y = 72 m

the distance traveled is

           r = (54 i + 72j) m / s

Answer:

D) 1.5032 x 10^-10 J

Explanation:

The rest energy of a proton, E₀, follows the equation:

E₀ = mp*rate²

Where mass of proton, mp = 1.6726 x 10^-27kg

rate = 2.9979 x 10^8 m/s (2.9979 x 10^9 m/s is not the speed light)

E₀ = 1.6726 x 10^-27kg * (2.9979 x 10^8 m/s)²

E₀ =1.5032 x 10^-10 J

Right answer is:

OPTION C is the correct answer.

Explanation:

V= IR

35=I×7

I=35/7

I=5amperes

pls give brainliest

Answer:

deductive reasoning usually follows steps .

Answer:

right one is more efficient

Explanation:

Answer:

10.66 m/s

Explanation:

Applying,

The law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').............. Equation 1

Where m = mass of the first puck, u = initial velocity of the first puck, m' = mass of the second puck, u' = initial velocity of the second puck, V = Velocity of the center of the two pucks

make V the subject of the equation

V = (mu+m'u')/(m+m').............. Equation 2

Given: m = 0.18 kg, u = 16 m/s, m' = 0.14 kg, u' = 3.8 m/s

Substitute these values into equation 2

V = [(0.18×16)+(0.14×3.8)]/(0.18+0.14)

V = (2.88+0.532)/(0.32)

V = 3.412/0.32

V = 10.66 m/s

Answer:

[tex]F_x = 100N[/tex]

[tex]F_y = 100\sqrt 3 \ N[/tex]

Explanation:

Given

[tex]F = 200N[/tex]

[tex]\theta = 60^o[/tex]

Required

The component of the force in F direction

To do this, we simply calculate the force in the vertical and horizontal direction.

This is calculated as:

[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal

[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical

So, we have:

[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal

[tex]F_x = 200N * \cos(60^o)[/tex]

[tex]F_x = 200N * 0.5[/tex]

[tex]F_x = 100N[/tex]

[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical

[tex]F_y = 200N * \sin(60^o)[/tex]

[tex]F_y = 200N * \frac{\sqrt 3}{2}[/tex]

[tex]F_y = 100\sqrt 3 \ N[/tex]

Answer:

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Explanation:

From the given information:

The elastic potential energy can be calculated by using the formula:

[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]

Making K the subject;

[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]

[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]

k = 2.3 × 10⁻² N/m

Now; the frequency of the small oscillation can be determined by using the formula:

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

where;

m = mass of each atom = 1.66 × 10⁻²⁶ kg

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Answer:

Explanation is given

Mark Me as Branliest

Answer:

6.28 m/s.

Explanation:

Given that,

The mass of the object, m = 0.2 kg

The radius of the circle, r = 0.5 m

It takes the object 0.500 s to complete one revolution.

We need to find the translational speed of the object. Let it is v. We know that,

[tex]v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.5}{0.5}\\\\v=6.28\ m/s[/tex]

So, the transalational speed of the object is 6.28 m/s.

Answer:

(a) The speed is 7.96 m/s

(b) The direction is 76 degree from positive X axis in counter clockwise direction.  

Explanation:

Width of river = 280 m

speed of river, vR = 4.7 m/s towards east

speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north

[tex]vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j[/tex]

(a) The velocity of boat with respect to ground is

[tex]\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s[/tex]

(b) The direction is given  by

[tex]tan\theta = \frac{6.4}{1.6} =4\\\\\theta = 76^o[/tex]

Answer: 3.32x10^-4

Explanation: Works for Acellus

Magnitude of magnetic force F= qvB Sin0®

q is the magnitude of charge moving with speed v in magnetic field B. Theta is the angle between velocity and magnetic field.

F=1.25×10⁻⁴C×5200m/s×8.49×10⁻⁴T(sin37deg).

F=3.32×10⁻⁴N.

An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge.

Matter is positively charged if it contains more protons than electrons, and negatively charged if it contains more electrons than protons.

Thus, the magnetic force on the charge is 3.32×10⁻⁴N.

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Answer:

the charge of the bees must be of the same sign

Explanation:

The electric field is given by the relation

            E = k q / r²

This electric field has outgoing direction if the charge is positive and incoming towards the charge if it is negative.

The force generated by this field on a test charge is

            F = q E

Since the charge is a scalar, the direction of the force is the same as the electric field.

In this case the two flowers are at a certain distance and the two charged bees land on them, so the force on a test charge is the vector sum of the force that each bee creates, so that this force is subtracted from the two bees must have the same charge sign.

The force created by the bee on the left goes to the right and the force created by the bee on the right goes to the left, so the forces are subtracted,

Consequently the charge of the bees must be of the same sign

Answer:

[tex]F=78.3hz[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=12[/tex]

Length [tex]l=80cm=0.8m[/tex]

Linear density [tex]\mu= 6.0g[/tex]

Generally the equation for Frequency is mathematically given by

 [tex]F=\frac{1}{2l}\sqrt{\frac{T}{K}}[/tex]

 [tex]F=\frac{1}{2(0.8)}\sqrt{\frac{12*9.8*0.8}{6*10^{-3}}}[/tex]

 [tex]F=78.3hz[/tex]