What is inertia?by Walter Levin..
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The final temperature of the tea cup is 100°C.

The Internal energy is the energy of a substance due to to the constant random motion of its particles.

The symbol for internal energy of a substance is U and it is measured in Joules.

ΔU = q + W

In conclusion,  the final temperature of the tea cup at room temperature of 24 °C which is heated until it has twice the internal energy is 100°C.

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Hi there!

To start, we can use the equation for the charge of a capacitor versus time.
[tex]q(t) = C\epsilon (1 - e^{-\frac{t}{\tau}})[/tex]

[tex]q(t)[/tex] = Charge on capacitor vs. time

[tex]\epsilon[/tex] = EMF of battery (40 V)

**Note: 'Cε' is equal to the MAXIMUM CHARGE on a capacitor.

Where the time constant is given as:
[tex]\tau = RC[/tex]
[tex]\tau[/tex] = Time Constant (s)
[tex]R[/tex] = Resistance (0.5 MΩ)
[tex]C[/tex] = Capacitance (90 μF)

Let's first convert the given units to the necessary 'J',  'Ω' and 'F' to make things easier:

[tex]J = 50.2 mJ[/tex]

[tex]50.2 mJ * \frac{0.001J}{1mJ} = 0.0502 J[/tex]

[tex]R = 0.5 M\Omega\\\\0.5M\Ohm * \frac{1000000\Omega}{1M\Ohm} = 500000\Omega[/tex]

[tex]F = 90 \mu F[/tex]

[tex]90 \mu F * \frac{0.000001 F}{1\mu F} = 0.00009 F[/tex]

Now, let's calculate the time constant.

[tex]\tau = RC = (500000)(0.00009) = 45 s[/tex]

Time to use the equation for energy stored in a capacitor:
[tex]E_C = \frac{1}{2}\frac{Q^2}{C}[/tex]

Our 'Q' varies with time, so:
[tex]E(t)= \frac{1}{2}\frac{( C\epsilon (1 - e^{-\frac{t}{\tau}}))^2}{C}\\\\E(t) = \frac{1}{2}\frac{( C^2\epsilon^2 (1 - e^{-\frac{t}{\tau}})^2)}{C}\\\\E(t) = \frac{1}{2}( C\epsilon^2 (1 - e^{-\frac{t}{\tau}})^2)[/tex]

Let's now plug values into our equation.

[tex]0.0502 = \frac{1}{2}(0.00009)(40^2) (1 - e^{-\frac{t}{45}})^2[/tex]

Simplifying:
[tex]0.697 = (1 - e^{-\frac{t}{45}})^2[/tex]

Take the square root of both sides.

[tex]0.835 = 1 - e^{-\frac{t}{45}}\\\\ e^{-\frac{t}{45}} = 0.165[/tex]

Take the natural log of both sides.

[tex]-\frac{t}{45} = ln(0.165) \\\\\frac{t}{45} = 1.802\\\\t = \boxed{81.08 s}[/tex]

The magnitude of the force on the left-hand pole is mathematically given as

f'=0.167N

[tex]Q=45 \textdegree[/tex]

Generally, the equation for Force is  mathematically given as

[tex]F=\frac{mg}{sin\theta}[/tex]

Therefore

[tex]F=\frac{17.1*10^{-3}*9.8}{sin45}[/tex]

F=0.237N

Considering horizontal

f'-fcos=0

Therefore

f'=0.237*cos45

f'=0.167N

In conclusion, the slope

[tex]tan\theta=lv/ln\\\\tan\theta=30/30\\\\\theta=tan^{-1}\\\\[/tex]

[tex]Q=45 \textdegree[/tex]

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The speed of the longitudinal waves along the spring is 1.5 m/s.

The wavelength of the longitudinal waves produced when the free end is moved to and fro is 0.6 m.

The wavelength is the distance between the adjacent crest or trough of the sinusoidal wave. The wavelength is the reciprocal of the frequency of the wave.

Wavelength λ = c/f

where c is the speed of light in vacuum = 3*10⁸ m/s

A slinky spring 3m long rests on a horizontal bench with one end fixed. When the free end is suddenly pushed forward, the compression pulse travels to the fixed end and back in 2 secs.

λ = v/f

v =fλ =λ /T where T is the time period = 2sec

Put the values we get

v = 3/2 =1.5 m/s

Thus, the speed of wave is 1.5 m/s.

Using the same expression for frequency 2.5Hz, wavelength is

λ = v/f

λ = 1.5 /2.5 = 0.6 m

Thus,  the wavelength is 0.6m.

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The tension in the rope B is determined as 10.9 N.

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

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Answer:

(D) any of these objects can be used as long as all points on the object lie farther away from the lens than the focal length of the lens.

Points lying between the focal point and the lens will be erect and will not form an object visible on a screen.

The speed of the roller coater at the bottom of the hill is 31 m/s.

Apply the principle of conservation of mechanical energy as follows;

K.E(bottom) = P.E(top)

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

v = √(2 x 9.8 x 49)

v = 31 m/s

Thus, the speed of the roller coater at the bottom of the hill is 31 m/s.

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The skin is NOT a reproductive organ and the spines attract predators are the incorrect statements.

The list is incorrect because skin is not a reproductive organ and the spines do not attract predators towards each other. Skin protects the body from the outer environment whereas the spines repel the predators from the plant body.

So we can conclude that the skin is NOT a reproductive organ and the spines attract predators are the incorrect statements.

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The total kinetic energy  of the sphere when it reaches the bottom of the incline is  44.57 J.

The total kinetic energy of the sphere = translational + rotational kinetic energy.

K.E = ¹/₂mv²

K.E(bottom) = P.E(top) = mgh

K.E = (2.35)(9.8)(1.29)

K.E = 29.71 J

K.E(rot) = ¹/₂Iω²

K.E(rot) = ¹/₂I(v²/R²)

where;

I = ¹/₂MR²

K.E(rot) = ¹/₂( ¹/₂MR²)(v²/R²)

K.E(rot) = ¹/₄Mv²

K.E(rot) = ¹/₄M(2gh)

K.E(rot) = ¹/₂Mgh

K.E(rot) = ¹/₂(29.71) = 14.855 J

Total kinetic energy = 29.71 + 14.855 = 44.57 J

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Answer:

the total number of atoms of each element present is the same on both sides of the equation. (if I remember correctly)

Answer:

[tex]1,\!000\; {\rm kg}[/tex].

Explanation:

The "[tex]g[/tex]" given in the question likely refers to the strength of the gravitational field. The standard unit of [tex]g[/tex] is [tex]{\rm m \cdot s^{-2}}[/tex], same as the standard unit of acceleration. However, since [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}[/tex], the standard unit of [tex]g[/tex] is equivalent to [tex]{\rm N \cdot kg^{-1}}[/tex].

For instance, the gravitational strength on the surface of the earth is approximately [tex]g = 10\; {\rm N \cdot kg^{-1}}[/tex]. However, the question did not specify the unit of this value.

If the mass of an object is [tex]m[/tex], the weight of that object would be [tex](\text{weight}) = m\, g[/tex] when the gravitational field strength around that object is [tex]g[/tex]. It is given that for the object in this question, [tex]m\, g = (\text{weight}) = 10,\!000\; {\rm N}[/tex] when [tex]g = 10\; {\rm N \cdot kg^{-1}}[/tex]. Therefore, the mass [tex]m\![/tex] of this object would be:

[tex]\begin{aligned}m &= \frac{(\text{weight})}{g} \\ &= \frac{10,\!000\; {\rm N}}{10\; {\rm N \cdot kg^{-1}}} \\ &= 1,\! 000\; {\rm kg}\end{aligned}[/tex].

Answer:

As of the 2019 redefinition of the SI base units, the kilogram is defined in terms of the second and the metre, both based on fundamental physical constants.

It is defined as the mass of a particular international prototype made of platinum-iridium and kept at the International Bureau of Weights and Measures. It was originally defined as the mass of one liter (10-3 cubic meter) of pure water. At the Earth's surface, a mass of 1 kg weighs approximately 2.20 pounds ( lb ).

Explanation:

The tension in the cable is mathematically given as

T =266.659 N

Generally, the equation for the angle of the boom with horizontal is  mathematically given as

A = tan-1(5 /10)

[tex]A= 26.57 \textdegree[/tex]

The angle of cable with horizontal

B = tan-1(4 / 10)

B= 21.80 degrees

Hence

175.5 * cos26.57 + 94.7 *cos26.57 * 1/2 = T (sin(26.57+21.80)) * 1

T =266.659 N

In conclusion, the tension in the cable.

T =266.659 N

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Lindsay should fly the plane in the direction [W 12.5° S] to get Hamilton.

Using Sine rule to solve this question

Sine rule => SinA/a = SinB/b = SinC/c = constant

The magnitude of wind is 50 with an angle of 60 degrees.

The magnitude of plane is 200 and the angle at which it should fly is unknown and should be θ.

One side is 50 km/hr at an angle of 60 degrees.

sin 60°/200 = sin θ / 50

50 × sin 60° = 200 × sin θ

√3/2 = 4 × sin θ

√3/8 = sin θ

sin θ = 0.2165

θ = sin⁻¹(0.2165)

θ = 12.5°

So Lindsay have to fly the plane in the direction of [W 12.5° S].

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The maximum height to be reached by the water at the given pressure is 13 m.

The maximum height to be reached by the water is calculated as follows;

P = ρgh

where;

1 kgf/cm² = 98066.5 Pa = 98066.5 N/m²

1.3 kgf/cm² = 127486 N/m²

h = P/ρg

h = (127486)/(1000 x 9.8)

h = 13 m

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The term "latency" describes the 27 seconds that pass after someone has finished texting or conversing on the phone before they can concentrate again on driving. The given statement is true.

A measure of delay is latency. The amount of time it takes for some data to travel across a network is known as latency.

The amount of time it takes for information to travel from its source to its destination and back is often quantified as a round trip delay.

The 27 seconds that pass after someone has done texting or talking on the phone before they can focus on driving again are referred to as "latency." The assertion is accurate.

Hence, the given statement is true.

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Magnets are materials which are able to attract others to themselves by producing a magnetic field.

Based on the properties of magnetic lines of forces;

In conclusion, the law of magnetism states that like poles repel whereas unlike poles attract.

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(a)The position at the end of 1 s of a point originally at the top of the wheel will be 9.425 radians.

(b)The magnitude and the direction of the acceleration at the end of 1 s will be 1.885 m/sec² clockwise.

Angular acceleration is defined as the pace of change of angular velocity concerning time.

Given data;

Diameter of the wheel,d = 0.2 m

Initial velocity,ω₀ = 0

Angular acceleration,α = c

n = 900 rev min¹

ω = 94.2478 rad /sec

Time of rolling,t = 5 s

From Newton's first equation of motion;

ω = ω₀ + αt

94.2478 rad /sec  = 0 + α(5 sec)

α  = 18.85 rad/sec²

The position at the end of 1 s of a point is found as;

Θ = ω₀t+1/2(αt²)

Θ = 0+1/2[(18.85 rad/sec²) × (1 sec)²]

Θ = 9.425 radian

The magnitude and the direction of the acceleration at the end of 1 s are found as;

a = αr

a= 18.85 rad/sec² × 0.1 m

a = 1.885 m/sec²

Hence the position at the end of 1 s and magnitude and the direction of the acceleration at the end of 1 s will be  18.85 rad/sec²and the magnitude and the direction of the acceleration at the end of 1 s will be 1.885 m/sec² clockwise.

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Answer:

Velocity: 4 m/s

Distance: 4 m

Explanation:

initial velocity: u = 0 m/s

Acceleration: a = 2 m/s²

Time taken: t = 2 seconds

Distance traveled: d = (what we need to find)

Final velocity: v = (what we need to find)

---------------------------------------

v = u + at

v = 0 + 2 × 2

v = 4 m/s

---------------------------------------

d = 0 × 2 + 1/2 × 2 × (2)²

d = 1/2 × 2 × 4

d = 4 m

B. The electric potential at the given point from the proton is 2.7 x 10¹ volts

The electric potential of the proton is calculated as follows;

V = kq/r

where;

V = (9 x 10⁹ x 1.6 x 10⁻¹⁹)/(5.3 x 10⁻¹¹)

V = 27.2 volts

V = 2.7 x 10¹ volts

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Answer:

E=kQ/r*2

=(9. 10*9)(11.10*-9)/(0,07)*2

=20204 N. C*-1

The field lines will point towards the source charge because it is negatively charged.

The horizontal component of the vector is determined as 6i.

The horizontal component of the vector is calculated as follows;

V = (0 i, 0 j) + (6i, 8j)

V = (6i, 8j)

where;

Thus, the horizontal component of the vector is determined as 6i.

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B. When the ball is rolling across the floor at a constant velocity, the change in its kinetic energy is zero.

The change in kinetic energy of an object is the dereference between the final kinetic energy and the initial kinetic energy.

ΔK.E = K.Ef - K.Ei

ΔK.E = 0.5m(vf² - vi²)

where;

At constant velocity, the initial velocity and final velocity are equal.

ΔK.E = 0.5m(0) = 0

Thus, when the ball is rolling across the floor at a constant velocity, the change in its kinetic energy is zero.

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The net force is the 10N force and its direction is towards F2 force.

The resultant force or the net force is the actual force that acts on a body. We are told that the two forces act in opposite directions hence the net force is F2 - F1  = 30 N - 20 N = 10 N.

ii) The direction of this net force is going to be towards the 30N force.

iii) If the body does not move under the application of these forces then the F1 force must be reinforced by another force which balances it against the F2 force.

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the mass of mars is

6.39 × 10^23 kg

A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.

It is the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

A metallic rod of mass 150 g (m) absorbs 82.5 cal of heat (Q) and its temperature raises from 20 °C to 25 °C. We can calculate the specific heat (c) of the metal using the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 82.5 cal / 150 g × (25 °C - 20 °C) = 0.11 cal/g.°C

A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.

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For the given question the answer to the first part is that frequency of sound will be 404.76 Hz. and answer to the second part will be that the initial frequency for which that place will have the highest level of sound intensity

Given, the speed of sound 340 m/s

a) L1 + L2 = 3.5 m

and Ld = | L1 - L2 | = ( 1.75 + 0.21 ) - (1.75 - 0.21 ) = 1.96 - 1.54 = 0.42 m

Ld = λ/2

λ = 2Ld = 2×0.42 = 0.84 m

and finally,

f = v/λ

f = 340/0.84

f = 404.76 hertz

Frequency came out to be 404.76 hertz in this case

b) For the first frequency

0.42 = λ

f = v/λ

f = 340 / 0.42

f = 809.52 Hertz

Frequency came out to be 809.52 Hertz in this case.

To conclude with we can say that the Frequency of the sound in case on came out to be 404.76 hertz which is approximately 405 Hz after applying all the concepts and calculations, in second case first frequency for which that location will be a maximum of sound intensity came out to be 809.52 Hertz after applying all the concepts and calculations.

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Hello and Good Morning/Afternoon:

Original Question: C₂H₅OH + __O₂ → __CO₂ + __ H₂O

To balance this equation:

⇒ must ensure that there is an equal number of elements on both sides of the equation at all times

Let's start balancing:

              ⇒ but only so far one on the right side

         C₂H₅OH + __O₂ →  2CO₂ + __ H₂O

               ⇒ but only so far two on the right side

         C₂H₅OH + __O₂ →  2CO₂ + 3H₂O

                ⇒ but only so far three on the left side

         C₂H₅OH + 3O₂ →  2CO₂ + 3H₂O

Let's check and make sure we got the answer:

                           C₂H₅OH + 3O₂ →  2CO₂ + 3H₂O

                2 Carbon                ⇔                    2 Carbon

                6 Hydrogen            ⇔                   6 Hydrogen

                7 Oxygen                ⇔                   7 oxygen

Thefore the coefficients in order are:

  ⇒ 1, 3, 2, 3

Answer: 1,3,2,3

Hope that helps!

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