Velocity is the
i. distance travelled per unit time
il. time taken per unit distance
a.
9.
ii. time taken per unit displacement
iv. displacement per unit time​

Answer:

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Explanation:

B:Molecule

Molecules usually consist of 2 atoms linked together,however some molecules  consist of thousands of atoms!

Here is a list of smalles to largest with the answers provided

Molecules

Atoms

Elements

Mixtures

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Explanation:

Speed - The rate at which something moves

Velocity - The speed of something in a specific direction

Velocity is kind of a specific type of speed.

The charge on the capacitor 2 (C₂) is 48μF.

In the circuit diagram, C₁ and C₂ are in series connection, while C₃ is parallel to C₁ and C₂.

\(Q_1 = Q_2 = Q\\\\V = \frac{Q}{C_1} + \frac{Q}{C_2} \\\\V = Q(\frac{1}{C_1} + \frac{1}{C_2} )\\\\Q = V (\frac{C_1C_2}{C_1 + C_2} )\\\\Q = 24 \times (\frac{6 \times 10^{-6} \times 3 \times 10^{-6}}{6\times 10^{-6} \ + \ 3 \times 10^{-6}} )\\\\Q = \frac{4.32 \times 10^{-10}}{9\times 10^{-6}} \\\\Q = 48 \ \mu C\)

Thus, the charge on the capacitor 2 (C₂) is 48μF.

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Answer: gravity of earth = 9.8 n/s , new weight of the body on other planet = mg = 1.913 ×6×9.8=112..5

Explanation:

Answer:

Explanation:

Initial kinetic energy of particle

= 1/2 m V²

= .5 x .208 x 1.26²

= .165 J

Work done by force = force x displacement

= .766 x .195

= .149 J

This energy will be added up .

Total final kinetic energy

= initial kinetic energy + work done on the particle

= .165 + .149 J

=  .314 J .

Mass = 804.6 g

The dimension of the rectangular copper = 10 cm wide, 45 cm long, and 0.2 cm thick

The volume of the rectangular copper = 0.1m x 0.45m x 0.002m

The volume of the rectangular copper = 0.00009 m³

The density of copper = 8.94 g/cm³ = 8.94 x 1000 kg/m³

The density of copper = 8940 kg/m³

Mass = Density x Volume

Mass = 8940 x 0.00009

Mass = 0.8046 kg

Mass = 804.6 g

Answer:

1885.2 ohms

Explanation:

Step one:

given data

L=5H

f=60Hz

Required

The inductive reactance of the inductor

Step two:

Applying the expression

XL= 2πfL

substitute

XL=2*3.142*60*5

XL=1885.2 ohms

Direct axis and quadrature axis components of armature current are 30.28 A and 46.92 A respectively, and the excitation voltage of the generator is 765.36 V.

Given:

Apparent power (S) = 10 KVA = 10,000 VA

Line voltage (V) = 380 V

Frequency (f) = 50 Hz

Xd = 12 ohms

Xq = 8 ohms

Ra = 1 ohm

Power factor (pf) = 0.8 lagging

Load angle (δ) = 16.15 degrees

(i) Armature current's direct axis and quadrature axis components

We know that the apparent power is given by S = 3VLILcos(φ), where VL is the line voltage, IL is the line current, and φ is the angle between them. For a star-connected alternator, line voltage is equal to phase voltage, so we can write:

S = 3Vphase Iphase cos(φ)

Iphase = S / (3Vphase cos(φ))

For a lagging power factor, cos(φ) = 0.8, so

Iphase = 10,000 / (3 x 380 x 0.8) = 10.46 A

The direct axis component (Id) and the quadrature axis component (Iq) make up the armature current.  Using the given values of Xd, Xq, and Ra, we can calculate these components as follows:

Id = (VL - IaRa) / Xd

Iq = (VL - IaRa) / Xq

where Ia is the magnitude of the armature current, which is equal to the magnitude of the line current divided by √3. Thus,

Ia = Iphase / √3 = 10.46 / √3 = 6.03 A

Substituting the given values:

Id = (380 - 6.03 x 1) / 12 = 30.28 A

Iq = (380 - 6.03 x 1) / 8 = 46.92 A

(ii) Excitation voltage of the generator:

The excitation voltage (E) of the generator is given by:

E = Vphase + IqXq

Substituting the given values:

E = 380 + 46.92 x 8 = 765.36 V

Therefore, the direct axis and quadrature axis components of armature current are 30.28 A and 46.92 A respectively, and the excitation voltage of the generator is 765.36 V.

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Answer:

a

Explanation:

a

a

a

a

a

a

a

a

a is the answer

Answer:7.58 m

Explanation:

The required wavelength of a sound wave of frequency 660 Hz in steel is approximately 7.58 meters.

Wavelength is a fundamental property of waves, including electromagnetic waves like light and radio waves, as well as other types of waves, such as sound waves or water waves.

Here,
The wavelength (λ) of a sound wave is given by the formula,

λ = v/f

where v is the speed of sound in the medium and f is the frequency of the wave.

In this case, the speed of sound in steel is v = 5000 m/s, and the frequency of the sound wave is f = 660 Hz. Substituting these values into the formula, we get:

λ = 5000 m/s / 660 Hz

λ = 7.58 meters (rounded to two decimal places)

Therefore, the wavelength of a sound wave of frequency 660 Hz in steel is approximately 7.58 meters.

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Answer: sound waves travel faster in solid than water or air.

In vacuum sound waves don't travel at all

Answer:

A: Sound travels faster through air

Explanation:

The speed of sound is the distance travelled per unit of time by a sound wave as it propagates through an elastic medium. At 20 °C, the speed of sound in air is about 343 metres per second, or a kilometre in 2.9 s or a mile in 4.7 s.

Answer:

C. Mass

Explanation:

Answer:

heat

Explanation:

The car is moving at approximately 12.5 meters per second.

The kinetic energy (KE) of an object can be calculated using the formula:

KE = 1/2 * m * \(v^2\)

where

KE = kinetic energy,

m =Mass of the object, and

v = velocity.

In this case, we are given the mass (m) of the car as 1600 kg and the kinetic energy (KE) as 125,000 J. To find the velocity .

Substituting the  values , we have:

125,000 J = 1/2 * 1600 kg *\(v^2\)

Now, we can solve for v by rearranging the equation:

\(v^2\) = (2 * 125,000 J) / 1600 kg

\(v^2\) = 156.25 \(m^2/s^2\)

Taking the square root, we find:

v = √156.25\(m^2/s^2\)

v ≈ 12.5 m/s

Therefore, the car is moving at approximately 12.5 meters per second.

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The mass spectrometer separates ions of different masses by utilizing the relationship between the strength of the magnetic field, the accelerating voltage, the charge-to-mass ratio of the ions, and the resulting radius of the circular path

From the formula in the question;

B is a symbol for the magnetic field's intensity as it is applied to the mass spectrometer.

The accelerating voltage used to move the ions is called Vaccelm.

The charge of the ion, specifically its charge-to-mass ratio (q/m), is represented by the letter q.

The mass spectrometer may selectively alter the radius of the circular route for various ions by varying the magnetic field's intensity (B). This makes it possible to spatially segregate ions with various masses based on their various radii. The ions' locations and masses can then be measured using detectors placed along the journey.

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Answer:

The gravitational force on the moon is approximately 1/6th of the gravitational force on Earth. This means that the value of gravity on the moon is 1.62 m/s2.

Answer:

on earth the value of gravity on moon is 1.625 m/s2

Explanation:

1.625 m/s2

Answer:

The rock will rise H/6 units high on earth

Explanation:

In order to find the height to which rock rises, we use 3rd equation of motion. The 3rd equation of motion is as follows:

2gh = Vf² - Vi²

h = (Vf² - Vi²)/2g

where,

h = height

Vf = Final Velocity

Vi = Initial Velocity

g = acceleration due to gravity

ON MOON:

On moon:

h = H

Vf = 0 m/s (rock stops at highest point for a moment)

Vi = Vi

g = -(1/6) g (negative sign due to upward motion)

Therefore,

H = (0² - Vi²)/[-(2)(1/6)(g)]

H = 3Vi²/g

H/3 = Vi²/g  ------ equation (1)

ON EARTH:

On earth:

h = ?

Vf = 0 m/s (rock stops at highest point for a moment)

Vi = Vi (same initial velocity)

g = - g (negative sign due to upward motion)

Therefore,

h = (0² - Vi²)/(-2g)

h = Vi²/2g

h = (1/2)(Vi²/g)

using equation (1), we get:

h = (1/2)(H/3)

h = H/6

Two rocks are thrown directly upward with the same initial speeds is :

Part 1:

If the rock on the moon rises to a height H are :

                 2gh = Vf² - Vi²

                  h = (Vf² - Vi²)/2g

where,

Therefore,

The rock rise on the moon are H/3.

Part 2:

Therefore,

Using equation (1), we get:

The rock rise on the earth are H/6.

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Answer:

combustion reaction

glucose,oxygen

carbon dioxide,water,energy

8

no. required balanced equation is:

C6H12O6+6O2--->6CO2+6H2O+energy

endothermic as it needs heat in the form of sunlight.

Approximately 0.123 kg of liquid water at 26 degrees Celsius would be needed to melt 41 grams of ice.

To calculate the minimum amount of liquid water required to melt 41 grams of ice at 0°C, we need to consider the energy required for the phase change from solid to liquid, which is known as the specific heat of fusion of ice.

The energy required to melt 1 kg of ice is 3.33×105 J/kg.

Therefore, the energy required to melt 41 grams of ice is (3.33×105 J/kg) × (41/1000) kg = 13653 J.

To calculate the amount of liquid water required, we use the specific heat capacity of water, which is 4180 J/kg/°C.

Assuming the initial temperature of water is 26°C, the amount of water needed can be calculated as (13653 J) ÷ (4180 J/kg/°C) ÷ (26°C) = 0.123 kg or approximately 123 ml of water.

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The distance travelled by the vehicle around the circular track is 1,052.4 m.

The distance travelled by the vehicle in one complete cycle is calculated by using the following equation as show below.

d = 2πr

d = πd

Where;

In one complete cycle, the  vehicle will travel the circular track only once.

d = π(335 m)

d = 1,052.4 m

Thus, the distance travelled by the vehicle around the circular track is a function of the diameter of the circular track.

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The complete question is below:

A vehicle, starting from rest, accelerates on a circular track with a 335m diameter. What is the distance travelled by the vehicle when it makes one complete cycle?

ANSWER

20m and 40m

EXPLANATION

A rectangular ballroom has two dimensions: length and width,

The perimeter is the sum of all the sides of the rectangle,

The area is the product of these two dimensions,

We know that the perimeter is 120m and the area is 800m². Replacing in the equations above, we have two equations with two variables L and W,

Solve the first equation for W. Subtract 2L from both sides,

And divide both sides by 2,

The next step is to replace W by this expression in the second equation,

Now we have to solve this equation for L. Apply the distributive property on the right side of the equation,

Subtract 800 from both sides,

We have a quadratic equation to find the zeros of the polynomial. To solve it, we can use the quadratic formula,

In this case x = L, a = -1, b = 60 and c = -800,

We have two possible results for L,

To find the width of the ballroom we have to replace L into the equation we found when we solved the first equation for W,

So we have two possible results for W as well,

Note that the pairs are 20m and 40m or 40m and 20m. Hence, we can say that the dimensions of the ballroom are 20m and 40m

Given,

The spring constant of the spring, k=27.4 N/m

The stretch in the spring, x=2.1 m

The magnitude of the restoring force of the spring is given by,

Where F is the force required to stretch the given spring 2.1 m

On substituting the known values,

The force required to stretch the spring is 57.54 N

Answer:

A

Explanation:

took the test today at school, got a 95.

O a. How fast an object falls depends on its mass.

IS FALSE

Explanation:

Mass does not affect the speed of falling objects, assuming there is only gravity acting on it.

The magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.320 revolutions will be 1.266 m/s

Mass = m = 43.6 Kg.

radius = 0.300 m

F = 29 n

So, value of Torque to the rim of disc will be = Fr = 30*0.3 = 9 Nm.

So, the value of angular acceleration will be: \($\alpha=\frac{\tau}{I}: I=$\) moment of inert of dice

\(=\frac{M R^2}{2}\\$\alpha=\frac{29 \mathrm{~N} \times 0.3 \mathrm{~m}}{\frac{43.6 \times \mathrm{kg} \times(0.3 \mathrm{~m})^2}{2}}\\\\=\frac{29 \times 0.3 \times 2}{43.6 \times(0.3)^2} \mathrm{rad} / \mathrm{s}^2$\\\\=4.4342 \mathrm{rad} / \mathrm{s}^2$\)

\($\quad \Delta \theta=0.320 \mathrm{rev}=0.320 \times 215 \mathrm{rad}$\)

An Angular velocity at that instant \($=\omega_f > $\)

\($$\begin{aligned}& \omega_i=0 \\& \text { using } \omega_f^2=\omega_i^2+2 \alpha \Delta \theta \\& \omega_f^2=0+2 \times 4.4342 \times 0.320 \times 2 \pi \\& \omega_f=\sqrt{2 \times 4.4342 \times 0.320 \times 2\Pi} \mathrm{rad} / \mathrm{s} \\& \omega_f=4.22 \mathrm{rad} / \mathrm{s}\end{aligned}$$\)

The tangential velocity of point on \($n m=\omega_f r$\)

\($$\begin{aligned}& V=4.22 \times 0.3 \mathrm{~m} \\& V=1.266 \mathrm{~m} / \mathrm{s}\end{aligned}\)

The complete question should be:

A uniform disk with mass 43.6 kg and radius 0.300 m is pivoted at its centre about a horizontal, frictionless axle that is stationary. the disk is initially at rest, and then a constant force 29.0 n is applied tangent to the rim of the disk. What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.320 revolutions?

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Answer: ok

Explanation: ok

A 66.1-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.59 m, and ends with a speed of 8.51 m/s. The nonconservative work done on the boy is approximately -42.7 kilojoules.

To find the nonconservative work done on the boy, we need to consider the change in the boy's mechanical energy during the process. Mechanical energy is the sum of the boy's kinetic energy (KE) and gravitational potential energy (PE).

The initial mechanical energy of the boy is given by the sum of his kinetic energy and potential energy when he catches the wave:

E_initial = KE_initial + PE_initial

The final mechanical energy of the boy is given by the sum of his kinetic energy and potential energy after he drops through the height:

E_final = KE_final + PE_final

The nonconservative work done on the boy is equal to the change in mechanical energy:

Work_nonconservative = E_final - E_initial

Let's calculate each term:

KE_initial = (1/2) * m * v_initial^2

= (1/2) * 66.1 kg * (1.60 m/s)^2

PE_initial = m * g * h_initial

= 66.1 kg * 9.8 m/s^2 * 1.59 m

KE_final = (1/2) * m * v_final^2

= (1/2) * 66.1 kg * (8.51 m/s)^2

PE_final = m * g * h_final

= 66.1 kg * 9.8 m/s^2 * 0

Since the boy ends at ground level, the final potential energy is zero.

Substituting the values into the equation for nonconservative work:

Work_nonconservative = (KE_final + PE_final) - (KE_initial + PE_initial)

Simplifying:

Work_nonconservative = KE_final - KE_initial - PE_initial

Calculating the values:

KE_initial = (1/2) * 66.1 kg * (1.60 m/s)^2

PE_initial = 66.1 kg * 9.8 m/s^2 * 1.59 m

KE_final = (1/2) * 66.1 kg * (8.51 m/s)^2

Substituting the values:

Work_nonconservative = [(1/2) * 66.1 kg * (8.51 m/s)^2] - [(1/2) * 66.1 kg * (1.60 m/s)^2 - 66.1 kg * 9.8 m/s^2 * 1.59 m]

Calculating the result:

Work_nonconservative ≈ -42.7 kJ

Therefore, the nonconservative work done on the boy is approximately -42.7 kilojoules. The negative sign indicates that work is done on the boy, meaning that energy is transferred away from the boy during the process.

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I think the highest concentration of solar energy would occur when the Sun is directly overhead, as the angle of incidence is at its highest.

Concentration is the ability to focus on a certain task or activity, while blocking out distractions or other competing activities. It is an important skill to possess in order to be successful in both academic and professional settings. Concentration involves the ability to pay attention to a specific task or activity, while still being able to remember the details and information associated with it.

I think the highest concentration of solar energy would occur when the Sun is directly overhead, as the angle of incidence is at its highest. When the Sun is hovering over the Tropic of Cancer, the angle of incidence would be lower as the Sun is further from the zenith. Similarly, when the Sun is hovering over the Tropic of Capricorn, the angle of incidence would be even lower as the Sun is further from the zenith. Parallax did play a role in my observations as it affected the angle of the Sun's position relative to my location. When making my moon measurements, I noticed that the readings changed depending on the time of day and the phase of the moon. This experience helped me to understand the concepts of azimuth and altitude, as well as how they relate to the angle of incidence of sunlight.

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We will have that the diagram that represents the forces that would act in the object is:

Here "T" is the tension experienced by the spring and "mg" the force that the mass of the object would have.