A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 2.0 cm. What will be the compression if the same block collides with the spring at a speed of 2v?

Answer:

After 1 second and after 3 seconds.

Explanation:

the amount of medicine in time t is given by A = -40t2 + 160t.

To know the time when the amount of medicine will be 120 micrograms, we need to make the expression A equals 120, and then find the value of t:

120 = -40t2 + 160t

-40t2 + 160t - 120 = 0

dividing everything by -40, we have

t2 - 4t + 3 = 0

we have a second order equation, and to find its roots, we can use the baskhara formula:

D = b2 -4ac = 16 - 12 = 4.

The square root of D is +2 or -2

t_1 = (4 + 2)/2 = 3

t_2 = (4 - 2)/2 = 1

We have two values of t, both acceptable. So, after 1 second and after 3 seconds of the inhalation, there will be 120 micrograms of medicine in the bloodstream.

The time after inhalation when there will be about 120 micrograms of medicine in the bloodstream is approximately 2 minutes.

The equation given is A = -40t^2 + 160t. To find the time, 't', when the amount of medicine in the bloodstream is 120 micrograms, we should set 'A' equal to 120 and solve for 't'. So, 120 = -40t^2 + 160t. Simplify this equation to 40t^2 - 160t + 120 = 0. Here, you can observe that all terms are divisible by 10, thus simplifying it further gives 4t^2 - 16t + 12 = 0. Splitting the middle term, we get 4t^2 - 8t - 8t + 12 = 0. On factorizing, we get 4t(t - 2) - 4(2t - 3) = 0. Hence, (4t-4)(t-2) = 0, implies that t = 1 or t = 2. Since 't' cannot be negative, t = 1 minute is invalid in this context (as the amount of medicine isn't enough), we have t = 2 minutes as the valid solution.

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Answer:

True

Explanation:

The number density of an ideal gas at stp is called the loschmidt number, being named after the Austrian physicist Johann Josef Loschmidt who first estimated this quantity in 1865

n = [tex]\frac{P}{K_{B}T }[/tex]

where [tex]K_{B}[/tex] is the Boltzman constant

T is the temperature

P is the pressure

Loschmidt Number is the number of molecules of gas present in one cubic centimetre of it at STP conditions.Loschmidt constant is also used to define the amagat, which is a practical unit of number density for gases and other substances:

   1 amagat = n = 2.6867811×1025 m−3,

Answer:

True.

Explanation:

The Loschmidt number, n is defined as the number of particles (atoms or molecules) of an ideal gas in a given volume (the number density).

It is also the number of molecules in one cubic centimeter of an ideal gas at standard temperature and pressure which is equal to 2.687 × 10^19.

Answer:

[tex]w_c[/tex] = 28 mm

Explanation:

Displacement from the central maximum to minimum is as deterrmined as follows;

[tex]y_m[/tex] = [tex]\frac{m \lambda D}{a}[/tex]

If first minimum (m) = 1

single- Slit width is 170 times the wavelength of the light.

i.e

a = 170 λ

[tex]y_1[/tex]  = [tex]\frac{(1) \lambda (2,.4m)}{170 \lambda}[/tex]

[tex]y_1[/tex]  = 0.014 m

width of the central maximum can now be determined as:

[tex]w_c[/tex] = [tex]2y_1[/tex]

[tex]w_c[/tex] = 2(0.014 m)

[tex]w_c[/tex] = 0.028 m

[tex]w_c[/tex] = 28 mm

Hence, the width (in mm) of the central maximum on a screen 2.4 m behind the slit

= 28 mm

The width of the central maximum in a single-slit experiment is calculated by deriving the angle of the first minimum using light wavelength and slit width, and then applying it to the distance to the screen. The resulting value is an estimation.

In a single-slit experiment, the width of the central maximum can be calculated with the use of physics and understanding of light's behavior. The primary formula that guides this phenomenon is given as sin θ = λ/a, where θ is the angle of the first minimum, λ is the wavelength of the light, and a is the width of the slit. Given that the slit width is 170 times the wavelength of the light, the angle of first minimum will be very small and approximated as θ = λ/a.

Now, the angular width of the central maximum would be twice this angle in radians (θ), as the central maximum extends θ on either side of the central point. To find the actual width in mm on a screen 2.4 m away, you would multiply the total angular width (2*θ) by the distance of the screen (L) behind the slit, i.e. W = 2*θ*L. Please note that this is an approximation that widely works when the angle θ is small.

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Answer: 110km/h

100+10=110km/h

Explanation:

Motion is defined as a change of position. The frame of reference is usually assumed to be at rest. If one is sitting on a bus, the road appears to be moving backwards relative to the observer. If he/she now walks to the front of the bus, he has a speed relative to the earth which is now greater than that of the bus hence the answer.

Your speed relative to the road, when walking from the back to the front of a bus moving at 100km/h, is 110km/h.

Your speed relative to the road outside in this scenario will be the sum of the speed of the bus and your own speed walking within the bus. The bus is moving at 100 km/h, and you're moving toward the front of the bus at 10 km/h. Because you're moving in the same direction as the bus, you add your speeds together. Hence, your speed relative to the road outside would be 100 km/h (bus) + 10 km/h (you) = 110 km/h.

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Answer:

Explanation:

Given

Spring force constant [tex]k=5\ N/m[/tex]

Relaxed length [tex]l_0=2.59\ m[/tex]

mass is attached to the end of spring and allowed to come at rest with relaxed length [tex]l_2=3.86\ m[/tex]

Potential energy stored in the spring

[tex]U=\frac{1}{2}k(\Delta x)^2[/tex]

[tex]\Delta x=l_2-l_0[/tex]

[tex]U=\frac{1}{2}\times 5\times (3.86-2.59)^2[/tex]

[tex]U=4.03\ J[/tex]

Answer:

left

Explanation:

The magnetic field lines always point outwards from the north pole and merges towards the south pole. Thus, the arrow, is from right to left.

A bar magnet is an object made of materials having permanent magnetism.  They have two poles namely south poles and north poles. The like poles of magnets will repel and unlike pole attracts each other.

Conventionally, it is assumed that the magnet's field flows from its north pole inward to its south pole. Ferromagnetic materials can be used to create permanent magnets.

The magnetic field is strongest inside the magnetic substance, as seen by the magnetic field lines. Near the poles, there are the strongest external magnetic fields. A magnetic north pole will pull a magnet's south pole toward it while repelling another magnet's north pole. Hence, the arrow is pointing  to left direction.

To find more on bar magnets, refer here;

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Answer: 87KN

Explanation: F= m(v-u)/t

V= h/t

V= 60/0.0100

v= 6000m/s

M=145g/1000=0.145kg

F= 0.145*6000/0 .001

F= 87000N

F= 87KN

Answer:40.19 N

Explanation:

Given

Force needed to hold the ball at [tex]\theta =4^{\circ}\ is\ F_1=20\ N[/tex]

From FBD we can write

[tex]F\cos \theta =mg\sin \theta [/tex]

[tex]\tan \theta =\dfrac{F}{mg}[/tex]

For [tex]\theta =4^{\circ}[/tex]

[tex]F_1=20\ N[/tex]

[tex]\tan (4)=\dfrac{20}{mg}----1[/tex]

for [tex]\theta =8^{\circ}[/tex]

Force is [tex]F_2[/tex]

[tex]\tan (8)=\dfrac{F_2}{mg}---2[/tex]

Divide 1 and 2 we get

[tex]\dfrac{\tan (4)}{\tan (8)}=\dfrac{F_1}{F_2}[/tex]

[tex]F_2=20\times \dfrac{\tan 8}{\tan 4}[/tex]

[tex]F_2=20\times 2.009[/tex]

[tex]F_2=40.19\ N[/tex]

Answer:

Explanation:

UR MOM SUCKQ

Answer:

Incomplete questions

This is the complete question

The half-life of a radioactive isotope is the time it takes for a quantity for the isotope to be reduced to half its initial mass. Starting with 150 grams of a radioactive isotope, how much will be left after 6 half-lives

Explanation:

Let analyse the question generally first,

The the mass of the radioactive element be M.

We want to know it mass after n half life

Then,

After first half life, it mass is

M1=M×½

After second half life, it mass is

M2= M×(½)²

After third half life, it mass is

M3= M×(½)³

But now we can see a pattern developing, because for each new half-life we are dividing the quantity by 2 to a power that increases as the number of half-lives.

Then we can take the original quantity and quickly compute for

nth half-lives:

So after nth half life will be

Mn= M × (½)ⁿ

Generally,

Now, let apply it to our questions

Give that the mass of the radioactive isotope is 150grams

It mass after 6th half life

Then, n=6

So applying the formula

Mn= M × (½)ⁿ

M6= 150 ×(½)^6

M6= 150×1/64

M6=2.34grams

The mass of the radioactive isotope after 6th half life is 2.34grams

Answer:

If the same volume of air is inhaled and exhaled, the air we breathe out normally weighs more than the air we breathe in.

Since the output from the body normally exceeds the input, breathing leads to weight loss.

Explanation:

If equal volumes of gas is inhaled and exhaled, the exhaled gas is heavier.

The inhaled gas contains Oxygen and majorly Nitrogen.

The exhaled gas contains CO₂, H₂O and a very large fraction of the unused inhaled air that goes into the lungs.

So, basically, the body exchanges O₂ with CO₂ and H₂O (and some other unwanted gases in the body) in a composition that CO₂, the heavier gas of the ones mentioned here, is prominent.

So, because the mass leaving the body is more than the mass entering, breathing leads to a loss of weight. This is one of the reasons why we need food for sustenance. Breathing alone will wear one out.

Final answer:

The air we exhale likely has less mass than the air we inhale due to the replacement of dense oxygen with less dense carbon dioxide and water vapor.

Explanation:

When we inhale, we take in air that is rich in oxygen, and when we exhale, we expel air with a higher concentration of carbon dioxide and water vapor. Given that a liter of oxygen weighs more than the same volume of water vapor, and we replace some of the oxygen with the less dense carbon dioxide, the air we exhale likely has less mass than the air we inhale.

The breathing process does result in weight loss, but only in small amount attributable to the exhaled carbon dioxide which comes from the metabolic processes in the body. As for the volume and density of our bodies, taking a deep breath increases the volume, but because the density of air is much smaller than that of the body, the overall density decreases.

During gas exchange, oxygen flows from the bloodstream into the body's cells, while carbon dioxide flows out of the cells into the bloodstream to be expelled. This process is driven by the differing partial pressures of oxygen and carbon dioxide in the blood and the cells, facilitating diffusion across the cellular membrane.

Answer:

The correct answer is

1. the boy 2.

Explanation:

Since Tangential acceleration = v²/r  then the

v = ωr and v² =  ω²r² then the tangential acceleration = ω²r²/r = ω²r

This shows that  since ω is the same for both of them, the boy with greater r has the most acceleration.

The tangential acceleration [tex]a_t =[/tex] α·r

The isochoric process involving a gas at constant volume undergoes a temperature change from 270 K to 378 K, resulting in a change in internal energy of approximately 40.4 kJ.

1. Identify the process:

This problem describes an isochoric process, meaning the volume of the gas remains constant throughout the pressure change. Since the volume doesn't change, we can utilize the pressure-temperature relationship for ideal gases.

2. Apply the pressure-temperature relationship:

The pressure-temperature relationship for an ideal gas at constant volume is:

P₁V₁ = P₂V₂

where:

P₁ is the initial pressure

V₁ is the initial volume (constant in this case)

P₂ is the final pressure

V₂ is the final volume (constant in this case)

Since the volume remains constant, we can rewrite the equation as:

P₁ = P₂ * (T₁ / T₂)

where:

T₁ is the initial temperature (270 K)

T₂ is the final temperature (unknown)

3. Solve for the final temperature:

We are given that the final pressure (P₂) is 1.4 times the initial pressure (P₁). Substitute this information into the equation:

P₁ = 1.4 * P₁ * (T₁ / T₂)

Simplify the equation:

1 = 1.4 * (T₁ / T₂)

Solve for T₂:

T₂ = 1.4 * T₁

T₂ = 1.4 * 270 K

T₂ = 378 K

4. Calculate the change in internal energy:

For an ideal gas at constant volume, the change in internal energy (ΔU) is given by:

ΔU = n * Cv * ΔT

where:

n is the number of moles (30 mol)

Cv is the heat capacity at constant volume (unknown)

ΔT is the change in temperature (T₂ - T₁)

5. Relate Cv and Cp:

We are given the heat capacity at constant pressure (Cp) as 32.0 J/(mol*K). However, we need Cv for the internal energy calculation. We can utilize the relationship between Cv and Cp for ideal gases:

Cv = Cp - R

where:

R is the gas constant (8.314 J/(mol*K))

Substitute the given value of Cp:

Cv = 32.0 J/(mol*K) - 8.314 J/(mol*K)

Cv = 23.686 J/(mol*K)

6. Calculate the change in internal energy:

Now, substitute all the known values into the equation for ΔU:

ΔU = 30 mol * 23.686 J/(mol*K) * (378 K - 270 K)

ΔU = 40429.8 J

7. Convert to kilojoules and round the answer:

Convert the answer to kilojoules:

ΔU = 40429.8 J / 1000 J/kJ

ΔU ≈ 40.4 kJ

Therefore, the change in internal energy of the gas is approximately 40.4 kJ.

Answer:

[tex]x=4.2\ mm[/tex]

Explanation:

Given:

Change in length due to temperature can be  given as:

[tex]\Delta l=l.\alpha.(T_u-T_l)[/tex]

[tex]\Delta l=12\times 10^{-5}\times (50-(-30))[/tex]

[tex]\Delta l=0.0096\ m=9.6\ mm[/tex]

Now at temperature 15°C:

[tex]\Delta l'=12\times 10^{-5}\times (15-(-30))[/tex]

[tex]\Delta l'=0.0054\ m=5.4\ mm[/tex]

Hence the expansion crack between the slabs at this temperature must be:

[tex]x=\Delta l-\Delta l'[/tex]

[tex]x=9.6-5.4[/tex]

[tex]x=4.2\ mm[/tex]

Final answer:

To calculate the final temperature after placing ice cubes into water, principles of thermodynamics are used, involving calculations for warming the ice, melting it, and adjusting the water temperature. The steps highlight the application of heat transfer and phase change concepts within a closed system.

Explanation:

The student's question involves the calculation of the final temperature of the water after two 20.0 g ice cubes at -20.0 °C are placed into 285 g of water at 25.0 °C. The principles of thermodynamics, specifically the concept of heat transfer and the conservation of energy, are essential to solving this problem. However, with the given information, an exact numerical solution cannot be provided without knowing the specific heats of ice and water, as well as the heat of fusion of ice. Generally, the solution involves several steps:

Calculate the amount of energy required to warm the ice from -20.0 °C to 0 °C using the specific heat of ice.

Calculate the energy needed to melt the ice into water at 0 °C using the enthalpy of fusion.

Calculate the energy released by the water as it cools down to the new equilibrium temperature.

Set the energy gained by the ice equal to the energy lost by the water to find the final temperature of the mixture.

Without the numerical values, the step-by-step method demonstrates how principles of heat transfer and phase change are applied to predict changes in temperature and state within a closed system.

Answer:

Explanation:

Their explanation is: We first need to determine the velocity of the water that comes out of the hole, using Bernoulli's equation.

P1+ρgy1 + ½ρv1² = P2+ρgy2+½ρv2²

The atmospheric pressure exerted on the surface of the water at the top of the tank and at the hole are essentially the same.

Then, P1-P2 = 0

. Additionally, since the opening at the top of the tank is so large compared to the hole on the side, the velocity of water at the top of the tank will be essentially zero. We can also set y1 as zero, simplifying the equation to:

0 = ρgy2 + ½ρv2²

Divide through by density

-gy2 = ½ v2²

y2= 5m and g =-9.81

-•-9.81×5 = ½v2²

9.81×5×2 = v2²

98.1 = v2²

v2 = √98.1

v2 = 9.9m/s

Using equation of motion to know the time of fall

We assume that the initial vertical velocity of the water is zero and the displacement of the water is -20 m

y-yo = ut + ½gt²

0-20 = 0•t -½ ×9.81 t²

-20 = -4.905t²

t² = -20÷-4.905

t² = 4.07747

t =√4.07747

t = 2.02secs

Using range formula

Then, R=Voxt

R= 9.9 × 2.02

R =19.99m

R ≈20m

Answer:

The distance from the base of the tank to the ground is 20 m

Explanation:

From Bernoulli equation where P is pressure, V is velocity, ρ is density, g is acceleration due to gravity and y is the height

P₁ + 1/2ρV₁² + ρgy₁ = P₂ + 1/2ρV₂² + ρgy₂

The pressure on the surface of the water on the top of the tank and that in the hole is the same (P₁ = P₂). Since the opening at the top of the tank is large compared to the hole at the side of the tank, the velocity at the top of the tank is 0 and y₁ is 0

Therefore: 0 =  1/2ρV₂² + ρgy₂

1/2ρV₂² = -ρgy₂      g = -10 m/s²

-10(5) = -1/2V₂²

V₂ = 10 m/s

The time it would take the water to fall on the ground is given as:

d = ut + 1/2gt²

u is the initial velocity = 0 , g = -10 m/s² and the displacement (d) = - 20 m

Therefore: -20 = 1/2(10)t²

t = 2 sec

The horizontal displacement (d) can be gotten from

d = V₂t

d = 10(2) = 20 m

The distance from the base of the tank to the ground is 20 m

Answer:

Therefore the average thermal power is = 19.61 W

Explanation:

Given that the mass of the rock = 20.0 kg

initial velocity (u) = 8.00m/s

Final velocity = 0

Kinetic fiction [tex](\mu_k)[/tex] = 0.20.

Friction force = Kinetic fiction× weight

                      =[tex]\mu_k mg[/tex]

Force = mass × acceleration

[tex]\Rightarrow acceleration =\frac{Force}{mass}[/tex]

                        [tex]=-\frac{\mu_kmg}{m}[/tex]

                        [tex]=-\mu_kg[/tex]

To find the time , we use the the following formula,

v=u+at

Here a [tex]=-\mu_kg[/tex]

[tex]\Rightarrow 0=8+(-\mu_kg)t[/tex]

[tex]\Rightarrow 0.20 \times 9.8\times t=8[/tex]

⇒t = 4.08 s

Now,

Thermal energy= work done by friction = change of kinetic energy

The change of kinetic energy

= [tex]\frac{1}{2} mu^2-\frac{1}{2} mv^2[/tex]

[tex]=\frac{1}{2}\times 20\times 8 -\frac{1}{2}\times 20\times 0[/tex]

=80 J

Thermal energy=80 J

Thermal power = Thermal energy per unit time

                          [tex]=\frac{80}{4.08}[/tex] w

                         =19.61 W

Therefore 19.61 W average thermal power is produced as the rock stops.

Answer:

0.467 Hz

Explanation:

Wave properties are related thus

v = fλ

v = velocity of the wave = ?

f = 0.30 Hz

λ = wavelength = 30 m

v = 0.3×30 = 9.0 m/s

But if the boat is now moving at 5.0 m/s in the direction of the oncoming wave,

The speed of the wave relative to the boat = 9 - (-5) = 14.0 m/s

f = (v/λ) = (14/30) = 0.467 Hz

Hope this helps!

Answer:

The answer to the question is;

The boat’s vertical oscillation frequency if you drive the boat at 5.0 m/s in the direction of the oncoming waves moving at 9 m/s in the opposite direction is  0.467 Hz.

Explanation:

We note the frequency of the water wave = 0.3 Hz

Distance between crests or wavelength of water wave  = 30 m

Speed v of a wave is given by Frequency f × Wavelength λ

Therefore the speed of the wave = f·λ = 0.3 × 30 = 9 m/s

(b)  Distance between crests = 30 m = wavelength λ

     Boat speed                       = 5.0 m/s v

Speed of the wave with respect to the boat = 5 + 9 = 14 m/s

From the wave speed relationship, we have

v = fλ    f = [tex]\frac{v}{\lambda}[/tex] =[tex]\frac{14}{30} = \frac{7}{15}[/tex] = 0.467 Hz which is high.

Answer:

Explanation:

mass of ice skater, m = 55 kg

velocity, v = 4.07 m/s

radius, r = 0.808 m

The force is centripetal force, the formula for this force is

[tex]F_{c}=\frac{mv^{2}}{r}[/tex]

[tex]F_{c}=\frac{55\times 4.07\times 4.07}{0.808}[/tex]

Fc = 1127.56 N

Weight of the person, W = mg

W = 55 x 9.8 = 539 N

The ratio of force to he weight

[tex]\frac{F_{c}}{W}=\frac{1127.56}{539}[/tex]= 2.09

The magnitude of the attractive force on each one will be F.

Explanation:

As two small balls are experiencing gravitational force between them ,then they will obey universal law of gravity. As per the universal law of gravity, the gravitational force acting between two objects will be directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between the two objects.

So if the mass of the two small balls A and B are considered as M and m, respectively, with the distance of separation be considered as r. Then the gravitational force of attraction acting between A and B will be

[tex]F = \frac{GMn}{r^{2} }[/tex] , This is the original or initial gravitational force between A and B.

Now, if the masses of A and B are doubled, then the new masses will be M' = 2 M and m' = 2m, respectively. Similarly, if the separation of the balls is also doubled then r' = 2r. So the new gravitational force exerting between A and B is

[tex]F' = \frac{GM'm'}{r'^{2} } = \frac{G*2M*2m}{(2r)^{2} } =\frac{4GMm}{4r^{2} } = \frac{GMm}{r^{2} }=F[/tex]

So after doubling the masses as well as the distance of separation, there will be no change in the gravitational force. So the magnitude of the attractive force on each one will be F.

Answer: MOTION

Explanation:

motion is defined as the displacement of an object with respect to time relative to a stationary object (reference point). A good example of an object that can serve as a reference point includes: a tree or a building. The movement of a body at constant speed towards a particular direction at regular intervals of time can be determined and it's called uniform motion.

There are different types of motion, these includes: simple harmonic motion,

linear motion,

circular motion,

Brownian motion,

Rotatory motion

Answer:rest

Explanation:

Answer:

(A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

Explanation:

Given that,

Mass of disk = 10 kg

Radius = 9.0 cm

Rotational speed = 320 m/s

(A). We need to calculate the rotational momentum of the flywheel.

Using formula of momentum

[tex]L=I\omega[/tex]

[tex]L=\dfrac{1}{2}mr^2\omega[/tex]

Put the value into the formula

[tex]L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320[/tex]

[tex]L=12.96\ kg m^2/s[/tex]

(B). Rotation momentum of sphere is same rotational momentum of the  flywheel

We need to calculate the magnitude of the rotational speed of sphere

Using formula of rotational momentum

[tex]L_{sphere}=L_{flywheel}[/tex]

[tex]I\omega_{sphere}=I\omega_{flywheel}[/tex]

[tex]\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}[/tex]

[tex]\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}[/tex]

Put the value into the formula

[tex]\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}[/tex]

[tex]\omega_{sphere}=400\ rad/s[/tex]

Hence, (A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

To find the rotational momentum of a flywheel, one must calculate the moment of inertia and then use it to ascertain the angular momentum by multiplying the moment of inertia with the angular speed. L = Iw. To compare this with another rotating object, for example a sphere, one would then utilize their common momentum and calculate the necessary angular speed the sphere would need in order to match the flywheel's momentum.

Part A: To determine the rotational momentum of the flywheel, we first need to find its moment of inertia. For a disk, the moment of inertia (I) is given by I = 1/2 MR² where M is the mass and R is the radius. Substituting values, I becomes = 1/2 * 10 Kg * (0.09m)² = 0.0405 kg • m². The rotational momentum can then be found by multiplying moment of inertia by the rotational speed w. Therefore the rotational momentum (L) of the flywheel is L = Iw = 0.0405 kg • m² * 320 rad/s = 12.96 Kg • m²/s.

Part B: Let's now find the speed a spherical object needs to achieve the same momentum. The moment of inertia of a solid sphere is given by I=2/5 MR². The angular momentum L (which should be the same as the flywheel's) equals Iw, so we solve for w giving w = L/I = 12.96 kg•m²/s / (2/5*10 kg (0.09 m)²) = 180 rad/s. Thus, a 10-kg solid sphere of 9.0 cm radius would have to rotate at a speed of 180 rad/s to have the same rotational momentum as the flywheel.

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Answer: Internetwork

Explanation:

Internetworking is the process of connecting different networks together by the use of intermediary devices such as routers or gateway devices. Internetworking helps data communication among networks owned and operated by different bodies.

Answer:

Internetwork

Explanation:

Internetwork involves having a central network system which is shared into other systems and location through the help of intermediary devices such as routers. This usually helps to increase the orderliness and precision by which data is shared between the various connections.

Explanation:

The work done on the body = force applied x displacement

in this case , the acceleration of body a = [tex]\frac{Force}{Mass}[/tex] = [tex]\frac{4.9}{14}[/tex] = 0.35 ms⁻²

The displacement in first second S₁ = u + [tex]\frac{1}{2}[/tex] a x t²

here u = 0 , because body was at rest

Thus S₁ = [tex]\frac{1}{2}[/tex] x 0.35 x ( 1 )² = = 0.175 m

The work done = 4.9 x 0.175 =  0.86 J

The displacement in 2 seconds = [tex]\frac{1}{2}[/tex] x 0.35 x ( 2 )² = 0.7 m

Work done in 2 seconds = 4.9 x 0.7 = 3.43 J

Work done in second second = 3.43 - 0.86 = 2.57 J

The displacement in three seconds = [tex]\frac{1}{2}[/tex] x 0.35 x ( 3 )² = 1.575 m

Work done in three seconds = 4.9 x 1.575 = 7.7 J

Work done in third second = 7.7 - 3.43 = 4.3 J

The power at the end of third second is 4.3 watt

Because 4.3 J of work is done in the last second .

Answer:

Work done = 100 J

Explanation:

Given:

Distance moved by the box (d) = 10 m

Parallel force acting on it [tex](F_1)[/tex] = 10 N

Perpendicular force acting on it [tex](F_2)[/tex] = 5 N

Work done is the product of force and displacement along the line of action of force.

So, the perpendicular force doesn't do any work and it's only the parallel force that does work as the displacement caused is along the parallel force direction.

So, work done is given as:

[tex]Work=F_1\times d\\\\Work=10\ N \times 10\ m \\\\W=100\ J[/tex]

Therefore, work done is 100 J.

Answer : 0.814 newton

Explanation:

force (magnetic) acting on the wire is given by

F= ? , I=2.2amp , B = 0.37 T

F = B i l sin (theta) = 0.37 x 2.2 x 2x 0.5 = 0.814N

A straight wire segment 2 m long makes an angle of 30degrees with a uniform magnetic field of 0.37 T. The magnitude of the force is

F= 0.814N

Generally, the equation for the magnitude of the force is  mathematically given as

[tex]F = B i l sin (\theta)[/tex]

Therefore

F= 0.37 x 2.2 x 2x 0.5

F= 0.814N

In conclusion, the magnitude of the force

F= 0.814N

Read more about the magnitude of the force

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Answer:

Moisture content

Explanation:

Air mass is a volume of air spread through a vast area may it cover hundred to thousand Kilometers and is identified on the basis of nearly equal temperature and water vapour content of the air thorough out the area.

Answer:

The four wavelengths of the problem are not given. Here they are:

a) [tex]Li^+,\lambda=671 nm[/tex]

b) [tex]Cs^+, \lambda=456 nm[/tex]

c) [tex]Ca^{2+}, \lambda=649 nm[/tex]

d) [tex]Na^+, \lambda=589 nm[/tex]

The relationship between wavelength and frequency of light wave is

[tex]f=\frac{c}{\lambda}[/tex]

where

f is the frequency

[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light

[tex]\lambda[/tex] is the wavelength

For case a), [tex]\lambda=671 nm = 6.71\cdot 10^{-7}m[/tex] (corresponds to red color), so its frequency is

[tex]f=\frac{3\cdot 10^8}{6.71\cdot 10^{-7}}=4.47\cdot 10^{14}Hz[/tex]

For case b), [tex]\lambda=456 nm = 4.56\cdot 10^{-7}m[/tex] (corresponds to blue color), so its frequency is

[tex]f=\frac{3\cdot 10^8}{4.56\cdot 10^{-7}}=6.58\cdot 10^{14}Hz[/tex]

For case c), [tex]\lambda=649 nm = 6.49\cdot 10^{-7}m[/tex] (corresponds to red color), so its frequency is

[tex]f=\frac{3\cdot 10^8}{6.49\cdot 10^{-7}}=4.62\cdot 10^{14}Hz[/tex]

For case d), [tex]\lambda=589 nm = 5.89\cdot 10^{-7}m[/tex] (corresponds to yellow color), so its frequency is

[tex]f=\frac{3\cdot 10^8}{5.89\cdot 10^{-7}}=5.09\cdot 10^{14}Hz[/tex]

In fireworks displays, the color of light is determined by its wavelength and frequency. The color red has the longest wavelength and the lowest frequency, while the color violet has the shortest wavelength and the highest frequency. Therefore, the order of the given colors from shortest wavelength to longest wavelength is blue, yellow, and red. Similarly, the order of the given colors from lowest frequency to highest frequency is also blue, yellow, and red.

The colors of light in a fireworks display indicate the presence of different elements. The wavelength and frequency of the light determines its color. Within the visible range, our eyes perceive radiation of different wavelengths as light of different colors. The color red corresponds to the longest wavelength and the lowest frequency, while the color violet corresponds to the shortest wavelength and the highest frequency. Therefore, the order of the given colors from shortest wavelength to longest wavelength would be blue, yellow, and red. Similarly, the order of the given colors from lowest frequency to highest frequency would also be blue, yellow, and red.

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Answer:

[tex]\Delta V=\Delta V_1+\Delta V_2+\Delta V_3[/tex]

Explanation:

We are given that three resistors R1, R2 and R3 are connected in series.

Let

Potential difference across [tex]R_1=\Delta V_1[/tex]

Potential difference across [tex]R_2=\Delta V_2[/tex]

Potential difference across [tex]R_3=\Delta V_3[/tex]

We know that in series  combination

Potential difference ,[tex]V=V_1+V_2+V_3[/tex]

Using the formula

[tex]\Delta V=\Delta V_1+\Delta V_2+\Delta V_3[/tex]

Hence, this is required expression for potential difference.

Answer: we can conclude that the wavelength is decreasing. This means that the star is moving towards the observer on earth.

Explanation:

Since light has a constant speed of 3 x 10^8m/s, and this speed is a product of its wavelength and its frequency c = f¥

Where f is the frequency and ¥ is the wavelnght.

For a decreasing wavelength, it is seen that the frequency is increasing.

According to the doppler's effect, a moving body that is a source of a wave of frequency f, moving relatively towards an observer, the frequency will increase as they move closer to a frequency f' which is greater than f. This is known as the doppler shift of light wave.