state any 3 properties of an ideal gas as assumed by the kinetic theory.​

Answers

Answer 1

Answer:

The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the walls of the container, but otherwise do not interact; and (3) the transfer of kinetic energy between molecules is heat.


Related Questions

any four difference between velocity and acceleration​

Answers

Answer:

https://physicsabout.com/acceleration-and-velcoity/

(a) What is the efficiency of an out-of-condition professor who does 1.90 ✕ 105 J of useful work while metabolizing 500 kcal of food energy? % (b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 25%? kcal

Answers

Answer:

a) The energy efficiency of the out-of-condition professor is 9.082 %.

b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Explanation:

a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:

[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)

Where:

[tex]W[/tex] - Useful work, in joules.

[tex]E[/tex] - Food energy, in joules.

If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:

[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]

[tex]\eta = 9.082\,\%[/tex]

The energy efficiency of the out-of-condition professor is 9.082 %.

b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:

[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]

[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]

[tex]E = 7.60\times 10^{5}\,J[/tex]

[tex]E = 181.644\,kcal[/tex]

The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

A falcon is hovering above the ground, then suddenly pulls in its wings and begins to fall toward the ground. Air resistance is not negligible.
Identify the forces on the falcon.
a. Kinetic friction
b. Weight w
c. Static friction
d. Drag D
e. Normal force n
f. Thrust
g. Tension T

Answers

Answer:

Explanation:

When a falcon is hovering, the force of up thrust is balanced by the weight.

When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.

A magnetic force acting on an electric charge in a uniform magnetic field what happend

Answers

Answer:

hgff

Explanation:

Answer:

The charge moves to equilibrium.

E.e = B.e.V

E is electric field force.

e is the charge.

B is magnetic field force.

V is acceleration voltage.

A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency is 890 Hz, the wavelength is .10m, and the amplitude is 6.5 mm. The tension in the line, in SI units, is closest to

Answers

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

[tex]\mu = \frac{m}{L}[/tex]

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

[tex]\mu = \frac{0.09\ kg}{1\ m}[/tex]

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

T = 712.9 N

A supertrain with a proper length of 100 m travels at a speed of 0.950c as it passes through a tunnel having a proper length of 50.0 m. As seen by a trackside observer, is the train ever completely within the tunnel? If so, by how much do the train’s ends clear the ends of the tunnel?

Answers

Answer:

19m

Explanation:

we have proper length L = 100m

the speed of the train v = 0.95

the speed of light is given as = 3x10⁸

length of the tunnel is given as = 50 meters

we can solve for the lenght contraction as

LX√1-v²/c²

= 100 * √1-(0.95*3x10⁸)²/(3x10⁸ )

= 31.22 metres

the train would be well seen at

50 - 31.22

= 18.78

= this is approximately 19 metres

we conclude tht the trains ends clears the ends of the tunnel by 19 meters.

thank you!

The Sun is a type G2 star. Type G stars (from G0 to G9) have a range of temperatures from 5200 to 5900. What is the range of log(T) for G stars? Show your work

Answers

Answer:

log T = 3.72 to 3.77

Explanation:

Temperature range is

T = 5200 to 5900

Take the log

So,

log T = log 5200 to log 5900

log T = 3.72 to 3.77

you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week. Your employer withhold 10% of your gross pay federal taxes, 7.65% for FICA taxes, and 5% for state taxes

Answers

I guess that we want to find how much money you get each week.

We know that the job pays $8.60 per hour.

We know that you work 20 hours per week.

Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:

20*$8.60 = $172.

Now we know that your employer witholds:

10% + 7.65% + 5% = 22.65%

Then your employer withholds 22.65% of your gross pay.

if the 100% of your gross pay is $172

Then the 22.65% will be:

(22.65%/100%)*$172 = 0.2265*$172 = $38.96

This means that your employer withholds $38.96 of your weekly gross pay.

Then each week you get:

$172 - $38.96 = $133.04

If you want to learn more, you can read:

https://brainly.com/question/6692050

A car of mass 2100 kg collides with a motorcycle of mass 290 kg. After the collision, the car and motorcycle stick and slide together. The car's velocity just before the collision was<30,-10>m/s , and that of the motorcycle was <10,10>m/s. Determine the velocity of the stuck-together car and motorcycle just after the collision.

Answers

Answer:

V = 29.49 m/s

Explanation:

Given that,

The mass of a car,[tex]m_c=2100\ kg[/tex]

The mass of a motorcycle, [tex]m_m=290\ kg[/tex]

The initial velocity of the car,[tex]v_c=30i-10j[/tex]

[tex]|v_c|=\sqrt{30^2+(-10)^2} =31.62\ m/s[/tex]

The initial velocity of the motorcycle,[tex]v_m=10i+10j[/tex]

[tex]|v_m|=\sqrt{10^2+10^2} =14.14\ m/s[/tex]

As they stick together. Let V is the speed. So, using the conservation of momentum,

[tex]m_cv_c+m_mv_m=(m_c+m_m)V\\\\V=\dfrac{m_cv_c+m_mv_m}{(m_c+m_m)}\\\\V=\dfrac{2100\times 31.62+290\times 14.14}{(2100+290)}\\\\V=29.49\ m/s[/tex]

So, the velocity of the stuck together car and the motorcycle after the collision is 29.49 m/s.

What is the maximum wavelength, in nm, of light that can eject an electron from a metal with Φ =4.50 x 10–19 J?

Answers

[tex]4.4×10^{-7}\:\text{m}[/tex]

Explanation:

The minimum energy needed to kick out an electron from a metal's surface is when the energy of the incident radiation is equal to the metal's work function [tex]\phi[/tex]:

[tex]E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi = 0[/tex]

or

[tex]\dfrac{hc}{\lambda} = \phi[/tex]

Solving for the wavelength [tex]\lambda[/tex],

[tex]\lambda = \dfrac{hc}{\phi}[/tex]

[tex]\:\:\:\:\:=\dfrac{(6.62×10^{-34}\:\text{J-s})(3.0×10^8\:\text{m/s})}{4.5×10^{-19}\:\text{J}}[/tex]

[tex]\:\:\:\:\:= 4.4×10^{-7}\:\text{m}[/tex]

Note that as the radiation's wavelength increases, its energy decreases. So a radiation whose wavelength is longer than this maximum will lose its ability to kick out an electron from this metal.

The maximum wavelength, in nm, of light that can eject an electron from the metal, given the data is 441.73 nm.

To find the wavelength, the given values are,

Energy (E) = 4.50×10¯¹⁹ J

What is wavelength?

The distance between two consecutive crests and troughs is called the wavelength of a wave.

Here, for the wavelength,

Energy (E) = 4.50×10¯¹⁹ J

Planck's constant (h) = 6.626×10¯³⁴ Js

Speed of light (v) = 3×10⁸ m/s

The wavelength of the light can be obtained as illustrated below:

E = hv / λ

Cross multiply λ,

E × λ = hv

Divide both sides by E,

λ = hv / E

Substituting all the values,

λ = (6.626×10¯³⁴ × 3×10⁸) / 4.50×10¯¹⁹

λ = 0.000000441733 m

λ = 441.73nm

λ - The maximum wavelength of light.

Thus, the wavelength of the light that can eject an electron from the metal is 441.73 nm

Learn more about wavelength,

https://brainly.com/question/13047641

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How does an airpump work? ​

Answers

The inlet and the outlet are used to direct the flow of air, while the piston is used to generate the flow of air. When the piston is pulled up, air gets sucked into the pump through the inlet. ... When the piston is forced down, the air becomes compressed and closes the inlet. Then the air flows out from the outlet.

The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. A

Answers

Complete Question

The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2

Answer:

[tex]h=1614m[/tex]

Explanation:

From the question we are told that:

Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]

Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]

Density [tex]\rho=1.20kg/m^2[/tex]

Generally the equation for Height climbed is mathematically given by

[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]

[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]

[tex]h=1614m[/tex]

A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constant acceleration of 3 r a d s 2 . What was the initial angular speed of the CD

Answers

Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, [tex]\theta=12\ rad[/tex]

Final angular speed, [tex]\omega_f=17\ rad/s[/tex]

The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]

Put all the values,

[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]

So, the initial angular speed of the CD is equal to 14.73 rad/s.

3. Define 1 standard kilogram?

Answers

Answer:

standard kilogram is the SI unit of mass

Answer:

The total mass of platinum-irridum cylinder whose diameter is equal to its height and stored at 0°C in the bureau of weight and measure in France is called 1 standard kilogram

The displacement x of a particle varies with time t as x = 4t 2 -15t + 25. Find the position,
velocity and acceleration of the particle at t = 0. When will the velocity of the particle becomes
zero? Can we call the motion of the particle as one with uniform acceleration?

Answers

Answer:

x = 4 t^2 - 15 t + 25        displacement of particle

dx / dt = 8 t - 15        velocity of particle

d^2x / dt^2 = 8       acceleration of particle

If 8 t -15 = o     then t = 8 / 15

Since acceleration is a constant 8 then motion has uniform acceleratkon

In what kind of reaction is water (H20) broken down into hydrogen gas (H2) and oxygen gas (O2)?

A. Combination
B. Decomposition
C. Displacement
D. Combustion ​

Answers

Answer:

Answer is B (Decomposition)

Sorry I really see ur questions but I don't know the answer but next time I will try to answer sorry:(

A soap bubble was slowly enlarged from a radius of 4cm to 6cm. The amount of work necessary for enlargement was 1.5 x 10^-4 joules. Calculate the surface tension of the soap bubble.​

Answers

Answer:

[tex]T=3*10^-3 N/m[/tex]

Explanation:

From the question we are told that:

Radius :

[tex]R_1=4=>0.04\\\\R_2=6=>0.06[/tex]

Work [tex]W=1.5 * 10^{-4}[/tex]

Generally the equation for Work done  is mathematically given by

[tex]W=TdA[/tex]

Where

[tex]dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)[/tex]

[tex]dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)[/tex]

[tex]dA=0.050m^2[/tex]

Therefore

[tex]W=TdA[/tex]

[tex]T=\frac{1.5 * 10^{-4}}{0.05m^2}[/tex]

[tex]T=3*10^-3 N/m[/tex]

A series LR circuit contains an emf source of having no internal resistance, a resistor, a inductor having no appreciable resistance, and a switch. If the emf across the inductor is of its maximum value after the switch is closed, what is the resistance of the resistor

Answers

Answer:

b. 1.9 Ω

Explanation:

Here is the complete question

A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? a. 1.5 ? b. 1.9 ? c. 5.0 ? d. 14 ?

Solution

The voltage across the inductor V is

[tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex] where V₀ = emf of source = 14 V, R = resistance, L = inductance = 34 H and t = time

Given that V = 80% of its maximum value after 4.0 s, this implies that V = 80 % of V₀ = 0.8V₀ and t = 4.0 s

Since [tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex] and V = 0.8V₀.

Since we need to find R, we make R subject of the formula, we have

[tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex]

[tex]V/V_{0}= e^{-\frac{Rt}{L} }[/tex]

taking natural logarithm of both sides, we have

㏑(V/V₀) = -Rt/L

R = -L㏑(V/V₀)/t

Substituting the values of the variables into the equation, we have

R = -34㏑(0.8V₀/V₀)/4.0 s

R = -34㏑(0.8)/4.0 s

R = -34 × -0.2231/4.0 s

R = 7.587/4

R = 1.896 Ω

R ≅ 1.9 Ω

So, B is the answer


Find the ratio of the Coulomb electric force Fe to the gravitational force Fo between two
electrons in vacuum.

Answers

Answer:

thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.

A mass of 5 kg stretches a spring 20 cm. The mass is acted on by an external force of 10 sint4 N (newtons) and moves in a medium that imparts a viscous force of 4 N when the speed of the mass is 2 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 7 cm/s.

Required:
Formulate the initial value problem describing the motion of the mass. Assume that g = 9.8 m/s^2.

Answers

Answer:Answer:

Initial value problem is:

u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0

u'(0) = 0.03m/s

Explanation:

The directions of Fd(t*) and U'(t*) are not specified in the question, so we'll take Fd(t*) to be negative and U'(t*) to be positive. This is due to the fact that the damping factor acts in the direction opposite the direction of the motion of the mass.

M = 5kg; L= 10cm or 0.1m;

F(t) = 10 sin(t/2) N ; Fd(t*) = - 2N

U'(t*) = 4cm/s or 0.04m/s

u(0) = 0

u'(0) = 3cm/s or 0.03m/s

Now, we know that W = KL.

Where K is the spring constant.

And L is the length of extension.

So, k = W/L

W= mg = 5 x 9.81 = 49.05N

So,k = 49.05/0.1 = 490.5kg/s^(2)

Now from spring damping, we know that; Fd(t*) = - γu'(t*)

Where,γ = damping coefficient

So, γ = - Fd(t*)/u'(t*)

So, γ = 2/0.04 = 50 Ns/m

Therefore, the initial value problem which describes the motion of the mass is;

5u'' + 50u' + 490u = (10 sin(t/2) N

Divide each term by 5 to give;

u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0

u'(0) = 0.03m/s

Explanation:

Two carts are involved in an inelastic collision. Cart A with mass 0.900 kg hits cart B with mass 0.550 kg (initially at rest). The two carts stick together after the collision and continue to move along together. Cart A has an initial velocity of 0.29 m/s.

a. What is the final velocity of the two-cart system?
b. What is the initial kinetic energy of cart A?
c. What is the initial kinetic energy of cart B?
d. What is the final kinetic energy of the system?
e. Is kinetic energy conserved for inelastic collisions?
f. Is momentum conserved for inelastic collisions?

Answers

Answer:

a)  v = 0.18 m / s, b)  K₀ₐ = 0.0378 J, c) K_{ob}= 0, d)  K = 0.02349 J,

Explanation:

a) For this exercise we must define a system formed by the two cars, so that the forces during the collision are internal and the moment is conserved

initial instant. Before the hole

         p₀ = ma v₀ₐ

final intnate. After the crash

         p_f = (mₐ + m_b) v

the moment is preserved

         p₀ = p_f

         mₐ v₀ₐ = (mₐ + m_b) v

         v = [tex]\frac{m_a}{m_a+m_b} \ v_{oa}[/tex]

       

let's calculate

         v = [tex]\frac{0.900}{0.900+0.550} \ 0.29[/tex]

         v = 0.18 m / s

in the same direction of the movement of carriage A

b) the initial kinetic energy car A

         K₀ₐ = ½ m  v₀ₐ²

         K₀ₐ = ½ 0.900 0.29²

         K₀ₐ = 0.0378 J

c) kinetic energy of carriage B

          k_{ob} = 0

because the car is stopped

d) the kinetic energy of the system

          K = ½ (mₐ + m_b) v²

           K = ½ (0.900 + 0.550) 0.18²

           K = 0.02349 J

E) we see that part of the kinetic energy is lost, therefore the scientific reeling is not conserved in inelastic collisions

F) and momentum is conserved since it is equal to the variation of the moment and this is conserved in all collisions

Two people, who have the same mass, throw two different objects at the same velocity. If the first object is heavier than the second, compare the velocities gained by the two people as a result of recoil.

a. The first person will gain more velocity as a result of recoll.
b. The second person will gain more velocity as a result of recoll.
c. Both people will gain the same velocity as a result of recoll.
d. The velocity of both people will be zero as a result of recoil

Answers

Answer:

The first person will gain more velocity as a result of recoil.

Explanation:

Let us recall that from Newton's third law of motion, action and reaction are equation and opposite. A consequence of this law is the proposition that ''momentum can neither be created nor destroyed.''

Hence, when two people who have the same mass, throw two different objects at the same velocity but the first object is heavier than the second, the first object possesses greater momentum than the second object hence the first person will gain more velocity as a result of recoil.

A frictionless spring with a 9-kg mass can be held stretched 1.8 meters beyond its natural length by a force of 80 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1.5 m/sec, find the position of the mass after tt seconds. meters

Answers

Answer:

the required solution is; x(t) = 0.675sin( 2.222t )

Explanation:

Given the data in the question;

Using both Newton's and Hooke's law;

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0, [tex]x[/tex](0) = 0, [tex]x^f[/tex](0) = 1.5

given that mass m = 9 kg

[tex]x[/tex] = 1.8 m

k is F / x

hence

k = F / x

given that, F = 80 N

we substitute

k = 80 / 1.8

k = 44.44

so

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0,

we input

9[tex]x^{ff[/tex] + 44.44[tex]x[/tex] = 0,

[tex]x^{ff[/tex] + 4.9377[tex]x[/tex] = 0

so auxiliary equation is,

r² + 4.9377 = 0

r² = -4.9377

r = √-4.9377

r = ±2.222i

hence, the solution will  be;

x(t) = A×cos( 2.222t ) + B×sin( 2.222t )

⇒ [tex]x^t[/tex](t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )

using initial conditions

x(0) = 0

⇒ 0 = A

[tex]x^t[/tex](t) = 1.5

1.5 = 2.222B

so

B = 1.5 / 2.222 = 0.675

Hence, the required solution is; x(t) = 0.675sin( 2.222t )

The best and most common way to measure the intensity of a cardiovascular exercise is to determine
O The person's heart rate
O The fatigue level of the person
O Amount of perspiration the person produces
The person's breathing rate

Answers

Answer:

the person's heart rate

The person’s heart rate

A particle, mass 0.25 kg is at a position () m, has a velocity () m/s, and is subject to a force () N. What is the magnitude of the torque on the particle about the origin

Answers

Question

A particle, mass 0.25 kg is at a position (-7i + 7j + 5k) m, has a velocity (6i - j + 4k) m/s, and is subject to a force (-5i + 0j - k) N. What is the magnitude of the torque on the particle about the origin?

Answer:

47.94Nm

Explanation:

The torque (τ) on a particle subject to a force (represented as force vector F) at a position (represented as position vector r) about the origin is given by the cross product of the position vector r for the point of application of a force and the force F. i.e

τ = r x F

Given:

r = (-7i + 7j + 5k) m

F = (-5i + 0j - k) N

                    |   i             j              k    |

r x F  =         |   -7            7              5  |

                    |  -5           0              -1   |

r x F  =       i(-7 - 0) - j(7+25) + k(0+35)

r x F  =       i(-7) - j(32) + k(35)

r x F  =       -7i - 32j + 35k

Therefore the torque τ = -7i - 32j + 35k

The magnitude of the torque is therefore;

|τ| = [tex]\sqrt{(-7)^2 + (-32)^2 + (35)^2}[/tex]

|τ| = [tex]\sqrt{49 + 1024 + 1225}[/tex]

|τ| = [tex]\sqrt{2298}[/tex]

|τ| = 47.94Nm

The magnitude of the torque on the particle about the origin is 47.94Nm

A 77 turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.18 T field, starting with the normal of the plane of the coil perpendicular to the field. Assume that the positive max emf is reached first.

a. What is the peak emf?
b. At what time is the peak emf first reached?
c. At what time is the emf first at its most negative?
d. What is the period of the AC voltage output?

Answers

Answer:

a) fem = 5.709 V,  b)  t = 0.196 s,  c)  t = 0.589 s, d)   T = 0.785 s

Explanation:

This is an exercise in Faraday's law

          fem= - N [tex]\frac{d \Phi _B}{dt}[/tex]

          fem = - N [tex]\frac{d \ (B A cos \theta)}{dt}[/tex]

The magnetic field and the area are constant

          fem = - N B A [tex]\frac{d \ cos \ \theta}{dt}[/tex]

          fem = - N B A (-sin θ)  [tex]\frac{d \theta}{dt}[/tex]

          fem = N B (π d² / 4) sin θ   w

          fem= [tex]\frac{\pi }{4}[/tex]  N B d² w sin θ

with this expression we can correspond the questions

a) the peak of the electromotive force

this hen the sine of the angle is 1

         sin θ = 1

         fem = [tex]\frac{\pi }{4}[/tex]   77  1.18  0.10² 8.0

         fem = 5.709 V

b) as the system has a constant angular velocity, we can use the angular kinematics relations

          θ = w₀ t

          t = θ/w₀

Recall that the angles are in radians, so the angle for the maximum of the sine is

           θ= π/2

           t = [tex]\frac{\pi }{2} \ \frac{1}{8}[/tex]

           t = 0.196 s

c) for the electromotive force to be negative, the sine function of being

            sin θ= -1

whereby

          θ = 3π/ 2

          t = [tex]\frac{3\pi }{2} \ \frac{1}{8}[/tex]  

          t = 0.589 s

d) This electromotive force has values ​​that change sinusoidally with an angular velocity of

          w = 8 rad / s

angular velocity and period are related

          w = 2π / T

          T = 2π / w

          T = 2π / 8

          T = 0.785 s

5 How does air get polluted?​

Answers

Answer:

Explanation:

- Pollution from cars

- Burning fossil fuels

what is the gravitational potential in a field produced by an object of mass 2000 kg at a distance of 10 km

Answers

Answer:

196 megajoules

Explanation:

Since you are talking about the gravitational potential I am assuming 10km is the height of the object in free fall.

PEg = mgh    2000kg×9.8m/s²×10000m = 196 megajoules

Hannah wants to create a record keeping system to track the inventory needed to efficiently run her lawn and landscape business, such as spare parts, gas cans, string trimmers, etc. Her crew manager will also be using the system. Hannah is considering whether to use Excel or Access. Which one of the following is NOT a benefit of using Access?

a. More data storage
b. Multiuser capability
c. Easier setup
d. Additional reporting features

Answers

Answer:

c). Easier setup

Explanation:

As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.



Which formula below correctly states Coulomb's Law?
A. F = kqq/r2
B. F = kqq/r
C. F = qq/kr2
D. F = kr2/qq

Answers

QUESTION:- Which formula below correctly states Coulomb's Law?

OPTIONS:-

[tex]A. \: \: \: \: F = kqq/r^ 2 \\B. \: \: \: \: F = kqq/r \\C. \: \: \: \: F = qq/kr^ 2 \\D. \: \: \: \: F = kr^2 /qq \\ [/tex]

ANSWER:-

F is directly proportional to the product of the charges

[tex]F∝qq[/tex]

F is inversely proportional to the square of the distance between them

[tex]F∝ \frac{1}{ {r}^{2} } [/tex]

from above 2 equation:-

we get:-

[tex]F∝ \frac{qq}{ {r}^{2} } [/tex]

To remove proportionality sign we use constant for this case we r using constant k

[tex]F = \frac{Kqq}{ {r}^{2} } [/tex]

So your answer is :-

OPTION A.

[tex]F = \frac{Kqq}{ {r}^{2} } [/tex]

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