A system can only achieve a lower energy state by:_______

Answer:

A) 3.367 × 10^(-6)

B) 2.97 × 10^(7) N/C

C) Upwards

Explanation:

We are given;

Mass of bee; m = 100 mg = 100 × 10^(-6) kg

Charge on bee;q=33 pC = 33 × 10^(-12)C

Electric field strength; E = 100 N/C

A) Formula for weight of bee; W = mg = 100 × 10^(-6) × 9.8 = 9.8 × 10^(-4) N

Electric force on Bee; F = qE = 33 × 10^(-12) × 100 = 33 × 10^(-10) N

ratio of the electric force on the bee to the bee's weight; F/W = (33 × 10^(-10))/(9.8 × 10^(-4)) = 3.367 × 10^(-6)

B) For the bee to be suspended in the air, it means the weight of the bee must be equal to the electric force. Thus;

mg = qE

100 × 10^(-6) × 9.8 = 33 × 10^(-12) × E

E = (100 × 10^(-6) × 9.8)/(33 × 10^(-12))

E = 2.97 × 10^(7) N/C

C) From Newton's law, sum of forces = 0.

Thus;

F_n + F + W = 0

Where F is the normal force.

Thus;

F_n = -(F + W)

F_n = - ((33 × 10^(-10)) + (9.8 × 10^(-4)))

F_n = -9.8 × 10^(-4) N

Thus, applied electric field is;

E_a = F_n/q = (-9.8 × 10^(-4))/(33 × 10^(-12)) = -2.97 × 10^(7) N/C

This is negative and so it means the direction will be opposite the Earth's electric filed which is upwards.

Answer:

cm/s

6

128.2

96.0

7

145.8

Table B

Answer:

belpw

Explanation:

The distance prior to the sliding friction dispersing her energy would be:

- The distance will remain unaffected by the sliding friction i.e. 354m

As we know, When Sliding friction dissolves her energy, leading her Kinetic Energy to turn 0 on coming to the state of rest. So,

[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]            (∵ Work in -ve denotes it is done opposite to friction)

Given that,

m(mass) [tex]= 50 kg[/tex]

v(velocity) [tex]= 30 km/hr[/tex] or [tex]8.33 m/s[/tex]

The coefficient of Kinetic Friction [tex]= 0.01[/tex]

g(gravitational force) [tex]= 9.8 m/s^2[/tex]

Initial Velocity(u) [tex]= 30[/tex] × [tex]1000/3600 m/s[/tex]

[tex]= 8.33 m/s[/tex]

Now by employing the provided values,

[tex]F =[/tex] μ[tex]mg[/tex]

[tex]= (0.01) (50) (9.8)[/tex]

[tex]= 4.9[/tex]

∵ [tex]F = 4.9 N[/tex]

By using the above expression, we will find the distance;

[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]

⇒ [tex]1/2 (50) (0)^2 - 1/2 (50) (8.33)^2 = -4.9(S)[/tex]

⇒ [tex]1734.7225 = 4.9S[/tex]

⇒ [tex]S = 1734.7225/4.9[/tex]

∵ [tex]S = 354 m[/tex]

Because [tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]  [tex]= -[/tex] μmgS

⇒ [tex]S = (u^2 - v^2)[/tex]/2μ[tex]g[/tex]

Thus, the distance will remain unaffected by the sliding friction i.e. 354m

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Answer:

Explanation:

From the given information:

the car's momentum = momentum of the truck

∴

(a) 816 kg × v = 2650 kg × 16.0 km/h

v = (2650 kg × 16.0 km/h) /  816 kg

v = 51.96 km/hr

(b) 816 kg × v = 9080 kg × 16.0 km/h

v = (9080 kg × 16.0 km/h) /  816 kg

v = 178.04 km/hr

Answer:

the correct answer is C   v = 60 cm / s

Explanation:

The speed of a wave is related to the frequency and the wavelength

         v = λ f

They indicate that the object performs 20 oscillations every second, this is the frequency

         f = 20 Hz

the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength

         λ = 3 cm = 0.03 m

let's calculate

       v = 20 0.03

       v = 0.6 m / s

       v = 60 cm / s

the correct answer is C

Answer:

C

Explanation:

The normal is the line which divides the angle between the incident ray (which is the ray of an object which strikes the mirror) and the reflected ray(the ray which is thrown back as the object hits the mirror surface) into two equal parts. The normal is always perpendicular to the surface. In the description agram Given , the Noa which is the line C, divides the reflected ray (line D) and the incident ray (line A) into two equal parts. The plane surface is line B and the other incident ray (line C) is perpendicular to B

Answer:

B

Explanation:

because it take to the thought of a situation

Answer:

d i think

Explanation:

and are u using IXL???

Answer:

[tex]{ \tt{initial \: velocity, \: u = 0}} \: (at \: rest) \\ { \tt{final \: velocity, \: v = 15 { {ms}^{ - 1} }}} \\ { \tt{time, \: t = 6s}} \\ { \bf{from \: first \: newtons \: equation \: of \: motion : }} \\ { \bf{v = u + at}} \\ { \tt{15 = 0 + (a \times 6)}} \\ { \tt{6a = 15}} \\ { \tt{acceleration, \: a = 2.5 \: {ms}^{ - 2} }}[/tex]

Answer:

(a) 1470 N/m

(b) 48.2 m

Explanation:

Applying,

(a) F = ke.................... Equation 1

Where F = force applied to the spring, k = spring constant, e = extension

make k the subject of the equation

k = F/e............... Equation 2

But,

F = mg............. Equation 3

Where m = mass, g = acceleration due to gravity

Substitute equation 3 into equation 2

k = mg/e.............. Equation 4

From the question,

Given: m = 6 kg, e = 42-38 = 4 cm = 0.04 m

Constant: g = 9.8 m/s²

Substitute these values into equation 4

k = (6×9.8)/0.04

k = 1470 N/m

(b) Consider the end of the spring to the left which exert a force to the right

Then,

e = F/k............. Equation 5

Given: F = 150 N, k = 1470 N/m

Substitute these values into equation 5

e = 150/1470

e = 0.102 m

Hence the length of the spring is

L = 0.38+0.102 = 0.482 cm = 48.2 m

Answer:

[tex]E=6.68\times 10^{-26}\ J[/tex]

Explanation:

Given that,

The frequency of FM ratio station, f = 100.8 MHz = 100.8 × 10⁶ Hz

We need to find the energy of the wave. We know that,

Energy, E = hf

Put all the values,

[tex]E=6.63\times 10^{-34}\times 100.8\times 10^6\\\\=6.68\times 10^{-26}\ J[/tex]

So, the energy of the wave is equal to [tex]6.68\times 10^{-26}\ J[/tex].

Air is classified as a mixture. Option D is the correct answer.

Air is a combination of different gases, primarily nitrogen (about 78%), oxygen (about 21%), and small amounts of other gases such as carbon dioxide, argon, and trace elements. These gases are not chemically bonded to each other, but rather exist together in the same space. Option D is the correct answer.

In a mixture, the substances involved retain their individual properties and can be separated by physical means. This is true for air as well. The gases in air can be separated through processes like fractional distillation or filtration. It's important to note that air also contains other components such as water vapor, dust particles, and pollutants, which can vary in concentration depending on the location and environmental conditions. These components further contribute to the complex nature of air as a mixture.

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Answer:

The range of the photon energies is between:

2.652 x 10⁻²⁵ J    to    4.973 x 10⁻²⁵ J

Explanation:

The energy of a photon is calculated using the following equation;

E = hf

where;

h is Planck's constant = 6.63 x 10⁻³⁴ Js

f is frequency of the photon

[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]

[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]

The range of the photon energies is between:

2.652 x 10⁻²⁵ J    to    4.973 x 10⁻²⁵ J

Explanation:

The weight of an object on the surface of the earth is equal to the gravitational force exerted by the earth on the object.

[tex]W=F_G[/tex]

[tex]mg = G \dfrac{mM}{R^2}[/tex]

which gives us an expression for the acceleration due to gravity g as

[tex]g = G\dfrac{M}{R^2}[/tex]

At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity [tex]g_h[/tex] at this height is

[tex]mg_h = G \dfrac{mM}{(2R)^2}= G \dfrac{mM}{4R^2}[/tex]

Simplifying this, we get

[tex]g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g[/tex]

Answer:

1. Friction enables us to walk freely.

2. It helps to support ladder against wall.

3. It becomes possible to transfer one form of energy to another.

4. Objects can be piled up without slipping.

Answer:

A. liquid and solid

Explanation:

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

Answer:

Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018

Answer:

I = 15A

Explanation:

V = I*R

120V = I*8.0ohms

I = 120V/8.0ohms

I = 15A

Answer:

I=15A

Explanation:

you know what to do.

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Give them the brain list.

Answer:

Explanation:

i am sorry i needed points

Maximum horizontal force that can be applied on the box is 300.32 N.

Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.

Kinetic friction is defined as the opposing force exerted by the surface on an object in contact with it, when there is relative motion between the two surfaces.

Here,

Mass of the box, m = 150 lb = 68.1 kg

Coefficient of kinetic friction, μ = 0.45

Maximum horizontal force that can be applied on the box is the kinetic frictional force. Frictional force,

F(k) = μmg

F(k) = 0.45 x 68.1 x 9.8

F(k) = 300.32 N

Now, the box sits on a ramp inclined at 60°

Coefficient of kinetic friction, μ = 0.45

The net force here acting on the box placed in the ramp is due to the kinetic frictional force and the weight of the box.

So,

Frictional force, F(k)' = μmgcosθ

F(k)' = 0.45 x M x 9.8 x cos 60

F(k)' = 2.2M

Weight of the box acting horizontally,

W = Mgsinθ

W = M x 9.8 x sin60

W = 8.5M

Therefore, net force,

Fn = W - F(k)'

Fn = 8.5M - 2.2M

Fn = 6.3M

The total force acting on the box is

F = F(k) - Fn

ma = 300.32 - 6.3M

Since, the box is moving with constant speed, the acceleration, a = 0

Therefore,

300.32 - 6.3M = 0

6.3M = 300.32

M = 300.32/6.3

M = 47.7 kg = 105.16 pound

Hence,

Maximum horizontal force that can be applied on the box is 300.32 N.

Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.

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Answer:

9.6352× 10²⁰ C atoms

Explanation:

From the given information,

The molar mass of Carbon = 12 g/mol

number of moles = 0.020g/ 12 g/mol

number of moles = 0.0016 mol

If 1 mole of C = 6.022 × 10²³ C atoms

∴

0.0016 mol of C = (6.022 × 10²³ C atoms/ 1 mol of C)×0.0016 mol of C

= 9.6352× 10²⁰ C atoms

Hence, the number of carbon atoms present in 0.020 g of carbon = 9.6352× 10²⁰ C atoms

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)  

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion  P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄    

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K  

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

Answer:

The surface tension is 190.2 N/m.

Explanation:

Initial radius, r = 4 cm

final radius, r' = 6 cm

Work doen, W = 15 J

Let the surface tension is T.

The work  done is given by

W = Surface Tension x change in surface area

[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]

Answer:

The answer is c

Thermal energy moves within the air from the flames to the marshmallow.

Explanation:

Hope it helps

You sit with friends around a campfire, roasting marshmallows. then the transfer of thermal energy involved in this system is an example of radiation Thermal energy moves within the air from the flames to the marshmallow. Hence option C is correct.

In physics and engineering, the phrase "thermal energy" is thrown around in a lot of different situations. It can relate to a variety of distinct physical notions. Included in this are the internal energy, or enthalpy, of a body of matter and radiation; heat, which is a form of energy transfer (as is thermodynamic work); and the characteristic energy of a degree of freedom in a system described in terms of its microscopic particulate constituents (where T denotes temperature and k denotes the Boltzmann constant.

Hence option C is correct.

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Answer:

the largest distance we can measure is 10¹⁴ km

Explanation:

Given the data in the question;

Threshold hearing = 10⁻²⁰

smallest distance measured = 1 mm

Largest distance measured will be;

⇒ ( threshold hearing )⁻¹ × smallest distance

= ( 1 / 10⁻²⁰ ) × 1 mm

= 10²⁰ × 1mm

= 10²⁰ mm

we know that; 1000 mm = 10⁶ km

Largest distance = ( 10²⁰ / 10⁶ ) km

= 10¹⁴ km

Therefore, the largest distance we can measure is 10¹⁴ km

Solution :

Given expression :

[tex]$\mu_k$[/tex]mgcosθ = mgsinθ − ma

Here, g = 9.8 [tex]m/s^2[/tex] , a = 3.60 [tex]m/s^2[/tex] , θ = 27°

Therefore,

[tex]$\mu_k mg \cos \theta = mg \sin \theta - ma$[/tex]

[tex]$\mu_k mg \cos \theta = m(g \sin \theta - a)$[/tex]

[tex]$\mu_k g \cos \theta = (g \sin \theta - a)$[/tex]

[tex]$\mu_k =\frac{(g \sin \theta-a)}{g \cos \theta}$[/tex]

Mow calculating the coefficient of kinetic friction as follows :

[tex]$\mu_k=\frac{g \sin \theta-a}{g \cos \theta}$[/tex]

[tex]$\mu_k=\frac{9.8 \times \sin 27^\circ-3.60}{9.8 \times \cos 27^\circ}$[/tex]

[tex]$\mu_k=0.097$[/tex]

Answer:

336.96m/s

Explanation:

answer is in photo above

The speed of sound on that day is  336.96 m/s.

Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.

given that:

frequency of the sound = 312 Hz.

2nd resonant length = 81.0 cm.

If the wavelength of sound is λ; the 2nd resonant length of  plastic pipe with an adjustable closed end = 3λ/4

Hence, wavelength of sound is = 81×(4/3) cm = 0.81 ×(4/3) m.

So,  the speed of sound on that day is = frequency × wavelength

= 312 Hz ×  0.81 ×(4/3) m.

= 336.96 m/s.

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