A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals during the time the flashlight is in operation, how long was the flashlight used?

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:

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:

a. 1 Newton = 100000 Dyne

b. a represents acceleration.

Explanation:

Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;

1 Newton = 10⁵ Dyne

1 Newton = 100000 Dyne

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Force = mass * acceleration

[tex] F = ma[/tex]

Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.

Two connection methods are used for brushless DC motors. One method is to connect the coils in a loop as we compared it with the rotor winding of DC motors in Fig. 2.27. This method is called a Δ (delta) connection.

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Answer:

A. liquid and solid

Explanation:

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:

Explanation:

i am sorry i needed points

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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Explanation:

time=Distance/speed

t=32/8

t=4 seconds

Answer:

[tex]d=79.9m[/tex]

Explanation:

From the question we are told that:

coefficient of static friction [tex]\mu=0.38[/tex]

Velocity [tex]v=87.9=>24.41667m/s[/tex]

Generally the equation for Conservation of energy is mathematically given by

 [tex]\mu*mgd = 0.5 m v^2[/tex]

 [tex]d=\frac{0.5*24.42^2}{0.38*9.8}[/tex]

 [tex]d=79.9m[/tex]

Answer:

c) The pitch of the two sirens sounds the same to you

Explanation:

The pitch does not depend on the distance of the object from the observer.

As per the given data

pitch = frequency

Frequency = [tex]f_{0}[/tex]  [tex]\frac{V +- V_{0}}{V +- V_{s}}[/tex]

[tex]f^{'}[/tex] = [tex]f_{0}[/tex]  [tex]\frac{V }{V - V_{s}}[/tex]

Hence, the pitch of the two sirens remains the same for the observer.

Answer:

c) The pitch of the two sirens sounds the same to you

Explanation:

Answer:

if 1 light year was one millimeter then 105,700 light years = 105,700 mm, (or 105.7 meters in case you needed to simplify or something)

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]

Work done by  man will be A. 640 J

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

according to work energy theorem

Work done = final Kinetic energy - initial kinetic energy

                   = KE (final) - KE (initial )

                   = 1/2 m ([tex]v^{2}[/tex])  - 1/2 m ([tex]u^{2}[/tex])

                  = 1/2 m ([tex]v^{2}[/tex] - [tex]u^{2}[/tex])

                  = 1/2 * 500 * ( [tex]2^{2}[/tex] - [tex]1.2^{2}[/tex])

                  = 250 * 2.56 = 640 J

correct answer is A. 640 J

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Answer:

[tex]P=217600Pa[/tex]

Explanation:

From the question we are told that:

Density [tex]\rho=1000kg/m^3[/tex]

Depth of Water [tex]d=12.0m[/tex]

Generally the equation for Pressure is mathematically given by

 [tex]P=\rho gh[/tex]

 [tex]P=1000*9.8*12[/tex]

 [tex]P=117600N/m^2[/tex]

Therefore

Absolute Pressure=P+P'

Where

P=Pressure under water

P'=Atmospheric Pressure

Therefore

 [tex]P_A=P+P'[/tex]

 [tex]P_A=117,600+10^5[/tex]

 [tex]P=217600Pa[/tex]

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: hello some data related to your question is missing attached below is the missing data and diagram related to the solution

answer:

P = 141.21 N

Explanation:

Given data:

Mass of crate = 50 kg

coefficient  of static friction ( μ ) = 0.25

Calculate minimum horizontal force ( P ) that holds the crate from sliding

∑fx = 0

     = P + Fcos θ - N*sinθ = 0

     = P + 0.25N cos 30° - Nsin30°  = 0

∴ P = 0.2835 N = 0

P - 0.2853 N = 0 ------- ( 1 )

∑fy = 0

     - 50g + Ncosθ + Fsinθ

     - 50*9.81 + Ncos30° + 0.25Nsin30°

∴ N = 494.942 N ----- ( 2 )

input 2 into 1

P - 0.2853 ( 494.942 ) = 0

P = 141.21 N

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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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:

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

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:

[tex]\mu=0.185[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=17kg[/tex]

Force [tex]F=33N[/tex]

Velocity [tex]v=1.6m/s[/tex]

Distance [tex]d= 9.8m[/tex]

Generally the equation for Work done is mathematically given by

 [tex]W=\triangle K.E+\triangle P.E[/tex]

Where

 [tex]\triangle K.E=(F-F_f)*2[/tex]

 [tex]F_f=F+\frac{\triangle K.E}{d}[/tex]

 [tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]

 [tex]F_f=30.8N[/tex]

Since

 [tex]f = \mu*m*g[/tex]

 [tex]\mu= 30.8/(m*g)[/tex]

 [tex]\mu= 30.8/(17*9.81)[/tex]

 [tex]\mu=0.185[/tex]

Answer:

1. Carpenter

2. True

Explanation:

While a mason was working concrete into the formwork, the formwork collapses. The best person to rectify this problem is CARPENTER.

This is because it is the job of the Carpenter to design and build formwork, most especially wooden formwork. Formwork is like casing built to receive concrete and reinforcement during construction. Hence, when formwork collapses either due to stress, tension, or improper construction, it is the job of Carpenter to reconstruct the formwork or rectify the problem.

It is TRUE that when a device made in a workplace had defects. To address this issue the workshop manager should communicate directly with the workshop​. However, this communication will be an instruction on what to do next, and it usually directs those responsible to take action where necessary. For example, a workshop manager communicates to a carpenter about the need to rectify a chair or table that has a defect.

Answer:

a) [tex]T=0.01s[/tex]

b) [tex]T=0.001s[/tex]

c) [tex]T=0.00001s[/tex]

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

[tex]T=\frac{1}{f}[/tex]

Therefore

a)

For

[tex]T=100 Hz[/tex]

[tex]T=\frac{1}{100}[/tex]

[tex]T=0.01s[/tex]

b)

For

[tex]F=1kHz[/tex]

[tex]T=\frac{1}{1000}[/tex]

[tex]T=0.001s[/tex]

c)

For

[tex]F=100kHz[/tex]

[tex]T=\frac{1}{100*100}[/tex]

[tex]T=0.00001s[/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:

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

Answer:Increasing

Explanation:

Given

Car is driven on the straight road with  instantaneous acceleration in the direction of car's motion.

If instanateneous acceleration is constant then speed of car is increasing at a constant pace. As there are no turns on the road, therefore speed of car is increasing.

The speed of the car is "decreasing". A further description is provided in the below paragraph.

Thus the above answer is correct.

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