How many minutes would be required for a 300.0 W (i.e., power=energy per second=300 J/s)
immersion heater to heat 0.25 kg of water from 20.0°C to 100.0°C? (Note: The specific heat of water
is 4184 J/kg°C)

Answer:

1680 seconds

Explanation:

[tex]28 hrs * \frac{60 s}{1 hr} =1680s[/tex]

Answer:

See below....the question is unclear....pick the answer you think they want

Explanation:

Car goes from 30 m/s    to   24 m/s   in 10 seconds

acceleration = change in velocity/change in time

                     =   -6 m/s   / 10 s   =  - 3/5   m/s^2  

The wording of the question is not very good.....perhaps they meant to say that after the 5 seconds the velocity is now 24 m/s ?

a1 *10   +   a2 *5   = -6 m/s  

            ( 6 m/s is the change in velocity from 30 to 24 m/s)    

    a1/a2 = 1.5    so   a2 = a1/1.5   (substitute this in)

10 a1   + 5 ( a1/1.5)  =-6

15 a1 + 5 a1  = -9

a1 =  - .45 m/s^2

The velocity of the car at the end of the initial ten-second interval, given that the car further slows down for the next five seconds, is 25.5 m/s

From the first statement, we have:

v₁ = u₁ + a₁t₁

v₁ = 30 + (a₁ × 10)

v₁ = 30 + 10a₁ ......(1)

From the second statement, we have:

v₂ = u₂ + a₂t₂

24 = u₂ + (a₂ × 5)

24 = u₂ + 5a₂   .......(2)

But,

u₂ = v₁ = 30 + 10a₁

Thus, we have:

24 = 30 + 10a₁ + 5a₂

But,

a₁/a₂ = 1.5

a₁ = 1.5a₂ ..... (3)

Thus, we have:

24 = 30 + 10a₁ + 5a₂

24 = 30 + 10(1.5a₂) + 5a₂

24 = 30 + 15a₂ + 5a₂

24 = 30 + 20a₂

Collect like terms

24 - 30 = 20a₂

-6 = 20a₂

Divide both sides by 20

a₂ = -6 / 20

= -0.3 m/s²

Substituting the value of a₂ into equation 3, we have

a₁ = 1.5a₂

= 1.5 × -0.3

= -0.45 m/s²

Substitute the value of a₁ into equation 1 to obtain the velocity, v₁ at the end of the initial to 10 s

v₁ = 30 + 10a₁

= 30 + (10 × -0.45)

= 30 - 4.5

= 25.5 m/s

Thus, velocity of the car at the end of the initial ten-second interval is 25.5 m/s

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This is an exercise of thermal scales.

T = 40 °C

T = °F ¿?

To convert degrees Celsius to degrees Fahrenheit, we have the formula:

[tex]\bf{^{\circ}F=\dfrac{9}{5} \ ^{\circ}C+32 \ \ \to \ \ \ Formula }[/tex]

We substitute our data into the formula.

[tex]\bf{^{\circ}F=\dfrac{9}{5} \times 40+32}[/tex]  

First we divide 9/5 multiplied by 40.

[tex]\bf{=75+32}[/tex]

Add 75 plus 32

[tex]\bf{=104 \ ^{\circ}F}[/tex]

Therefore, the temperature change of 40 °C on the °F scale is equal to 104.

[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Piscis04}}}}}}[/tex]

Answer:

See below

Explanation:

Alcohol first goes slightly down because of the expansion of the glass of thermometer before doing up.

[tex]\rule[225]{225}{2}[/tex]

Weight is compared with the gravitational force exerted on each of them by the Earth.

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

When mass increases and distance reduces, gravity rises. Gravity also lowers when the distance between two points grows and the mass decreases.

The gravitational force is balanced by weight;

[tex]\rm F = W \\\\ G\frac{mM}{r^2} =mg[/tex]

Hence, weight is compared with the gravitational force exerted on each of them by the Earth.

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The force between the molecules involved in the bond is 6. 426 *10^-11 Newton

Using the formula:

F = K[q1 x q2]/D^2

where K is coulombs constant =9 *10 ^9 Nm^2/C^2.

q1  and q2 = charges  =  1.60x10 -20C

d = distance between the charges = 2x10 -10 m

Substitute the values into the formula

F =  [tex]9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }[/tex]

F = [tex]9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}[/tex]

F = [tex]9* 10^9 * 7. 14* 10^-21[/tex]

F = [tex]6. 426 * 10^-11[/tex] Newton

Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton

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Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

                  = 64°C

And now we have to write the temperature in kelvins.

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

Let us find it now.

64°C + 273 = 337k

Therefore,

64°C ⇒ 337k

Answer:

v(t)=s'(t)=5t⁴-sin(4).4, for velocity function at time t

a(t)=v'(t)=20t³, for acceleration function at time t

Answer:

42.9°C

Explanation:

Resistance of a conducting material is directly proportion to Temperature, that is, when being heated, their is more of collisions between the molecules of the material which as a result offers more opposition to the flow of current through it.

therefore R=kT

where R=resistance of the wire

T=Temperature

k= proportionality constant

when R=1.05 T=30°C

so k=1.05/30

k=0.035

when R=1.5 k=0.035

so T=1.5/0.035

T=42.9°C. or 43°C approximated

The loops must the coil have to generate a maximum emf of 2500 will be 439.

According to  Faraday's law of electromagnetic induction, the rate of change of magnetic flux linked with the coil is responsible for generating emf in the coil resulting in the flow of amount of current.

Given data;

Area,A = 0.239 m²

Angular velocity,ω=373 rad/sec

Magnetic field,B=0.0639 T

Maximum emf,E= 2500V

The formula for the maximum induced voltage is;

E{max} = N ×  B × A × ω

2500 = N × 0.639 × 0.0239 × 373

N = 438.66

N = 439 \ turns

Hence, loops must the coil have to generate a maximum emf of 2500  will be 439.

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Answer: 11392.99

Explanation: I had the same problem and here are the steps:

Values:

N (Number of loops) = 2000

A (area) = 0.239

ω (angular velocity) = 373

B (magnitude) = 0.0639

ε = NABω

= (2000)(0.239)(373)(0.0639)

ε = 11392.99 (FINAL ANSWER)

Answer: 1.32 x 10^-9 m

Explanation:

The formula for the radius of nth orbital is:

[tex]r_{n}=\frac{n^{2}}{Z} \times 0.528 \text { Angstorms }[/tex]

For hydrogen, Z= 1, so we have:

[tex]$$\begin{aligned}r_{n} &=5^{2} \times 0.528 \text { Angstorm } \\&=13.2 \text { Angstorms } \\&=13.2 \times 10^{-10} \mathrm{~m} \\&=1.32 \times 10^{-9} \mathrm{~m}\end{aligned}$$[/tex]

Answer: 1.32*10^-9

Explanation: I had the same question and here's what I got:

rn = n2(r1)

n = 5

r = 5.29*10^-11

rn = n^2(r1)

= 5^2(5.29*10^-11)

r = 1.32*10^-9

Physical science explains the physical world and not supernatural events and an idea is considered true by science if it can be observed, explained and reproduced by others.

Physical science is a systematic study of the of the physical world and phenomena based on observation and experiments.

The scientific method involves:

Science explains the physical world and not supernatural events.

Science studies physical phenomena such as the stars, the planets, living things, etc.

Science do not study about angels or demons or supernatural events.

An idea is considered true by science if it can be observed, explained and reproduced by others.

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Answer:

A

Explanation:

F = m * a

100 = 5 * a

a = 100/5 = 20 m/s^2

The allowable ampacity of a No. 12 copper wire with type THHN insulation installed in a raceway with 35 wires is approximately 12 amperes. This is best represented by option D.

According to the National Electrical Code Table (NEC), a derating factor must be applied for any raceway with more than three current-carrying conductors (in our case, copper wires). In a raceway with 31-40 current-carrying conductors, the derating factor is 50%. Because the ampacity of a No. 12 copper wire usually is 25 amperes (A), we can multiply this by our derating factor to find that:

25 • 0.50 = 12.5

The exact allowable ampacity of this copper wire is 12.5 amperes; however, we can round down to find that the closest answer is 12 A.

Hence, option D represents the best answer.

Answer:

  D)  12 A

Explanation:

You want to know the allowable ampacity of AWG 12 type THHN copper wire in a raceway with 35 wires.

Table 310.15(B)(16) of the 2017 National Electrical Code gives the nominal rating of AWG 12 type THHN copper wire as 30 A when installed in an environment with a temperature of 30 °C or less.

Table 310.15(B)(3)(a) gives the adjustment factor when there are more than three conductors in a raceway or cable. For 31–40 conductors, the ampacity must be derated to 40% of its nominal value.

In a raceway with 35 wires, the allowable ampacity of THHN AWG 12 wire is ...

  (30 A)(0.40) = 12 A

__

Additional comment

For environmental temperatures above 30 °C, Table 310.15(B)(2)(a) gives additional derating factors. The adjustment factor for temperatures up to 50 °C (122 °F) is 0.82, which would bring the allowable ampacity down to about 9.8 A.

Answer:

Define universal gravitational constant. Why is newton's law of gravitation called universal law?

The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move will be 81.03 N,

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

It is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration. Mathematically aion in the different components and balancing the equation gets. Components in the x-direction.

The normal force is balanced by weight;

N = mg

The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move is;

F = μN

F= μmg

F=0.14 ×  59 kg × 9.81 m/s²

F = 81.03 N

Hence, the magnitude of the maximum force will be 81.03 N,

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(a) The compression of the spring is 7.4 cm.

(b) When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.

The compression of the spring is calculated by applying Hooke's law as follows;

F = kx

x = F/k

x = (mg)/k

x = (3 x 9.8)/400

x = 0.074 m

x = 7.4 cm

When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.

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Explanation:

Power= 40 W

cost=21¢/kWh

t=1 month=30 days = 720h

Energy= power*time

E= 40W* 720h = 28800kWh

Cost= Energy*rate

cost= 28800kWh*21¢/kWh

cost=604800¢

Answer:

1.163 seconds

Explanation:

see the attached. solve the quadratic equation.

Answer:

Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].

Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)

Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].

Explanation:

Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.

Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.

Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.

Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].

Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.

Rearrange this equation to obtain an expression for [tex]v[/tex]:

[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].

For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

In the opposite figure, the work done by the force F to push the box upwards from the ground to the top of the inclined plane is 600 J.Option c is correct.

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

The displacement value from the triangle is;

[tex]\rm sin 30^0 = \frac{P}{H} \\\\ H=S =\frac{3 \ m}{sin 30^0} \\\\ S = 6\ m[/tex]

The work done is found as ;

[tex]\rm W= Fd \\\\ W = 100 \ N \times 6 \\\\ W= 600 \ J[/tex]

The force F in the opposing figure does 600 J of effort to lift the box from the ground to the top of the inclined plane.

Hence option c is correct.

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The x and the y coordinates of the center of the mass of this system will be 43.1 m and 3.12 m respectively.

A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.

Given data;

There are three little objects that are densely spaced out in the x-y plane.;

The value of the masses

m₁=1.41 kg

m₂=2.55 kg

m₃=3.19 kg

[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3x_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 6 +3.19 \times 4}{1.41+2.55+3.19} \\\\ X_{cm} =4.31 \ m[/tex]

[tex]\rm Y_{cm} =\frac{ (W_1y_1 + W_2y_2 + W_3y_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 4 +3.19 \times 7}{1.41+2.55+3.19} \\\\ X_{cm} =3.12 \ m[/tex]

Hence, the x and the y coordinates of the center of the mass of this system will be 4.31 m and 3.12 m respectively.

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We will use the design process to meet the requirements of the competition by the given explained in the main answer.

The initiative was taken by two or more parties operating independently to win the business of a third party by competing to hire contractors on the best terms possible.

The complete question is

"A local businessperson placed an advert in the local newspaper for a competition for girls only from different schools to participate in the competition. The competition is about drawing ideas for a shopping complex for the local community. Girls from the different schools are excited about the competition. you are a group of boys in the local schools who saw the advert and feel it is unfair for you to be excluded from the competition.

consider yourself as one of the learners in the local schools.

Explain why the project should be given to boys also who have better skills for this project than the girls do.

1.2 think about the design process skills, investigates, design, make, evaluate and communicate."

Explain how you will use the design process to meet the requirements of the competition.

[ 1.1 ]Boys should also be given the project because some of them have more artistic talent than girls do when it comes to sketching designs for local shopping centers.

Boys can also present a clear picture of how local communities can expand their shopping options.

For instance, they can encourage local merchants to offer seasonal items because doing so will benefit their businesses.

They can also suggest that retailers encourage customers to bring a friend to participate in competitions, which will be highly regarded.

[1.2] In order to fulfill the needs of the competition, I would make sure to explore how a shopping center for the neighborhood functions, for instance, by making sure I am aware of the strategies one may use to enhance his or her firm in the neighborhood.

In order to win the competition, I would make sure that I was confident in what I was saying and that I was audible enough to be heard during the competition.

Finally, I would make sure I met the competition's conditions by staying on-topic and solely addressing the issues raised in the advertisement

Hence will use the design process to meet the requirements of the competition.

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a) Mirrors

Mirror requires the high luster of metals in order to work.

When light shine on to the surface of a metal, its electrons absorb small amounts of energy and become excited into one of its many empty orbitals.

The electrons immediately fall back down to lower energy levels and emit light.

This process is responsible for the high luster of metals.

Here,

Mirrors have metal coating on its back side, because of high reflective property , high lustrous mirror is preferred . When metal is more lustrous, the less light reflect into it and bounces back.

Coating one side of a piece of glass with shiny metals can turn it into a mirror, reflecting light coming toward it.

Window glass can reflect only eight percent of light hitting it, while mirrors can reflect 95 percent of light hitting them.

The glass in a mirror is usually coated with a layer of silver or aluminum.

Mirrors are shiny because when photons (rays of light) coming from an object , it strike the smooth surface of a mirror, then they bounce back at the same angle. Our eyes see these reflected photons as a mirror image.

Therefore,

Mirror requires the high luster of metals in order to work.

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Answer:A) mirrors

Explanation:

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612

(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) angular frequency = velocity / radius

ωf = 10/2 = 5 rad/s and ωi = 20 rad/s

Angular acceleration, α = ωf - ωi /t

Put the values, we get

α = -15/5 = -3 rad/s²

Coefficient of friction, μ = a/g = rα/g

Plug the values, we get

μ = 0.612

Thus, the coefficient of friction corresponding to the frictional force is 0.612.

(c) The energy lost = heat generated

 energy lost = 1/2 Iω² + 1/2 Mv²

 energy lost = 1/2 MR²ω² + 1/2 Mv²

Plug the values, we get

 energy lost = 7500 J

Thus, the  increase in the thermal energy is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands

s= (v- 2gH) sin2θ /g

s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81

s = 7.78m

Thus, the horizontal distance is  7.78 m

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A. The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 4 i m/s.

B. The velocity and acceleration at any time t is  4t i +1 j and 4i.

C. The average acceleration in the time interval given in part (a) is 4

Acceleration can be defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

Given the following equation shows the position of a particle in time t, x=at² i + bt j

where t is in second and x is in meter. a=2m/s², b=1m/s.

A. Velocity is time rate of change of displacement.

V = dx/dt

V = d/dt (2t² i + 1t j)

V = 4t i +1 j

Velocity in time interval,

At  t₁=2sec , V₂= 4x2 i + 1 j = 8 i + 1 j

At  t₂=3sec, V₃ = 4x3 i + 1 j = 12 i + 1 j

Average velocity in time interval t₁=2sec and t₂=3sec is

, V₃ - V₂ = (12 i + 1 j) - ( 8 i + 1 j )

Average velocity =  4 i

B. Velocity at time t, is

V(t) = V = 4t i +1 j

Acceleration a = dV/dt

a(t) = d/dt (4t i +1 j)

a(t) = 4 i +0 j

C. Average acceleration in time interval t₁=2sec and t₂=3sec is

a = 4 m/s²

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Answer:

0.5 m/s²

Explanation:

according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:

F = m×a

F for force

m for mass

a for acceleration

we use the given data and get:

20 = 40×a

we find a=20/40=0.5m/s²

acceleration of the box is 20m/s^2

given

mass (m) = 40 kg

force (f) = 20n

acceleration (a) =?

we know,

f = m × a

20 = 40 × a

40 - 20 = a

20m/s^2 = a

By using Newton's Second law of motion, the largest mass that can be lifted by this input force is equal to 2,244 kg.

Based on the information provided, we can logically deduce that the output to input force ratio is equal to 22:1. This ultimately implies that, the input force is equal to 1000 N, which would result in an output force of 22,000 N.

Next, we would calculate the largest mass by using Newton's Second law of motion:

Mass = Force/Acceleration

Mass = 22,000/9.8

Mass = 2,244 kg.

Note: Acceleration due to gravity is equal to 9.8 m/s².

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