What is the rms current flowing through a light bulb that uses an average power of 60.0 W when it is plugged into a wall receptacle supplying an rms voltage of 120.0 V?

Answer: Final temperature is 34.15°C.

Explanation: When two objects have different temperature, they will exchange heat energy until there is no more net energy transfer between them. At that state, the objects are in thermal equilibrium.

So, when in equilibrium, the total heat flow must be zero, i.e.:

[tex]Q_{1}+Q_{2}=0[/tex]

In our case, there will be a change in state of ice into water, so total heat flow will be:

[tex]m_{1}c_{1}(T_{f}-T_{i})+m_{2}c_{2}(T_{f}-T_{i})+mL=0[/tex]

where

m₁ is mass of ice

m₂ is mass of water

c₁ is specific heat of ice

c₂ is specific heat of water

[tex]T_{f}[/tex] is final temperature

[tex]T_{i}[/tex] is initial temperature

L is latent heat fusion

Temperature is in Kelvin so the transformation from Celsius to Kelvin:

For ice:

T = -15 + 273 = 258K

For water:

T = 48 + 273 = 321K

Solving:

[tex]21(2.09)(T_{f}-258)+158(4.186)(T_{f}-321)+21(333)=0[/tex]

[tex]43.89T_{f}-11323.62+661.4T_{f}-212305.55+6993=0[/tex]

[tex]705.3T_{f}=216636.17[/tex]

[tex]T_{f}=[/tex] 307.15K

In Celsius:

[tex]T_{f}=[/tex] 34.15°C

Final temperature of the system when in equilibrium is 34.15°C

Answer:

The sun is the ultimate source of energy for all food chains. Through the process of photosynthesis, plants use light energy from the sun to make food energy. Energy flows, or is transferred through the system as one organism consumes another.

Answer:

the answer is true.......

Answer:

d

In the Physics models, we do not take into account the energy change by the friction of moving components

Answer:

In the Physics models, we do not take into account the energy change by the friction of moving components

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 60 × 2

We have the final answer as

Hope this helps you

Answer:

Explanation:

for rolling motion down the plane acceleration is given by the following expression

a = g sinθ / (1 + k² / R²)

here k is radius of gyration and R is radius of the object rolling down .

for cylinder I = 1/2 m R²

so k² = R² / 2

k² / R² = 1/2

a = g sinθ /( 1 + 1 / 2 )

= 2 / 3 x  g sinθ

v = √ 2 a s

= √ (2 x  2 / 3 x  g sinθ s )

= √ (4  / 3 x  g h  )

= √ (4  / 3 x  g x .5  )

= √ 2g / 3

for sphere  I = 2/5  m R²

so k² = 2/5 R²

k² / R² = 2 / 5  

a = g sinθ / (1 + 2 / 5)  

= 5 / 7  x  g sinθ

v = √ 2 a s

= √ (2 x  5 / 7  x  g sinθ s )

= √ (10/7  x  g h  )

Given

√ (10/7  x  g h  ) = √ 2g / 3

10/7  x  g h  = 2g / 3

h = 14 / 30 m

= .47 m .

Answer:

bc it was a universal explosion and It started the future

Explanation:

FACTS

Answer:

i wouldn't believe so.

Explanation:

because there was no room or air for the sound to move through. this is because of immense heat and the amount of hyperactive neutrons, electrons and protons clouding everywhere. This would mean that even if there was sound it would a. not travel far or b. go in a completely different direction than expected.

Answer: I got an increase of 306 kJ in internal energy.

Explanation: I used the 1st Law and the sign convention to get the answer in the screenshot.

(sorry if i'm wrong)

:(

The efficiency of the machine is 96.07%.

Efficiency can be defined as the ratio of work output to work input expressed in percentage.

To calculate the efficiency of the machine, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the efficiency of the machine is 96.07%.

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Answer:

Sensitivity of biosensor

The biosensor showed good linear correlation in the wide detection range of 0.001–2000 ng/mL with good sensitivity. In addition, it retained its biosensing property for seven days with high reproducibility

Explanation:

Answer:

The force is  [tex]F = 18.33 \ N[/tex]

Explanation:

From the question we are told that

    The number of liquids is  n =  3

      The volume of the first liquid is [tex]V_1 = 0.50 L = 0.0005 \ m^3[/tex]

      The density of the first liquid is  [tex]\rho_1 = 2.6 \ g/cm^3[/tex]

      The volume of the second  liquid is [tex]V_2 = 0.25 L = 250\ cm^3[/tex]

      The density of the second liquid is  [tex]\rho_2 = 1.0 \ g/cm^3[/tex]

      The volume of the third  liquid is [tex]V_3 = 0.40 L = 400\ cm^3[/tex]

      The density of the  third  liquid is  [tex]\rho_3 = 0.80 \ g/cm^3[/tex]

Generally the force at the bottom of the container is mathematically represented  as

    [tex]F = m_t * g[/tex]

Here [tex]g = 980.665 \ cm/s^2[/tex]

Here  [tex]m_t[/tex]  is the total mass of all the liquid which is mathematically represented as

             [tex]m_t = ( V_1 * \rho_1 )+ ( V_2 * \rho_2)+ ( V_3 * \rho_3)[/tex]

=>         [tex]m_t = ( 500 * 2.6)+ ( 250 * 1.0 )+ ( 400 * 0.80 )[/tex]

=>         [tex]m_t = 1870 \ g[/tex]

So

       [tex]F = 1870 * 980.66[/tex]

=>   [tex]F = 1833843.55 \ g \cdot cm /s^2[/tex]

=>    [tex]F = 1833843.55 \ g \cdot cm /s^2 = \frac{1833843.55}{1000 * 100} kg \cdot m /s^2[/tex]

=>    [tex]F = 18.33 \ N[/tex]

Answer:

1.0 m/s^2

Explanation: happy to help :)

Answer: [tex]1\ m/s^2[/tex]

Explanation:

Given

Masses of the block are [tex]m_1=1\ kg[/tex]  and

[tex]m_2=2\ kg[/tex]

Force applied by [tex]1\ kg[/tex] block on [tex]2\ kg[/tex] block is [tex]2\ N[/tex]

From the free body diagram of [tex]2\ kg[/tex] block, the net force on

[tex]\therefore m_2a=2\\\\\Rightarrow 2\times a=2\\\\\Rightarrow a=\dfrac{2}{2}\\\\\Rightarrow a=1\ m/s^2[/tex]

Thus, the acceleration of two blocks is [tex]1\ m/s^2[/tex]

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Answer:

Explanation:

According to the coulombs law, the force existing vetween the ions is expressed as;

F = kQq/r² .... 1

Q and q are the ions

r is the distance between the ions

If the distance between the ion is quadrupled, then;

F2 =  kQq/(4r)²

F2 =  kQq/16r² ... 2

Divide equation 2 by 1;

F2/F = kQq/16r² ÷ kQq/r²

F2/F = kQq/16r² × r²/kQq

F2/F = 1/16

F2 = 1/16 F

Therefore the coulombic force between two ions is reduced to 1/16 of its original strength when the distance between them is quadrupled.

Answer:

Answer:

m = 80[kg]

Explanation:

The kinetic energy can be calculated by means of the following equation.

[tex]E_{k}=\frac{1}{2}*m*v^{2}[/tex]

where:

m = mass [kg]

v = velocity = 5 [m/s]

Ek = kinetic energy = 1000 [J]

Now replacing:

[tex]1000 = \frac{1}{2} *m*5^{2}\\2000 = 25*m\\m=80[kg][/tex]

The mass of a ball will be "80 kg".

According to the question,

Kinetic energy, [tex]E_k[/tex] = 1000 Joules

Velocity, V = 5 m/s

As we know the formula,

→ Kinetic energy = [tex]\frac{1}{2}[/tex] × mass × (velocity)²

→      [tex]E_k = \frac{1}{2}[/tex] × m × v²

By substituting the values,

     [tex]1000= \frac{1}{2}\times m\times (5)^2[/tex]

[tex]1000\times 2=25\times m[/tex]      

     [tex]2000=25\times m[/tex]

         [tex]m = \frac{2000}{25}[/tex]

             [tex]= 80[/tex] kg

Thus the above answer is correct.

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Answer:

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

Explanation:

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

A collision is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.

The following are a few instances of physical encounters that scientists might classify as collisions. Legs of an insect are said to collide with a leaf when it falls on one.

Every contact of a cat's paws with the ground while it strides across a lawn is seen as a collision, as is every brush of its fur with a blade of grass.

Therefore, Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

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Answer:

E = 8.26*10⁻¹⁰ N/C, due south.

Explanation:

       [tex]F = -eE =ma (1)[/tex]

       [tex]E =\frac{m*a}{e} = \frac{9.1e-31 kg*145m/s2}{1.6e-19C} = 8.26e-10 N/C (1)[/tex]

Answer:

walking to school

Explanation:

Driving a car to school

, and taking the bus to school​ both take up energy, unlike walking to school.

unless ur talking about energy, counting energy you produce and use to complete things, then it would be the 3rd one, taking the bus to school.

Answer:  

Explanation:

Step one:

given data

period T= 3 milliseconds= 0.003

velocity v= 175m/s

wave lenght λ=?

Step two:

we know that f=1/T

the expression relating period and wave lenght is

v=λ/T

λ=v*T

λ=175*0.002

λ=0.525m

to cm= 0.525*100

=52.5cm

The wavelength of the wave is E) 52.5 cm

Answer:

a) h(g) = 358,53 m

b) t = 8,16 s

c) t(t) = 16,71 s

Explanation:

Equations for vertical shooting are:

Vf = V₀ -  g * t   ;     h  =  V₀*t -  (1/2)*g*t²  ;    Vf² = V₀² - 2*g*h

And at maximum heigt Vf = 0 then

0 = V₀ - g * t

t = V₀/g                     V₀  = 80 m/s      and   g = 9,8 m/s²

t  =  80 / 9,8   (s)

t = 8,16 s

Then 8,16 s is the time to get maximum height

If we plug t = 8,16 (s) in equation  h  =  V₀*t -  (1/2)*g*t²

we get:     h (max)  =  (80)*8,16 - 0,5*9,8*(8,16)² (m)

h (max) = 652,8  -  326,27 m

h (max) = 326,53 m

Then relative to ground that height becomes

h(g) = 326,53 + 32

h(g) = 358,53 m

In order to get the time the arrow is in the air we proceed as follows:

a) for the arrow to be at the launched point will take the same time that from the launched point to the maximum height, and after that we have to find out the time the arrow takes from 32 m down to the ground level

Then  

t(t) = 8,16 + 8,16 + tₓ     (2)

Where tₓ  is the time from 32 m height to ground

h  =  V₀*tₓ -  (1/2)*g*tₓ²  but since the arrow now is going down then we change the sign of the second term on the right side of the equation

32 = (80)*tₓ  +  0,5 * 9,8 * tₓ²     Note that when the arrow is at 32 m height the speed is again V₀ = 80 m/s

32 = 80*tₓ  + 4,9*tₓ²

A second-degree equation for tₓ, solving it

4,9*tₓ² + 80*tₓ - 32 = 0

t₁,₂ = -80 ± √ 6400 + 627,2 / 9,8

t₁,₂ =( - 80 ± 83,8 ) / 9,8

there is not a negative time therefore we dismiss such solution and

t₁ = 3,8 / 9,8

t₁ = 0,39 s

And

t(t) = 8,16 + 8,16 + 0,39  s

t(t) = 16,71 s

Answer:

The ratio is  KE : TM  =  0.75

Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        [tex]TM = \frac{1}{2} * k * A^2[/tex]

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        [tex]TM = KE + PE[/tex]

=>     [tex]KE = TM - PE[/tex]

Here the potential  energy of the mass is mathematically represented as

     [tex]PE = \frac{1}{ 2} * k * [ x ]^2[/tex]

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

      [tex]x = \frac{A}{2}[/tex]

So

     [tex]PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2[/tex]

So

      [tex]KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2[/tex]

=>    [tex]KE = \frac{1}{2} * k * A^2 - \frac{1}{8} * k * A ^2[/tex]

=>    [tex]KE = 0.375 * k * A^2[/tex]

So the ratio of  [tex]KE : TM[/tex] is  mathematically represented as

       [tex]\frac{KE}{TM} = \frac{0.375 k A^2 }{0.5 k A^2}[/tex]

=>    [tex]\frac{KE}{TM} = 0.75[/tex]

Answer:

Lifting a bag of groceries

Answer:

paddeling a canoe down a river :D or throwing a ball across a field

Explanation:

We know, [tex]1\ rpm = \dfrac{2\pi}{60} \ rad/s[/tex] .

[tex]64\ rpm\ is = \dfrac{2\pi}{60}\times 64\ rad/s\\\\= \dfrac{32\pi}{15}\ rad/s[/tex]

We know, kinetic energy is given by :

[tex]K.E = \dfrac{I\omega^2}{2}\\\\I = \dfrac{2(K.E)}{\omega^2}\\\\I = \dfrac{2\times 10^6}{\dfrac{32}{15}\times \pi}\\\\I = 298415.52 \ kg \ m^2[/tex]

Hence, this is the required solution.

Answer:

The one with equal forces on both sides, D is your answer

Explanation:

A, 5 and 10 No

B 13 and 8 no

C 5 and 8 no

D 6 and 6 YES

Object 4 have steady speed. Object 4 have same force on both the sides of object. Same force is applied on both the sides.

Uniform or constant velocity describes a body's motion when there is no acceleration. A body's velocity is its rate of movement in a specific direction.

When running steadily, your exhaustion should be caused by the length of the run, not by your speed. A steady-state run pace, according to the McMillan Running website, is halfway between your current 30k and half-marathon pace. This speed should allow you to run for 25 to 75 minutes.

Therefore, Object 4 have steady speed. Object 4 have same force on both the sides of object. Same force is applied on both the sides.

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Answer:

3g/mL

Explanation:

The density of a substance can be calculated using the formula;

Density = mass/volume

Where;

Density = g/mL

mass = grams (g)

volume = (mL)

According to this question, 45.0g of a sample placed in a graduated cylinder causes the water level to rise from

25.0mL to 40.0mL. This means that the volume of the sample is 40mL - 25mL = 15mL

Using D = m/v

D = 45/15

D = 3g/mL

Hence, the density of the sample is 3g/mL

Answer:

The bell has 8,550 Joule energy.

Explanation:

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object because of its height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where:

m = mass of the object

h  = height

g  = acceleration of gravity, or [tex]9.8 m/s^2[/tex]

Since the weight of an object of mass m can be calculated as:

W = m.g

The gravitational potential energy is:

U = W.h

The bell of weight W=190 N at the top of a tower is h=45 m high. Thus its energy is:

U = 190 N . 45 m

U = 8,550 Joule

The bell has 8,550 Joule energy.

Answer:

I think it's most likely true.

Explanation:

any organism has the properties of a living thing, which includes cells, whether it has one cell or many

Answer:

False

Explanation:

An organism is a living thing that is a single-celled life form

Answer:

[tex]-1.398\ \text{m/s}^2[/tex]

Decelerating or slowing down

Explanation:

F = Force = 50 N

[tex]\theta[/tex] = Angle force is being applied = [tex]42^{\circ}[/tex]

[tex]\mu[/tex] = Coefficient of friction = 0.64

m = Mass of wagon = 12 kg

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Normal force is given by

[tex]N=mg-F\sin\theta[/tex]

Frictional force is given by

[tex]f=\mu N\\\Rightarrow f=\mu (mg-F\sin\theta)[/tex]

The force balance is given by

[tex]F\cos\theta-f=ma\\\Rightarrow \dfrac{F\cos\theta-\mu (mg-F\sin\theta)}{m}=a\\\Rightarrow a=\dfrac{50\times \cos42^{\circ}-0.64(12\times 9.81-50\times\sin42^{\circ})}{12}\\\Rightarrow a=-1.398\ \text{m/s}^2[/tex]

The acceleration of the wagon is [tex]-1.398\ \text{m/s}^2[/tex]. The negative sign indicates that the wagon is decelerating or slowing down.

The acceleration of the wagon is  [tex]-1.398 \;\rm m/s^{2}[/tex].

Given data:

The magnitude of pulling force is, F = 50 N.

The angle of inclination is, [tex]\theta = 42^{\circ}[/tex].

The mass of wagon wheel is, m = 12 kg.

Coefficient of friction between wagon and grass is, [tex]\mu =0.64[/tex].

The given problem is based on the concept of frictional force. The standard expression for the frictional force is,

[tex]f= \mu \times N[/tex]

Here, N is the normal force and its value is,

[tex]N=mg-Fsin \theta[/tex]

And the net force acting on wagon is,

[tex]F' = Fcos\theta -f\\\\ma = Fcos\theta -(\mu(mg-Fsin \theta))\\\\a = \dfrac{Fcos\theta -(\mu(mg-Fsin \theta))}{m}[/tex]

Here, a is the acceleration of wagon.

Solving as,

[tex]a = \dfrac{50 \times cos42 -(0.64(12 \times 9.8-(50 \times sin42)))}{12}\\\\a=-1.398 \;\rm m/s^{2}[/tex]

Thus, we can conclude that the acceleration of the wagon is  [tex]-1.398 \;\rm m/s^{2}[/tex].

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