3kg of water at 80degree celcius is added to 8 kg of water at 25 degree celcius. find the temperature of final mixture provided there is no loss of heat in the surrounding. the specific heat capacity is 4200j/kg​

Answer:

A = 2.4 A

[tex]R_{eq} = 5 \ \Omega[/tex]

Explanation:

The voltage in the circuit, V = 12 V

The given circuit shows four resistors with R₁ and R₂ arranged in series with both in parallel to R₃ and R₄ which are is series to each other

R₁ = 4 Ω

R₂ = 6 Ω

R₄ = 5 Ω

The voltage across R₃ = 6 V

Voltage across parallel resistors are equal, therefore;

The total voltage across R₃ and R₄ = 12 V

The total voltage across R₁ and R₂ = 12 V

The voltage across R₃ + The voltage across R₄ = 12 V

∴ The voltage across R₄ = 12 V - 6 V = 6 V

The current flowing through  R₄ = 6V/(5 Ω) = 1.2 A

The current flowing through R₃ = The current flowing through R₄ = 1.2 A

The resistor, R₃ = 6 V/1.2 A = 5 Ω

Therefore, we have;

The sum of resistors in series are R₁ + R₂ and R₃ + R₄, which gives;

[tex]R_{series \, 1}[/tex] = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω

[tex]R_{series \, 2}[/tex] = R₃ + R₄ = 5 Ω + 5 Ω = 10 Ω

The sum of the resistors in parallel is given as follows;

[tex]\dfrac{1}{R_{eq}} = \dfrac{1}{R_{series \, 1}} + \dfrac{1}{R_{series \, 2}} = \dfrac{R_{series \, 2} + R_{series \, 1}}{R_{series \, 1} \times R_{series \, 2}}[/tex]

Therefore;

[tex]R_{eq} = \dfrac{R_{series \, 1} \times R_{series \, 2}}{R_{series \, 1} + R_{series \, 2}}[/tex]

Therefore;

[tex]R_{eq} = \dfrac{10\times 10}{10 + 10} \ \Omega = 5 \ \Omega[/tex]

[tex]R_{eq} = 5 \ \Omega[/tex]

The value of the current, A, in the circuit, I = V/[tex]R_{eq}[/tex]

A = I = 12 V/(5 Ω) = 2.4 A

A = 2.4 A

Answer:

1) Force is the action of pull or push that changes or tends to change the position of a body. Its SI unit is newton.

2) The types of force are:

• Pulling and Pushing force

• Frictional force

• Gravitational force

•Muscular force

• Centripetal and Centrifugal force

Answer:

I HOPE IT WILL HELP YOU.

:D

Answer:

The person feels cool at first because the swimming pool water is usually cool and he/she has that water on his body. But when it evaporates, the cool air directly touches his body and that's why he/she feels cold.

Thank You! Please mark me Brainliest!

This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.

The acceleration of the bird is  "a = 26.13 m/s²".

First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:

[tex]h = v_it + \frac{1}{2}gt^2[/tex]

where,

h = height fall = 3 m

vi = initial velocity = 0 m/s

g = acceleration due to gravity = 9.8 m/s²

t = time taken = ?

Therefore,

[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]

The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:

[tex]s = v_it + \frac{1}{2}at^2[/tex]

where,

s = distance covered by the bird = 5 m + 3 m = 8 m

vi = initial velocity of the bird = 0 m/s

a = acceleration of the bird = ?

t = time taken = 0.78 s

Therefore, using these values we get:

[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]

a = 26.13 m/s²

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Answer:

When a tree is vigorously shaken, the branches of the tree come in motion but the leaves tend to continue in their state of rest due to inertia of rest. As a result of this, leaves get separated from the branches of the tree and hence fall down.

Answer:

When a tree is vigorously shaken, the branches of the tree come in motion moving left and right and shake the leaves off, that's why some of the tree got detached when we vigorously shake its branch.

Explanation:

Like I did in the answer section.

Answer:

The net force is 91780.8 N.

Explanation:

mass, m = 8950 kg

Radius, R = 9.33 miles = 15015.2 m

Time, T = 0.123 h = 442.8 s

There are two forces acting on the plane.

Horizontal force is the centripetal force and the vertical force is the weight.

[tex]Fx =m R w^2\\\\Fx = m R \frac{4\pi^2}{T^2}\\\\Fx = 8950\times 15015.2\times \frac{4\times 3.14\times 3.14}{442.8\times 442.8}\\\\Fx = 27030.8 N \\\\Fy = m g \\\\ Fy = 8950\times 9.8 \\\\Fy = 87710 N[/tex]

The net force is

[tex]F = \sqrt{Fx^2 + Fy^2}\\\\F = \sqrt {27030.8^2 + 87710^2}\\\\F = 91780.8 N[/tex]

Answer:

a) v = 0.4799 m / s,  b)  K₀ = 1600.92 J,    K_f = 5.46 J

Explanation:

a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.

Initial instant. Before the crash

        p₀ = m v₁ + M v₂

where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive

Final moment. After the crash

       p_f = (m + M) v

as the system is isolated, the moment is preserved

       p₀ = p_f

       m v₁ + M v₂ = (m + M) v

       v =[tex]\frac{m v_1 + M v_2}{m+M}[/tex]

let's calculate

        v = [tex]\frac{ -95 \ 3.75 \ + 111 \ 4.10}{95+111}[/tex]

        v = 0.4799 m / s

b) let's find the initial kinetic energy of the system

         K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2

         K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2

         K₀ = 1600.92 J

the final kinetic energy

         K_f = ½ (m + M) v ^ 2

         k_f = ½ (95 + 111) 0.4799 ^ 2

         K_f = 5.46 J

Answer:

3 to 7 seconds.

Explanation:

The question is incomplete, the complete question is;

Water has a specific heat of 4.186 J/g°C, and ethanol has a specific heat of 2.450 J/g°C. Based on this information, which best compares water and ethanol?

It requires more heat to raise the temperature of a gram of ethanol by 1°C.

There are more molecules in a gram of water.

Ethanol has a lower formula mass.

Water has more protons and neutrons in its nuclei.

Answer:

There are more molecules in a gram of water.

Explanation:

Let us recall that the molar mass of water is 18g/mol while the molar mass of ethanol is 46 g/mol.

Hence;

1 g of water contains 18 g

There are 6.02 × 10^23/18g = 3.34 × 10^22 molecules in 1 g of water

1 g of ethanol contains 46 g

There are 6.02 × 10^23/46 g = 1.31 ×10^22 molecules

Hence, there are more molecules in 1 g of water than in 1g of ethanol.

Answer:

there are more molecules in 1 g of water than in 1g of ethanol.

Explanation:

edge 23

D. The density of the sir increases and the air rises.

When coming in contact with the kettle, it becomes hot, it expands and rises up, due to lesser density as compared to the surrounding air.

Explanation:

And keep changing gradually.

Eat a variety of foods. ...

Base your diet on plenty of foods rich in carbohydrates. ...

Replace saturated with unsaturated fat. ...

Enjoy plenty of fruits and vegetables. ...

Reduce salt and sugar intake. ...

Eat regularly, control the portion size. ...

Drink plenty of fluids. ...

Maintain a healthy body weight

Answer:

force=mass×acceleration

hence

acceleration is given by force÷mass

(500÷200)*5=12.5

Answer:

A ball rolls off the table and free falls to the ground. Which statement accurately describes its acceleration? Its acceleration is downward but remains a constant value as it falls.

Answer:

  an object sliding down hill

Explanation:

On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.

Answer:

I. Vs = 8.0 Volts.

II. Is = 4.5 Amperes.

Explanation:

Given the following data;

Np = 1.5Ns = [tex] \frac {N_{P}}{N_{S}} = 1.5 [/tex]  ..... equation 1

Ip = 3.0 A

Vp = 12 V

To find the voltage and current on the secondary coil;

I. For the voltage in the secondary coil (Vs), we would use the following formula;

[tex] \frac {V_{P}}{V_{S}} = \frac {N_{P}}{N_{S}} [/tex]  ...... equation 2.

Substituting eqn 1 into eqn 2, we have;

[tex] \frac {V_{P}}{V_{S}} = 1.5 [/tex]

[tex] \frac {12}{V_{S}} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] V_{S} * 1.5 = 12 [/tex]

[tex] V_{S} = \frac {12}{1.5} [/tex]

Vs = 8.0 V

II. For the current in the secondary coil (Is), we would use the following formula;

[tex] \frac {I_{S}}{I_{P}} = \frac {N_{P}}{N_{S}} [/tex]  .... equation 3

Substituting eqn 1 into eqn 3, we have;

[tex] \frac {I_{S}}{I_{P}} = 1.5 [/tex]

[tex] \frac {I_{S}}{3.0} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] I_{S} = 1.5 * 3.0 [/tex]

Is = 4.5 A

Explanation:

initial velocity(u)=30km/hr = 30*1000/60*60=8.33 m/s

final velocity(v) =60km/hr = 60*1000/60*60 =16.67 m/s

time taken(t) = 1 minutes

= 60 seconds

Now,

Average velocity = u+v/2

= 8.33 m/s + 16.67m/s÷2

=12.5 m/s

Answer:

Explanation:

The heat from the Sun gets to the surface of the Earth by radiation and not by convection because:

i. Radiation does not require material medium for its propagation: Since the space between the Sun and Earth is a vacuum ( i.e it is an empty space), then the only method for the heat from to Sun to get to the Earth is by radiation. This method does not require a material medium, while convection is a method that requires material medium for its propagation.

ii. Radiation occurs without heating the intervening medium: During the propagation of the heat from the Sun to the Earth, the space between the two heavenly bodies is not heated up. Some fraction of the heat after hitting the surface of the Earth get reflected back into the Earth's atmosphere. While in convection, the medium would have been heated up in the process.

Answer:

The carbon dioxide produced by fossil fuel into the air is a greenhouse gas just as is nitrous oxide, methane, water vapor, and synthetic gases that contain fluorine. Compared to the other greenhouse gases, carbon dioxide is the most abundantly present greenhouse gas.

The greenhouse gases cause greenhouse effect which is the  observed trapping of heat in the atmosphere, similar to the glass roof of a greenhouse, thereby causing the average global temperature to rise, and changing the climate through global warming

There is presently a 1°C rise in the average global temperature as part of the Earth reflected heat from the Sun is returned back to Earth by the increasing greenhouse gas presence in the atmosphere which cause the greenhouse effect

Explanation:

Answer:

The carbon dioxide produced by fossil fuels in the air is a greenhouse gas, as is nitrous oxide, methane, water vapor, and synthetic gases containing fluorine. Compared to the other greenhouse gases, carbon dioxide is the most abundant greenhouse gas.

The greenhouse gases cause the greenhouse effect which is the observed trapping of heat in the atmosphere, similar to the glass roof of a greenhouse, thereby causing the average global temperature to rise, and changing the climate through global warming

There is presently a 1°C rise in the average global temperature as part of the Earth's reflected heat from the Sun is returned to Earth by the increasing greenhouse gas presence in the atmosphere which causes the greenhouse effect

Explanation:

I got a 100% on this assignment copy it and put it on your notes then go sentence by sentence copy but dont highlight any "." you will not be able to paste it into the box add the "." after you paste it, it one sentence.

Answer:

sorry I don't understand this language

Answer:

301.6 N

Explanation:

The length of the wire L₀ = 8 m and its diameter, d = 4 mm = 4 × 10⁻³ m. Since its temperature drops by  10°C, it will have a change in length ΔL = L₀αΔθ where α = linear expansivity of steel, a 12 × 10⁻⁶ /K, and Δθ = temperature change = -10°C = -10 K(negative since it is a drop)

So, the strain, ε = ΔL/L₀ = αΔθ = 12 × 10⁻⁶ /K × 10 K = 12 × 10⁻⁵

Now the Young's modulus of steel, Y = σ/ε where σ = stress = T/A where T = increase in tension in steel wire and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 4 × 10⁻³ m and ε = strain = 12 × 10⁻⁵

So, σ = Yε

Since Y = 2 × 10¹¹ N/m².

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

σ = Yε

σ = 2 × 10¹¹ N/m² × 12 × 10⁻⁵

σ = 24 × 10⁶ N/m²

Since σ = T/A

T = σA

T = σπd²/4

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

T = σπd²/4

T = 24 × 10⁶ N/m² × π × (4 × 10⁻³ m)²/4

T = 24 × 10⁶ N/m² × π × 16 × 10⁻⁶ m²/4

T = 24 × 10⁶ N/m² × π × 4 × 10⁻⁶ m²

T = 96 N × π

T = 301.59 N

T ≅ 301.6 N

So, the increase in tension in the steel wire is 301.6 N

Answer:

V1 = 4/3 pi R^3 = pi D^3 / 6     D = 2 R       volume of sphere

V2 = pi r^2 L = pi d^2 L / 4       volume of wire

V2 / V1 = 1 = 3/2 d^2 L / D^3     since volumes are equal

L = 2/3 D^3 / d^2 = 2/3 * 6^3 / .02^2 = 360,00 cm = 3600 m

Answer:

[tex]8.4 \times 10^{5}\; \rm J[/tex], assuming that there's no heat exchange between the washing machine and the environment.

Explanation:

Let [tex]m[/tex] denote the mass of water and [tex]c[/tex] the specific heat capacity of water. The energy required to raise the temperature of that much water by [tex]\Delta T[/tex] would be:

[tex]Q = c \cdot m \cdot \Delta T[/tex].

Washing at [tex]30\; \rm ^{\circ} C[/tex] would require a temperature change of [tex]\Delta T = 30\; \rm ^{\circ} C - 16\; ^{\circ} \rm C = 14\; \rm K[/tex].

Washing at [tex]50\; \rm ^{\circ} C[/tex] would require a temperature change of [tex]\Delta T = 50\; \rm ^{\circ} C - 16\; ^{\circ} \rm C = 34\; \rm K[/tex].

In both situations, [tex]c = 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1}[/tex] while [tex]m = 10\; \rm kg[/tex].

Calculate the energy required in either situation:

Washing at [tex]30\; \rm ^{\circ} C[/tex]:

[tex]\begin{aligned}& Q({30\; ^{\circ} {\rm C}}) \\ &= c \cdot m \cdot \Delta T \\ &= 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1} \times 10\; \rm kg \times 14\; \rm K \\ &= 588000 \times 10^{5}\; \rm J\end{aligned}[/tex].

Washing at [tex]50\; ^{\circ} {\rm C}[/tex]:

[tex]\begin{aligned}& Q({50\; ^{\circ} {\rm C}}) \\ &= c \cdot m \cdot \Delta T \\ &= 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1} \times 10\; \rm kg \times 34\; \rm K \\ &= 1428000 \; \rm J\end{aligned}[/tex].

[tex]1428000\; \rm J - 588000\; \rm J = 8.4 \times 10^{5}\; \rm J[/tex].

The three dimensional space of a body is called volume

Answer:

a. Acceleration, a = 1.47 * 10^{-9} m/s²

b. Mass = 4.57 * 10^{10} kilograms

c. Force = 67.12 Newton

Explanation:

Given the following data;

Distance = 93 cm to meters = 93/100 = 0.93 meters

Weight = 67 N

Time = 9.89 hours to seconds = 35604 seconds

Initial velocity = 0 m/s (since it's starting from rest)

Acceleration due to gravity, g = 9.8 m/s

a. To find the acceleration, we would use the second equation of motion;

[tex] S = ut + \frac{1}{2} at^{2} [/tex]

Where;

S is the distance covered or displacement of an object.

u is the initial velocity.

a is the acceleration.

t is the time.

Substituting the values into the equation, we have;

[tex] 0.93 = 0*35604 + \frac{1}{2} * a*35604^{2} [/tex]

[tex] 0.93 = 0 + \frac{1}{2} * 1267644816a [/tex]

[tex] 0.93 = 633822408a [/tex]

[tex] Acceleration, a = \frac{0.93}{633822408} [/tex]

Acceleration, a = 1.47 * 10^{-9} m/s²

b. To find the mass

Weight = mass * acceleration due to gravity

67 = mass * 1.47 * 10^{-9}

[tex] Acceleration, a = \frac{67}{1.47 * 10^{-9}} [/tex]

Mass = 4.57 * 10^{10} kilograms

c. To find the force;

Force = mass * acceleration

Force = 4.57 * 10^{10} * 1.47 * 10^{-9}

Force = 67.12 Newton

We know

[tex]\boxed{\sf W=Fscos\theta}[/tex]

[tex]\\ \sf\longmapsto W=25(20)cos30°[/tex]

[tex]\\ \sf\longmapsto W=500\times \dfrac{\sqrt{3}}{2}[/tex]

[tex]\\ \sf\longmapsto W=250\sqrt{3}[/tex]

[tex]\\ \sf\longmapsto W=250(1.73)[/tex]

[tex]\\ \sf\longmapsto W\approx 430J[/tex]

Option a) 430 J work done he do after walking 20.0 m.

Work, in physics, is a measure of energy transfer that happens when an object exists moved over a distance by an external force at least part of which exists involved in the direction of the displacement.

Displacement exists as the distance moved in a straight line, in a provided direction, from the starting point. Displacement exists as a vector quantity as it includes size and direction.

An angle stands the figure formed by two rays, named the sides of the angle, sharing a standard endpoint, named the vertex of the angle.

⇒[tex]W=Fcos\theta[/tex]

⇒[tex]W=25(20)cos 30^{0}[/tex]

Then,

⇒[tex]W=50*\sqrt{3} /2[/tex]

⇒[tex]W=250\sqrt{3}[/tex]

Hence,

⇒[tex]W=250(1.73)[/tex]

[tex]W[/tex] approximately 430J.

Hence, Option a) 430 J work done he do after walking 20.0 m.

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The frequency of the other car is 669HZ

Let start off by writing out the parameters given in the question

frequency= 595 HZ

Speed of sound(v)= 343m/s

Speed of the car(vs)= 20m/s

Speed of the second car(vo) = 20m/s

The frequency of the other car can be calculated as follows

(v+vo/v-vs) f

= (343+20/343-20)595

= (363/323)595

= 1.1238(595)

= 669 HZ

Hence the frequency of the other car is 669 HZ

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Answer:

Wavelength = 5 * 10^{-11} meters

Explanation:

Given the following data;

Frequency = 6.0 x 10^18 Hz

Speed = 3 * 10⁸ m/s

To find the wavelength of the wave;

Mathematically, the wavelength of a wave is given by the formula;

[tex] Wavelength = \frac {speed}{frequency} [/tex]

Substituting into the formula, we have;

[tex] Wavelength = \frac {3 * 10^{8}}{6.0 x 10^{18}} [/tex]

Wavelength = 5 * 10^{-11} meters

Answer:

136 m.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 0.80 s

Speed of sound (v) = 340 m/s

Distance (x) =?

The distance of the player from the wall can be obtained as follow:

v = 2x / t

340 = 2x / 0.80

Cross multiply

2x = 340 × 0.80

2x = 272

Divide both side by 2

x = 272 / 2

x = 136 m

Thus, the distance of the player from the wall is 136 m